Problem 3

2.3.3 Problem 3

2.3.3.1 Solved using ﬁrst_order_ode_poly
2.3.3.2 Solved using ﬁrst_order_ode_exact
2.3.3.3 Solved using ﬁrst_order_ode_dAlembert
2.3.3.4 Solved using ﬁrst_order_ode_homog_type_maple_C
2.3.3.5 Solved using ﬁrst_order_ode_abel_second_kind_case_5
2.3.3.6 Solved using ﬁrst_order_ode_abel_second_kind_solved_by_converting_to_ﬁrst_kind
2.3.3.7 Solved using ﬁrst_order_ode_LIE
2.3.3.8 ✓Maple
2.3.3.9 ✓Mathematica
2.3.3.10 ✗Sympy

Internal problem ID [19718]
Book : Elementary Diﬀerential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter IV. Methods of solution: First order equations. section 29. Problems at page 81
Problem number : 3
Date solved : Friday, January 30, 2026 at 12:10:51 AM
CAS classiﬁcation : [[_homogeneous, `class C`], _exact, _rational, [_Abel, `2nd type`, `class A`]]

2.3.3.1 Solved using ﬁrst_order_ode_poly

3.269 (sec)

Entering ﬁrst order ode poly solver

\begin{align*} 2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime }&=g \\ \end{align*}

This is ODE of type polynomial. Where the RHS of the ode is ratio of equations of two lines. Writing the ODE in the form

\[ y^{\prime }= \frac {a_1 x + b_1 y + c_1}{ a_2 x + b_2 y + c_3 } \]

Where \(a_1=-2 a, b_1=-b, c_1 =g, a_2=b, b_2=2 c, c_2=e\). There are now two possible solution methods. The ﬁrst case is when the two lines \(a_1 x + b_1 y + c_1\),\( a_2 x + b_2 y + c_3\) are not parallel and the second case is if they are parallel. If they are not parallel, then the transformation \(X=x-x_0\), \(Y=y-y_0\) converts the ODE to a homogeneous ODE. The values \( x_0,y_0\) have to be determined. If they are parallel then a transformation \(U(x)=a_1 x + b_1 y\) converts the given ODE in \(y\) to a separable ODE in \(U(x)\). The ﬁrst case is when \(\frac {a_1}{b_1} \neq \frac {a_2}{b_2}\) and the second case when \(\frac {a_1}{b_1} = \frac {a_2}{b_2}\). From the above we see that \(\frac {a_1}{b_1}\neq \frac {a_2}{b_2}\). Hence this is case one where lines are not parallel. Using the transformation

\begin{align*} X &=x-x_0 \\ Y &=y-y_0 \end{align*}

Where the constants \(x_0,y_0\) are obtained by solving the following two linear algebraic equations

\begin{align*} a_1 x_0 + b_1 y_0 + c_1 &= 0\\ a_2 x_0 + b_2 y_0 + c_2 &= 0 \end{align*}

Substituting the values for \(a_1,b_1,c_1,a_2,b_2,c_2\) gives

\begin{align*} -2 a x_{0} -b y_{0} +g &= 0 \\ b x_{0} +2 c y_{0} +e &= 0 \\ \end{align*}

Solving for \(x_0,y_0\) from the above gives

\begin{align*} x_0 &= \frac {b e +2 c g}{4 a c -b^{2}} \\ y_0 &= -\frac {2 a e +b g}{4 a c -b^{2}} \end{align*}

Therefore the transformation becomes

\begin{align*} X &=x-{| \frac {b e +2 c g}{4 a c -b^{2}}|} \\ Y &=y+{| \frac {2 a e +b g}{4 a c -b^{2}}|} \end{align*}

Using this transformation in \(2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime } = g\) result in

\begin{align*} \frac {dY}{dX} &= \frac {-2 X a -Y b}{X b +2 Y c} \end{align*}

This is now a homogeneous ODE which will now be solved for \(Y(X)\). In canonical form, the ODE is

\begin{align*} Y' &= F(X,Y)\\ &= \frac {-2 X a -Y b}{X b +2 Y c}\tag {1} \end{align*}

An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if

\[ f(t^n X, t^n Y)= t^n f(X,Y) \]

In this case, it can be seen that both \(M=-2 X a -Y b\) and \(N=X b +2 Y c\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence

\[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \]

Applying the transformation \(Y=uX\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u &= \frac {-b u -2 a}{2 c u +b}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {-b u \left (X \right )-2 a}{2 c u \left (X \right )+b}-u \left (X \right )}{X} \end{align*}

\[ \frac {d}{d X}u \left (X \right )-\frac {\frac {-b u \left (X \right )-2 a}{2 c u \left (X \right )+b}-u \left (X \right )}{X} = 0 \]

\[ 2 u \left (X \right ) \left (\frac {d}{d X}u \left (X \right )\right ) X c +2 u \left (X \right )^{2} c +\left (\frac {d}{d X}u \left (X \right )\right ) X b +2 b u \left (X \right )+2 a = 0 \]

\[ X \left (2 c u \left (X \right )+b \right ) \left (\frac {d}{d X}u \left (X \right )\right )+2 u \left (X \right )^{2} c +2 b u \left (X \right )+2 a = 0 \]

Which is now solved as separable in \(u \left (X \right )\).

The ode

\begin{equation} \frac {d}{d X}u \left (X \right ) = -\frac {2 \left (u \left (X \right )^{2} c +b u \left (X \right )+a \right )}{X \left (2 c u \left (X \right )+b \right )} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d X}u \left (X \right )&= -\frac {2 \left (u \left (X \right )^{2} c +b u \left (X \right )+a \right )}{X \left (2 c u \left (X \right )+b \right )}\\ &= f(X) g(u) \end{align*}

Where

\begin{align*} f(X) &= -\frac {2}{X}\\ g(u) &= \frac {u^{2} c +b u +a}{2 c u +b} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\ \int { \frac {2 c u +b}{u^{2} c +b u +a}\,du} &= \int { -\frac {2}{X} \,dX} \\ \end{align*}

\[ \ln \left (u \left (X \right )^{2} c +b u \left (X \right )+a \right )=\ln \left (\frac {1}{X^{2}}\right )+c_2 \]

Taking the exponential of both sides the solution becomes

\[ u \left (X \right )^{2} c +b u \left (X \right )+a = \frac {c_2}{X^{2}} \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \frac {u^{2} c +b u +a}{2 c u +b}=0 \]

for \(u \left (X \right )\) gives

\begin{align*} u \left (X \right )&=\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\\ u \left (X \right )&=-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c} \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} u \left (X \right )^{2} c +b u \left (X \right )+a &= \frac {c_2}{X^{2}} \\ u \left (X \right ) &= \frac {-b +\sqrt {-4 a c +b^{2}}}{2 c} \\ u \left (X \right ) &= -\frac {b +\sqrt {-4 a c +b^{2}}}{2 c} \\ \end{align*}

Converting \(u \left (X \right )^{2} c +b u \left (X \right )+a = \frac {c_2}{X^{2}}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right )^{2} c +b Y \left (X \right ) X +X^{2} a = c_2 \end{align*}

Converting \(u \left (X \right ) = \frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = \frac {X \left (-b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \end{align*}

Converting \(u \left (X \right ) = -\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = -\frac {X \left (b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \end{align*}

The solution is implicit \(Y \left (X \right )^{2} c +b Y \left (X \right ) X +X^{2} a = c_2\). Replacing \(Y=y-y_0, X=x-x_0\) gives

\[ \left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right )^{2} a +\left (y+\frac {2 a e +b g}{4 a c -b^{2}}\right ) b \left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right )+\left (y+\frac {2 a e +b g}{4 a c -b^{2}}\right )^{2} c = c_2 \]

The solution is

\[ Y \left (X \right ) = \frac {X \left (-b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \]

Replacing \(Y=y-y_0, X=x-x_0\) gives

\[ y+\frac {2 a e +b g}{4 a c -b^{2}} = \frac {\left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right ) \left (-b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \]

\[ y = \frac {\left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right ) \left (-b +\sqrt {-4 a c +b^{2}}\right )}{2 c}-\frac {2 a e +b g}{4 a c -b^{2}} \]

The solution is

\[ Y \left (X \right ) = -\frac {X \left (b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \]

Replacing \(Y=y-y_0, X=x-x_0\) gives

\[ y+\frac {2 a e +b g}{4 a c -b^{2}} = -\frac {\left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right ) \left (b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \]

\[ y = -\frac {\left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right ) \left (b +\sqrt {-4 a c +b^{2}}\right )}{2 c}-\frac {2 a e +b g}{4 a c -b^{2}} \]

Simplifying the above gives

\begin{align*} \frac {\left (4 a \,c^{2}-b^{2} c \right ) y^{2}+4 \left (b x +e \right ) \left (-\frac {b^{2}}{4}+a c \right ) y+4 a^{2} c \,x^{2}+\left (-b^{2} x^{2}-4 c g x +e^{2}\right ) a +g \left (b^{2} x +b e +c g \right )}{4 a c -b^{2}} &= c_2 \\ y &= \frac {\left (-b^{2} x -b e +4 \left (a x -\frac {g}{2}\right ) c \right ) \sqrt {-4 a c +b^{2}}-4 \left (-\frac {b^{2}}{4}+a c \right ) \left (b x +e \right )}{8 \left (-\frac {b^{2}}{4}+a c \right ) c} \\ y &= \frac {\left (b^{2} x +b e -4 \left (a x -\frac {g}{2}\right ) c \right ) \sqrt {-4 a c +b^{2}}-4 \left (-\frac {b^{2}}{4}+a c \right ) \left (b x +e \right )}{8 \left (-\frac {b^{2}}{4}+a c \right ) c} \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= \frac {\left (-b^{2} x -b e +4 \left (a x -\frac {g}{2}\right ) c \right ) \sqrt {-4 a c +b^{2}}-4 \left (-\frac {b^{2}}{4}+a c \right ) \left (b x +e \right )}{8 \left (-\frac {b^{2}}{4}+a c \right ) c} \\ y &= \frac {\left (b^{2} x +b e -4 \left (a x -\frac {g}{2}\right ) c \right ) \sqrt {-4 a c +b^{2}}-4 \left (-\frac {b^{2}}{4}+a c \right ) \left (b x +e \right )}{8 \left (-\frac {b^{2}}{4}+a c \right ) c} \\ y &= \frac {-4 a b c x +b^{3} x -4 a c e +b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x +64 c_2 \,a^{2} c^{3}-32 c_2 a \,b^{2} c^{2}+4 c_2 \,b^{4} c -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}}}{2 c \left (4 a c -b^{2}\right )} \\ y &= -\frac {4 a b c x -b^{3} x +4 a c e -b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x +64 c_2 \,a^{2} c^{3}-32 c_2 a \,b^{2} c^{2}+4 c_2 \,b^{4} c -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}}}{2 c \left (4 a c -b^{2}\right )} \\ \end{align*}

Summary of solutions found

2.3.3.2 Solved using ﬁrst_order_ode_exact

1.334 (sec)

Entering ﬁrst order ode exact solver

\begin{align*} 2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime }&=g \\ \end{align*}

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisﬁes the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisﬁed, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisﬁed. If this condition is not satisﬁed then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The ﬁrst step is to write the ODE in standard form to check for exactness, which is

\[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]

Therefore

\begin{align*} \left (b x +2 c y +e\right )\mathop {\mathrm {d}y} &= \left (-2 a x -b y +g\right )\mathop {\mathrm {d}x}\\ \left (2 a x +b y -g\right )\mathop {\mathrm {d}x} + \left (b x +2 c y +e\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(x,y) &= 2 a x +b y -g\\ N(x,y) &= b x +2 c y +e \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisﬁed

\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (2 a x +b y -g\right )\\ &= b \end{align*}

And

\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (b x +2 c y +e\right )\\ &= b \end{align*}

Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)

\begin{align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end{align*}

Integrating (1) w.r.t. \(x\) gives

\begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int 2 a x +b y -g\mathop {\mathrm {d}x} \\ \tag{3} \phi &= x \left (a x +b y -g \right )+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = b x+f'(y) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = b x +2 c y +e\). Therefore equation (4) becomes

\begin{equation} \tag{5} b x +2 c y +e = b x+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = 2 c y +e \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( 2 c y +e\right ) \mathop {\mathrm {d}y} \\ f(y) &= c \,y^{2}+e y+ c_1 \\ \end{align*}

\[ \phi = x \left (a x +b y -g \right )+c \,y^{2}+e y+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = x \left (a x +b y -g \right )+c \,y^{2}+e y \]

Simplifying the above gives

\begin{align*} c y^{2}+y b x +a \,x^{2}+e y-g x &= c_1 \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= \frac {-b x -e +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}+2 b e x +4 c g x +4 c_1 c +e^{2}}}{2 c} \\ y &= -\frac {b x +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}+2 b e x +4 c g x +4 c_1 c +e^{2}}+e}{2 c} \\ \end{align*}

Summary of solutions found

2.3.3.3 Solved using ﬁrst_order_ode_dAlembert

3.721 (sec)

Entering ﬁrst order ode dAlembert solver

\begin{align*} 2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime }&=g \\ \end{align*}

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} 2 a x +b y +\left (b x +2 c y +e \right ) p = g \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= -\frac {\left (b p +2 a \right ) x}{2 c p +b}-\frac {e p -g}{2 c p +b} \\ \end{align*}

This has the form

\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {-b p -2 a}{2 c p +b}\\ g &= \frac {-e p +g}{2 c p +b} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {-b p -2 a}{2 c p +b} = \left (-\frac {x b}{2 c p +b}+\frac {2 x c b p}{\left (2 c p +b \right )^{2}}+\frac {4 x c a}{\left (2 c p +b \right )^{2}}-\frac {e}{2 c p +b}+\frac {2 c e p}{\left (2 c p +b \right )^{2}}-\frac {2 c g}{\left (2 c p +b \right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p -\frac {-b p -2 a}{2 c p +b} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\\ p_{2} &=-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c} \end{align*}

Substituting these in (1A) and keeping singular solution that veriﬁes the ode gives

\begin{align*} y = \frac {-b x \sqrt {-4 a c +b^{2}}-4 a c x +b^{2} x -e \sqrt {-4 a c +b^{2}}+b e +2 c g}{2 c \sqrt {-4 a c +b^{2}}}\\ y = \frac {-b x \sqrt {-4 a c +b^{2}}+4 a c x -b^{2} x -e \sqrt {-4 a c +b^{2}}-b e -2 c g}{2 c \sqrt {-4 a c +b^{2}}} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {-b p \left (x \right )-2 a}{2 c p \left (x \right )+b}}{-\frac {x b}{2 c p \left (x \right )+b}+\frac {2 x c b p \left (x \right )}{\left (2 c p \left (x \right )+b \right )^{2}}+\frac {4 x c a}{\left (2 c p \left (x \right )+b \right )^{2}}-\frac {e}{2 c p \left (x \right )+b}+\frac {2 c e p \left (x \right )}{\left (2 c p \left (x \right )+b \right )^{2}}-\frac {2 c g}{\left (2 c p \left (x \right )+b \right )^{2}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (x \right ) = \frac {2 \left (2 c p \left (x \right )+b \right ) \left (p \left (x \right )^{2} c +b p \left (x \right )+a \right )}{4 a c x -b^{2} x -b e -2 c g} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {2 \left (2 c p \left (x \right )+b \right ) \left (p \left (x \right )^{2} c +b p \left (x \right )+a \right )}{4 a c x -b^{2} x -b e -2 c g}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {2}{4 a c x -b^{2} x -b e -2 c g}\\ g(p) &= \left (2 c p +b \right ) \left (c \,p^{2}+b p +a \right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {1}{\left (2 c p +b \right ) \left (c \,p^{2}+b p +a \right )}\,dp} &= \int { \frac {2}{4 a c x -b^{2} x -b e -2 c g} \,dx} \\ \end{align*}

\[ \frac {-\ln \left (p \left (x \right )^{2} c +b p \left (x \right )+a \right )+2 \ln \left (2 c p \left (x \right )+b \right )}{4 a c -b^{2}}=\frac {2 \ln \left (x \left (4 a c -b^{2}\right )-b e -2 c g \right )}{4 a c -b^{2}}+c_1 \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \left (2 c p +b \right ) \left (c \,p^{2}+b p +a \right )=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=-\frac {b}{2 c}\\ p \left (x \right )&=\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c} \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {-\ln \left (p \left (x \right )^{2} c +b p \left (x \right )+a \right )+2 \ln \left (2 c p \left (x \right )+b \right )}{4 a c -b^{2}} &= \frac {2 \ln \left (x \left (4 a c -b^{2}\right )-b e -2 c g \right )}{4 a c -b^{2}}+c_1 \\ p \left (x \right ) &= -\frac {b}{2 c} \\ p \left (x \right ) &= \frac {-b +\sqrt {-4 a c +b^{2}}}{2 c} \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} \text {Expression too large to display} \\ y &= \frac {\sqrt {-4 a c +b^{2}}\, \left (b x \sqrt {-4 a c +b^{2}}+4 a c x -b^{2} x +e \sqrt {-4 a c +b^{2}}-b e -2 c g \right )}{2 c \left (4 a c -b^{2}\right )} \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= \frac {-b x \sqrt {-4 a c +b^{2}}-4 a c x +b^{2} x -e \sqrt {-4 a c +b^{2}}+b e +2 c g}{2 c \sqrt {-4 a c +b^{2}}} \\ y &= \frac {-b x \sqrt {-4 a c +b^{2}}+4 a c x -b^{2} x -e \sqrt {-4 a c +b^{2}}-b e -2 c g}{2 c \sqrt {-4 a c +b^{2}}} \\ y &= -\frac {16 \left (-\frac {b^{2} x}{4}-\frac {b e}{4}+\left (a x -\frac {g}{2}\right ) c \right )^{2} {\mathrm e}^{\frac {c_1 \left (4 a c -b^{2}\right )}{2}}-4 \,{\mathrm e}^{-2 \left (-\frac {b^{2}}{4}+a c \right ) c_1} c +\left (b x +e \right ) \sqrt {-64 \left (-\frac {b^{2}}{4}+a c \right ) \left (\left (\left (a x -\frac {g}{2}\right ) c -\frac {b \left (b x +e \right )}{4}\right )^{2} {\mathrm e}^{c_1 \left (4 a c -b^{2}\right )}-\frac {c}{4}\right )}}{2 \sqrt {-64 \left (-\frac {b^{2}}{4}+a c \right ) \left (\left (\left (a x -\frac {g}{2}\right ) c -\frac {b \left (b x +e \right )}{4}\right )^{2} {\mathrm e}^{c_1 \left (4 a c -b^{2}\right )}-\frac {c}{4}\right )}\, c} \\ y &= \frac {-b x \sqrt {-4 a c +b^{2}}-4 a c x +b^{2} x -e \sqrt {-4 a c +b^{2}}+b e +2 c g}{2 c \sqrt {-4 a c +b^{2}}} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= -\frac {16 \left (-\frac {b^{2} x}{4}-\frac {b e}{4}+\left (a x -\frac {g}{2}\right ) c \right )^{2} {\mathrm e}^{\frac {c_1 \left (4 a c -b^{2}\right )}{2}}-4 \,{\mathrm e}^{-2 \left (-\frac {b^{2}}{4}+a c \right ) c_1} c +\left (b x +e \right ) \sqrt {-64 \left (-\frac {b^{2}}{4}+a c \right ) \left (\left (\left (a x -\frac {g}{2}\right ) c -\frac {b \left (b x +e \right )}{4}\right )^{2} {\mathrm e}^{c_1 \left (4 a c -b^{2}\right )}-\frac {c}{4}\right )}}{2 \sqrt {-64 \left (-\frac {b^{2}}{4}+a c \right ) \left (\left (\left (a x -\frac {g}{2}\right ) c -\frac {b \left (b x +e \right )}{4}\right )^{2} {\mathrm e}^{c_1 \left (4 a c -b^{2}\right )}-\frac {c}{4}\right )}\, c} \\ y &= \frac {-b x \sqrt {-4 a c +b^{2}}-4 a c x +b^{2} x -e \sqrt {-4 a c +b^{2}}+b e +2 c g}{2 c \sqrt {-4 a c +b^{2}}} \\ y &= \frac {-b x \sqrt {-4 a c +b^{2}}+4 a c x -b^{2} x -e \sqrt {-4 a c +b^{2}}-b e -2 c g}{2 c \sqrt {-4 a c +b^{2}}} \\ \end{align*}

2.3.3.4 Solved using ﬁrst_order_ode_homog_type_maple_C

1.559 (sec)

Entering ﬁrst order ode homog type maple C solver

\begin{align*} 2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime }&=g \\ \end{align*}

Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)

\[ \frac {d}{d X}Y \left (X \right ) = -\frac {2 a \left (X +x_{0} \right )+b \left (Y \left (X \right )+y_{0} \right )-g}{2 c \left (Y \left (X \right )+y_{0} \right )+b \left (X +x_{0} \right )+e} \]

Solving for possible values of \(x_{0}\) and \(y_{0}\) which makes the above ode a homogeneous ode results in

\begin{align*} x_{0}&=\frac {b e +2 c g}{4 a c -b^{2}}\\ y_{0}&=\frac {-2 a e -b g}{4 a c -b^{2}} \end{align*}

Using these values now it is possible to easily solve for \(Y \left (X \right )\). The above ode now becomes

\begin{align*} \frac {d}{d X}Y \left (X \right ) = -\frac {2 a X +b Y \left (X \right )+\frac {2 a \left (b e +2 c g \right )}{4 a c -b^{2}}+\frac {b \left (-2 a e -b g \right )}{4 a c -b^{2}}-g}{b X +2 c Y \left (X \right )+\frac {\left (b e +2 c g \right ) b}{4 a c -b^{2}}+\frac {2 c \left (-2 a e -b g \right )}{4 a c -b^{2}}+e} \end{align*}

In canonical form, the ODE is

\begin{align*} Y' &= F(X,Y)\\ &= -\frac {2 a X +b Y +\frac {2 a \left (b e +2 c g \right )}{4 a c -b^{2}}+\frac {b \left (-2 a e -b g \right )}{4 a c -b^{2}}-g}{b X +2 c Y +\frac {\left (b e +2 c g \right ) b}{4 a c -b^{2}}+\frac {2 c \left (-2 a e -b g \right )}{4 a c -b^{2}}+e}\tag {1} \end{align*}

\[ f(t^n X, t^n Y)= t^n f(X,Y) \]

In this case, it can be seen that both \(M=-\left (4 a c -b^{2}\right ) \left (8 X \,a^{2} c -2 X a \,b^{2}+4 Y a b c -Y \,b^{3}\right )\) and \(N=\left (4 X a b c -X \,b^{3}+8 Y a \,c^{2}-2 Y \,b^{2} c \right ) \left (4 a c -b^{2}\right )\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence

\[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \]

Applying the transformation \(Y=uX\) to the above ODE in (1) gives

\[ \frac {d}{d X}u \left (X \right )-\frac {\frac {-b u \left (X \right )-2 a}{2 c u \left (X \right )+b}-u \left (X \right )}{X} = 0 \]

\[ 2 \left (\frac {d}{d X}u \left (X \right )\right ) u \left (X \right ) X c +\left (\frac {d}{d X}u \left (X \right )\right ) X b +2 u \left (X \right )^{2} c +2 b u \left (X \right )+2 a = 0 \]

\[ X \left (2 c u \left (X \right )+b \right ) \left (\frac {d}{d X}u \left (X \right )\right )+2 u \left (X \right )^{2} c +2 b u \left (X \right )+2 a = 0 \]

Which is now solved as separable in \(u \left (X \right )\).

The ode

\begin{equation} \frac {d}{d X}u \left (X \right ) = -\frac {2 \left (u \left (X \right )^{2} c +b u \left (X \right )+a \right )}{X \left (2 c u \left (X \right )+b \right )} \end{equation}

is separable as it can be written as

Where

\begin{align*} f(X) &= -\frac {2}{X}\\ g(u) &= \frac {u^{2} c +b u +a}{2 c u +b} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\ \int { \frac {2 c u +b}{u^{2} c +b u +a}\,du} &= \int { -\frac {2}{X} \,dX} \\ \end{align*}

\[ \ln \left (u \left (X \right )^{2} c +b u \left (X \right )+a \right )=\ln \left (\frac {1}{X^{2}}\right )+c_1 \]

Taking the exponential of both sides the solution becomes

\[ u \left (X \right )^{2} c +b u \left (X \right )+a = \frac {c_1}{X^{2}} \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \frac {u^{2} c +b u +a}{2 c u +b}=0 \]

for \(u \left (X \right )\) gives

\begin{align*} u \left (X \right )&=\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\\ u \left (X \right )&=-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c} \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} u \left (X \right )^{2} c +b u \left (X \right )+a &= \frac {c_1}{X^{2}} \\ u \left (X \right ) &= \frac {-b +\sqrt {-4 a c +b^{2}}}{2 c} \\ u \left (X \right ) &= -\frac {b +\sqrt {-4 a c +b^{2}}}{2 c} \\ \end{align*}

Converting \(u \left (X \right )^{2} c +b u \left (X \right )+a = \frac {c_1}{X^{2}}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right )^{2} c +b Y \left (X \right ) X +X^{2} a = c_1 \end{align*}

Converting \(u \left (X \right ) = \frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = \frac {X \left (-b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \end{align*}

Converting \(u \left (X \right ) = -\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = -\frac {X \left (b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right )^{2} c +b Y \left (X \right ) X +X^{2} a = c_1\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x +x_{0} \end{align*}

\begin{align*} Y &= y +\frac {-2 a e -b g}{4 a c -b^{2}}\\ X &= x +\frac {b e +2 c g}{4 a c -b^{2}} \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} \left (y-\frac {-2 a e -b g}{4 a c -b^{2}}\right )^{2} c +b \left (y-\frac {-2 a e -b g}{4 a c -b^{2}}\right ) \left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right )+\left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right )^{2} a = c_1 \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = \frac {X \left (-b +\sqrt {-4 a c +b^{2}}\right )}{2 c}\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x +x_{0} \end{align*}

\begin{align*} Y &= y +\frac {-2 a e -b g}{4 a c -b^{2}}\\ X &= x +\frac {b e +2 c g}{4 a c -b^{2}} \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} y-\frac {-2 a e -b g}{4 a c -b^{2}} = \frac {\left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right ) \left (-b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = -\frac {X \left (b +\sqrt {-4 a c +b^{2}}\right )}{2 c}\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x +x_{0} \end{align*}

\begin{align*} Y &= y +\frac {-2 a e -b g}{4 a c -b^{2}}\\ X &= x +\frac {b e +2 c g}{4 a c -b^{2}} \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} y-\frac {-2 a e -b g}{4 a c -b^{2}} = -\frac {\left (x -\frac {b e +2 c g}{4 a c -b^{2}}\right ) \left (b +\sqrt {-4 a c +b^{2}}\right )}{2 c} \end{align*}

Simplifying the above gives

\begin{align*} \frac {\left (4 a \,c^{2}-b^{2} c \right ) y^{2}+4 \left (b x +e \right ) \left (-\frac {b^{2}}{4}+a c \right ) y+4 a^{2} c \,x^{2}+\left (-b^{2} x^{2}-4 c g x +e^{2}\right ) a +g \left (b^{2} x +b e +c g \right )}{4 a c -b^{2}} &= c_1 \\ y+\frac {2 a e +b g}{4 a c -b^{2}} &= -\frac {\left (4 a c x -b^{2} x -b e -2 c g \right ) \left (b -\sqrt {-4 a c +b^{2}}\right )}{8 \left (-\frac {b^{2}}{4}+a c \right ) c} \\ y+\frac {2 a e +b g}{4 a c -b^{2}} &= -\frac {\left (4 a c x -b^{2} x -b e -2 c g \right ) \left (b +\sqrt {-4 a c +b^{2}}\right )}{8 \left (-\frac {b^{2}}{4}+a c \right ) c} \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= \frac {-4 a b c x +b^{3} x -4 a c e +b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x +64 c_1 \,a^{2} c^{3}-32 c_1 a \,b^{2} c^{2}+4 c_1 \,b^{4} c -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}}}{2 \left (4 a c -b^{2}\right ) c} \\ y &= -\frac {4 a b c x -b^{3} x +4 a c e -b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x +64 c_1 \,a^{2} c^{3}-32 c_1 a \,b^{2} c^{2}+4 c_1 \,b^{4} c -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}}}{2 \left (4 a c -b^{2}\right ) c} \\ y &= \frac {4 \sqrt {-4 a c +b^{2}}\, a c x -\sqrt {-4 a c +b^{2}}\, b^{2} x -4 a b c x +b^{3} x -\sqrt {-4 a c +b^{2}}\, b e -2 \sqrt {-4 a c +b^{2}}\, c g -4 a c e +b^{2} e}{2 \left (4 a c -b^{2}\right ) c} \\ y &= -\frac {4 \sqrt {-4 a c +b^{2}}\, a c x -\sqrt {-4 a c +b^{2}}\, b^{2} x +4 a b c x -b^{3} x -\sqrt {-4 a c +b^{2}}\, b e -2 \sqrt {-4 a c +b^{2}}\, c g +4 a c e -b^{2} e}{2 \left (4 a c -b^{2}\right ) c} \\ \end{align*}

Summary of solutions found

Entering ﬁrst order ode abel second kind solver

\begin{align*} 2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime }&=g \\ \end{align*}

2.3.3.5 Solved using ﬁrst_order_ode_abel_second_kind_case_5

1.288 (sec)

Abel ﬁrst order ode of the second kind has the form

\begin{align} (y+ g)y' &= f_0 + f_1 y+ f_2 y^2 + f_3 y^3\tag {1} \end{align}

Comparing the given ode

\[ 2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime } = g \]

To the form in (1) shows that

\begin{align*} g &=\frac {b x +e}{2 c}\\ f_0 &=\frac {-2 a x +g}{2 c}\\ f_1 &=-\frac {b}{2 c}\\ f_2 &=0\\ f_3 &=0 \end{align*}

When the condition \(f_1 = 2 f_2 g - g'\) is satisﬁed, then this ode has direct solution given by

\begin{align} y &= -g \pm \triangle \tag {2} \end{align}

Where

\begin{align} \triangle = U \sqrt { 2 \int { \frac {f_0+g g' - f_2 g^2}{U^2} \,dx}+ c_1 } \tag {3}\end{align}

And \(U\) is given by

\begin{align} U &= e^{\int {f_2 \,dx}} \tag {4}\end{align}

But \(f_1=-\frac {b}{2 c}\) and \(2 f_2 g - g'=-\frac {b}{2 c}\). Hence the condition is satisﬁed. Calcuating \(U\) from (4) gives

\begin{align*} U &= e^{\int {f_2 \,dx}}\\ U &= e^{\int {0\,dx}}\\ U &= 1 \end{align*}

Substituting the above in (3) gives

\begin{align*} \triangle &= U \sqrt { 2 \int { \frac {f_0+g g' - f_2 g^2}{U^2} \,dx}+ c_1 }\\ &= 1\sqrt { 2 \int { \frac {\left (\frac {-2 a x +g}{2 c}\right )+\left (\frac {b x +e}{2 c}\right ) \left (\frac {b}{2 c}\right ) - \left (0\right ) \left (\frac {\left (b x +e \right )^{2}}{4 c^{2}}\right )}{1} \,dx}+ c_1 }\\ &= \frac {\sqrt {\frac {-4 a c \,x^{2}+b^{2} x^{2}+2 b e x +4 c g x}{c^{2}}+4 c_1}}{2} \end{align*}

Hence from (2) the solution is

\begin{align*} y &= -g \pm \triangle \\ y &= -\frac {b x +e}{2 c}+\frac {\sqrt {\frac {-4 a c \,x^{2}+b^{2} x^{2}+2 b e x +4 c g x}{c^{2}}+4 c_1}}{2} \\ y &= -\frac {b x +e}{2 c}-\frac {\sqrt {\frac {-4 a c \,x^{2}+b^{2} x^{2}+2 b e x +4 c g x}{c^{2}}+4 c_1}}{2} \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= \frac {\sqrt {\frac {-4 a c \,x^{2}+b^{2} x^{2}+4 c_1 \,c^{2}+2 b e x +4 c g x}{c^{2}}}\, c -b x -e}{2 c} \\ y &= -\frac {\sqrt {\frac {-4 a c \,x^{2}+b^{2} x^{2}+4 c_1 \,c^{2}+2 b e x +4 c g x}{c^{2}}}\, c +b x +e}{2 c} \\ \end{align*}

Summary of solutions found

2.3.3.6 Solved using ﬁrst_order_ode_abel_second_kind_solved_by_converting_to_ﬁrst_kind

2.417 (sec)

This is Abel second kind ODE, it has the form

\[ \left (y+g\right )y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \]

Comparing the above to given ODE which is

\begin{align*}2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime } = g\tag {1} \end{align*}

Shows that

\begin{align*} g &= \frac {b x +e}{2 c}\\ f_0 &= \frac {-2 a x +g}{2 c}\\ f_1 &= -\frac {b}{2 c}\\ f_2 &= 0\\ f_3 &= 0 \end{align*}

Applying transformation

\begin{align*} y&=\frac {1}{u(x)}-g \end{align*}

Results in the new ode which is Abel ﬁrst kind

\begin{align*} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{3} \left (4 a c x -b^{2} x -b e -2 c g \right )}{4 c^{2}} \end{align*}

Which is now solved. Entering ﬁrst order ode separable solverThe ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{3} \left (4 a c x -b^{2} x -b e -2 c g \right )}{4 c^{2}} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right )^{3} \left (4 a c x -b^{2} x -b e -2 c g \right )}{4 c^{2}}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {4 a c x -b^{2} x -b e -2 c g}{4 c^{2}}\\ g(u) &= u^{3} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u^{3}}\,du} &= \int { \frac {4 a c x -b^{2} x -b e -2 c g}{4 c^{2}} \,dx} \\ \end{align*}

\[ -\frac {1}{2 u \left (x \right )^{2}}=\frac {x \left (4 a c x -b^{2} x -2 b e -4 c g \right )}{8 c^{2}}+c_4 \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u^{3}=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {1}{2 u \left (x \right )^{2}} &= \frac {x \left (4 a c x -b^{2} x -2 b e -4 c g \right )}{8 c^{2}}+c_4 \\ u \left (x \right ) &= 0 \\ \end{align*}

Simplifying the above gives

\begin{align*} -\frac {1}{2 u \left (x \right )^{2}} &= -\frac {x \left (\left (-a x +g \right ) c +\frac {b \left (\frac {b x}{2}+e \right )}{2}\right )}{2 c^{2}}+c_4 \\ u \left (x \right ) &= 0 \\ \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= -\frac {2 c}{\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}} \\ u \left (x \right ) &= \frac {2 c}{\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}} \\ \end{align*}

Now we transform the solution \(u \left (x \right ) = -\frac {2 c}{\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}}\) to \(y\) using \(u \left (x \right )=\frac {1}{y+\frac {b x +e}{2 c}}\) which gives

\[ y = -\frac {b x +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}+e}{2 c} \]

Now we transform the solution \(u \left (x \right ) = \frac {2 c}{\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}}\) to \(y\) using \(u \left (x \right )=\frac {1}{y+\frac {b x +e}{2 c}}\) which gives

\[ y = \frac {-b x +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}-e}{2 c} \]

Summary of solutions found

\begin{align*} y &= \frac {-b x +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}-e}{2 c} \\ y &= -\frac {b x +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}-8 c_4 \,c^{2}+2 b e x +4 c g x}+e}{2 c} \\ \end{align*}

2.3.3.7 Solved using ﬁrst_order_ode_LIE

1.971 (sec)

Entering ﬁrst order ode LIE solver

\begin{align*} 2 a x +b y+\left (2 c y+b x +e \right ) y^{\prime }&=g \\ \end{align*}

Writing the ode as

\begin{align*} y^{\prime }&=-\frac {2 a x +b y -g}{b x +2 c y +e}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coeﬃcients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}-\frac {\left (2 a x +b y -g \right ) \left (b_{3}-a_{2}\right )}{b x +2 c y +e}-\frac {\left (2 a x +b y -g \right )^{2} a_{3}}{\left (b x +2 c y +e \right )^{2}}-\left (-\frac {2 a}{b x +2 c y +e}+\frac {\left (2 a x +b y -g \right ) b}{\left (b x +2 c y +e \right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {b}{b x +2 c y +e}+\frac {2 \left (2 a x +b y -g \right ) c}{\left (b x +2 c y +e \right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {4 a^{2} x^{2} a_{3}-2 a b \,x^{2} a_{2}+2 a b \,x^{2} b_{3}+4 a b x y a_{3}+4 a c \,x^{2} b_{2}-8 a c x y a_{2}+8 a c x y b_{3}-4 a c \,y^{2} a_{3}-2 b^{2} x^{2} b_{2}+2 b^{2} y^{2} a_{3}-4 b c x y b_{2}-2 b c \,y^{2} a_{2}+2 b c \,y^{2} b_{3}-4 c^{2} y^{2} b_{2}+4 a c x b_{1}-4 a c y a_{1}-4 a e x a_{2}+2 a e x b_{3}-2 a e y a_{3}-4 a g x a_{3}-b^{2} x b_{1}+b^{2} y a_{1}-3 b e x b_{2}-b e y a_{2}-b g x b_{3}-3 b g y a_{3}-4 c e y b_{2}-2 c g x b_{2}+2 c g y a_{2}-4 c g y b_{3}-2 a e a_{1}-b e b_{1}-b g a_{1}-2 c g b_{1}-e^{2} b_{2}+e g a_{2}-e g b_{3}+g^{2} a_{3}}{\left (b x +2 c y +e \right )^{2}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -4 a^{2} x^{2} a_{3}+2 a b \,x^{2} a_{2}-2 a b \,x^{2} b_{3}-4 a b x y a_{3}-4 a c \,x^{2} b_{2}+8 a c x y a_{2}-8 a c x y b_{3}+4 a c \,y^{2} a_{3}+2 b^{2} x^{2} b_{2}-2 b^{2} y^{2} a_{3}+4 b c x y b_{2}+2 b c \,y^{2} a_{2}-2 b c \,y^{2} b_{3}+4 c^{2} y^{2} b_{2}-4 a c x b_{1}+4 a c y a_{1}+4 a e x a_{2}-2 a e x b_{3}+2 a e y a_{3}+4 a g x a_{3}+b^{2} x b_{1}-b^{2} y a_{1}+3 b e x b_{2}+b e y a_{2}+b g x b_{3}+3 b g y a_{3}+4 c e y b_{2}+2 c g x b_{2}-2 c g y a_{2}+4 c g y b_{3}+2 a e a_{1}+b e b_{1}+b g a_{1}+2 c g b_{1}+e^{2} b_{2}-e g a_{2}+e g b_{3}-g^{2} a_{3} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \{x, y\} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \{x = v_{1}, y = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} -4 a^{2} a_{3} v_{1}^{2}+2 a b a_{2} v_{1}^{2}-4 a b a_{3} v_{1} v_{2}-2 a b b_{3} v_{1}^{2}+8 a c a_{2} v_{1} v_{2}+4 a c a_{3} v_{2}^{2}-4 a c b_{2} v_{1}^{2}-8 a c b_{3} v_{1} v_{2}-2 b^{2} a_{3} v_{2}^{2}+2 b^{2} b_{2} v_{1}^{2}+2 b c a_{2} v_{2}^{2}+4 b c b_{2} v_{1} v_{2}-2 b c b_{3} v_{2}^{2}+4 c^{2} b_{2} v_{2}^{2}+4 a c a_{1} v_{2}-4 a c b_{1} v_{1}+4 a e a_{2} v_{1}+2 a e a_{3} v_{2}-2 a e b_{3} v_{1}+4 a g a_{3} v_{1}-b^{2} a_{1} v_{2}+b^{2} b_{1} v_{1}+b e a_{2} v_{2}+3 b e b_{2} v_{1}+3 b g a_{3} v_{2}+b g b_{3} v_{1}+4 c e b_{2} v_{2}-2 c g a_{2} v_{2}+2 c g b_{2} v_{1}+4 c g b_{3} v_{2}+2 a e a_{1}+b e b_{1}+b g a_{1}+2 c g b_{1}+e^{2} b_{2}-e g a_{2}+e g b_{3}-g^{2} a_{3} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} \left (-4 a^{2} a_{3}+2 a b a_{2}-2 a b b_{3}-4 a c b_{2}+2 b^{2} b_{2}\right ) v_{1}^{2}+\left (-4 a b a_{3}+8 a c a_{2}-8 a c b_{3}+4 b c b_{2}\right ) v_{1} v_{2}+\left (-4 a c b_{1}+4 a e a_{2}-2 a e b_{3}+4 a g a_{3}+b^{2} b_{1}+3 b e b_{2}+b g b_{3}+2 c g b_{2}\right ) v_{1}+\left (4 a c a_{3}-2 b^{2} a_{3}+2 b c a_{2}-2 b c b_{3}+4 c^{2} b_{2}\right ) v_{2}^{2}+\left (4 a c a_{1}+2 a e a_{3}-b^{2} a_{1}+b e a_{2}+3 b g a_{3}+4 c e b_{2}-2 c g a_{2}+4 c g b_{3}\right ) v_{2}+2 a e a_{1}+b e b_{1}+b g a_{1}+2 c g b_{1}+e^{2} b_{2}-e g a_{2}+e g b_{3}-g^{2} a_{3} = 0 \end{equation}

Setting each coeﬃcients in (8E) to zero gives the following equations to solve

\begin{align*} -4 a b a_{3}+8 a c a_{2}-8 a c b_{3}+4 b c b_{2}&=0\\ 4 a c a_{3}-2 b^{2} a_{3}+2 b c a_{2}-2 b c b_{3}+4 c^{2} b_{2}&=0\\ -4 a^{2} a_{3}+2 a b a_{2}-2 a b b_{3}-4 a c b_{2}+2 b^{2} b_{2}&=0\\ 4 a c a_{1}+2 a e a_{3}-b^{2} a_{1}+b e a_{2}+3 b g a_{3}+4 c e b_{2}-2 c g a_{2}+4 c g b_{3}&=0\\ -4 a c b_{1}+4 a e a_{2}-2 a e b_{3}+4 a g a_{3}+b^{2} b_{1}+3 b e b_{2}+b g b_{3}+2 c g b_{2}&=0\\ 2 a e a_{1}+b e b_{1}+b g a_{1}+2 c g b_{1}+e^{2} b_{2}-e g a_{2}+e g b_{3}-g^{2} a_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=\frac {2 a c e a_{3}-b^{2} e a_{3}-b c e b_{3}-b c g a_{3}-2 c^{2} g b_{3}}{c \left (4 a c -b^{2}\right )}\\ a_{2}&=\frac {b a_{3}+c b_{3}}{c}\\ a_{3}&=a_{3}\\ b_{1}&=\frac {a b e a_{3}+2 a c e b_{3}+2 a c g a_{3}+b c g b_{3}}{c \left (4 a c -b^{2}\right )}\\ b_{2}&=-\frac {a a_{3}}{c}\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= \frac {4 a c x -b^{2} x -b e -2 c g}{4 a c -b^{2}} \\ \eta &= \frac {4 a c y -b^{2} y +2 a e +b g}{4 a c -b^{2}} \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= \frac {4 a c y -b^{2} y +2 a e +b g}{4 a c -b^{2}} - \left (-\frac {2 a x +b y -g}{b x +2 c y +e}\right ) \left (\frac {4 a c x -b^{2} x -b e -2 c g}{4 a c -b^{2}}\right ) \\ &= \frac {2 a \left (2 c y +e \right )+b \left (-b y +g \right )}{4 a c -b^{2}}+\frac {2 \left (-2 a x -b y +g \right ) \left (\frac {b^{2} x}{2}+\frac {b e}{2}+\left (-2 a x +g \right ) c \right )}{\left (b x +2 c y +e \right ) \left (4 a c -b^{2}\right )}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to ﬁnd the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the ﬁrst pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {2 a \left (2 c y +e \right )+b \left (-b y +g \right )}{4 a c -b^{2}}+\frac {2 \left (-2 a x -b y +g \right ) \left (\frac {b^{2} x}{2}+\frac {b e}{2}+\left (-2 a x +g \right ) c \right )}{\left (b x +2 c y +e \right ) \left (4 a c -b^{2}\right )}}} dy \end{align*}

Which results in

\begin{align*} S&= \frac {\left (2 a c -\frac {b^{2}}{2}\right ) \ln \left (4 a^{2} c \,x^{2}-a \,b^{2} x^{2}+4 a b c x y +4 a \,c^{2} y^{2}-b^{3} x y -b^{2} c \,y^{2}+4 a c e y -4 a c g x -b^{2} e y +b^{2} g x +a \,e^{2}+b e g +c \,g^{2}\right )}{4 a c -b^{2}} \end{align*}

Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= -\frac {2 a x +b y -g}{b x +2 c y +e} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {\left (2 a x +b y -g \right ) \left (4 a c -b^{2}\right )}{8 a^{2} c \,x^{2}+\left (-2 b^{2} x^{2}+8 c x y b +8 c^{2} y^{2}+\left (8 e y -8 g x \right ) c +2 e^{2}\right ) a -2 \left (b^{2} x +\left (c y +e \right ) b +c g \right ) \left (b y -g \right )}\\ S_{y} &= \frac {\left (b x +2 c y +e \right ) \left (4 a c -b^{2}\right )}{8 a^{2} c \,x^{2}+\left (-2 b^{2} x^{2}+8 c x y b +8 c^{2} y^{2}+\left (8 e y -8 g x \right ) c +2 e^{2}\right ) a -2 \left (b^{2} x +\left (c y +e \right ) b +c g \right ) \left (b y -g \right )} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= 0\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= 0 \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {0\, dR} + c_2 \\ S \left (R \right ) &= c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \frac {\ln \left (4 a^{2} c \,x^{2}+\left (-b^{2} x^{2}+4 y b c x -4 c g x +4 \left (c y+\frac {e}{2}\right )^{2}\right ) a -\left (b^{2} x +\left (c y+e \right ) b +c g \right ) \left (b y-g \right )\right )}{2} = c_2 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= \frac {-4 a b c x +b^{3} x -4 a c e +b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}+16 \,{\mathrm e}^{2 c_2} a \,c^{2}-4 \,{\mathrm e}^{2 c_2} b^{2} c}}{2 c \left (4 a c -b^{2}\right )} \\ y &= -\frac {4 a b c x -b^{3} x +4 a c e -b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}+16 \,{\mathrm e}^{2 c_2} a \,c^{2}-4 \,{\mathrm e}^{2 c_2} b^{2} c}}{2 c \left (4 a c -b^{2}\right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {-4 a b c x +b^{3} x -4 a c e +b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}+16 \,{\mathrm e}^{2 c_2} a \,c^{2}-4 \,{\mathrm e}^{2 c_2} b^{2} c}}{2 c \left (4 a c -b^{2}\right )} \\ y &= -\frac {4 a b c x -b^{3} x +4 a c e -b^{2} e +\sqrt {-64 a^{3} c^{3} x^{2}+48 a^{2} b^{2} c^{2} x^{2}-12 a \,b^{4} c \,x^{2}+b^{6} x^{2}+32 a^{2} b \,c^{2} e x +64 a^{2} c^{3} g x -16 a \,b^{3} c e x -32 a \,b^{2} c^{2} g x +2 b^{5} e x +4 b^{4} c g x -4 a \,b^{2} c \,e^{2}-16 a b \,c^{2} e g -16 a \,c^{3} g^{2}+b^{4} e^{2}+4 b^{3} c e g +4 b^{2} c^{2} g^{2}+16 \,{\mathrm e}^{2 c_2} a \,c^{2}-4 \,{\mathrm e}^{2 c_2} b^{2} c}}{2 c \left (4 a c -b^{2}\right )} \\ \end{align*}

2.3.3.8 ✓ Maple. Time used: 0.065 (sec). Leaf size: 88

ode:=2*a*x+b*y(x)+(2*c*y(x)+b*x+e)*diff(y(x),x) = g; 
dsolve(ode,y(x), singsol=all);

\[ y = \frac {\sqrt {-64 \left (a c -\frac {b^{2}}{4}\right ) \left (\left (a x -\frac {g}{2}\right ) c -\frac {b \left (b x +e \right )}{4}\right )^{2} c_1^{2}+4 c}+\left (-4 a b c x +b^{3} x -4 a c e +b^{2} e \right ) c_1}{8 \left (a c -\frac {b^{2}}{4}\right ) c c_1} \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying homogeneous C 
trying homogeneous types: 
trying homogeneous D 
<- homogeneous successful 
<- homogeneous successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 a x +b y \left (x \right )+\left (2 c y \left (x \right )+b x +e \right ) \left (\frac {d}{d x}y \left (x \right )\right )=g \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \square & {} & \textrm {Check if ODE is exact}\hspace {3pt} \\ {} & \circ & \textrm {ODE is exact if the lhs is the total derivative of a}\hspace {3pt} C^{2}\hspace {3pt}\textrm {function}\hspace {3pt} \\ {} & {} & \frac {d}{d x}G \left (x , y \left (x \right )\right )=0 \\ {} & \circ & \textrm {Compute derivative of lhs}\hspace {3pt} \\ {} & {} & \frac {\partial }{\partial x}G \left (x , y\right )+\left (\frac {\partial }{\partial y}G \left (x , y\right )\right ) \left (\frac {d}{d x}y \left (x \right )\right )=0 \\ {} & \circ & \textrm {Evaluate derivatives}\hspace {3pt} \\ {} & {} & b =b \\ {} & \circ & \textrm {Condition met, ODE is exact}\hspace {3pt} \\ \bullet & {} & \textrm {Exact ODE implies solution will be of this form}\hspace {3pt} \\ {} & {} & \left [G \left (x , y\right )=\mathit {C1} , M \left (x , y\right )=\frac {\partial }{\partial x}G \left (x , y\right ), N \left (x , y\right )=\frac {\partial }{\partial y}G \left (x , y\right )\right ] \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} G \left (x , y\right )\hspace {3pt}\textrm {by integrating}\hspace {3pt} M \left (x , y\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} x \\ {} & {} & G \left (x , y\right )=\int \left (2 a x +b y -g \right )d x +\textit {\_F1} \left (y \right ) \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & G \left (x , y\right )=a \,x^{2}+b y x -g x +\textit {\_F1} \left (y \right ) \\ \bullet & {} & \textrm {Take derivative of}\hspace {3pt} G \left (x , y\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} y \\ {} & {} & N \left (x , y\right )=\frac {\partial }{\partial y}G \left (x , y\right ) \\ \bullet & {} & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & b x +2 c y +e =b x +\frac {d}{d y}\textit {\_F1} \left (y \right ) \\ \bullet & {} & \textrm {Isolate for}\hspace {3pt} \frac {d}{d y}\textit {\_F1} \left (y \right ) \\ {} & {} & \frac {d}{d y}\textit {\_F1} \left (y \right )=2 c y +e \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_F1} \left (y \right ) \\ {} & {} & \textit {\_F1} \left (y \right )=c \,y^{2}+e y \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_F1} \left (y \right )\hspace {3pt}\textrm {into equation for}\hspace {3pt} G \left (x , y\right ) \\ {} & {} & G \left (x , y\right )=a \,x^{2}+b y x +c \,y^{2}+e y -g x \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} G \left (x , y\right )\hspace {3pt}\textrm {into the solution of the ODE}\hspace {3pt} \\ {} & {} & a \,x^{2}+b y x +c \,y^{2}+e y -g x =\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \left \{y \left (x \right )=\frac {-b x -e +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}+2 b e x +4 c g x +4 \mathit {C1} c +e^{2}}}{2 c}, y \left (x \right )=-\frac {b x +\sqrt {-4 a c \,x^{2}+b^{2} x^{2}+2 b e x +4 c g x +4 \mathit {C1} c +e^{2}}+e}{2 c}\right \} \end {array} \]

2.3.3.9 ✓ Mathematica. Time used: 20.463 (sec). Leaf size: 132

ode=(2*a*x+b*y[x])+(2*c*y[x]+b*x+e)*D[y[x],x]==g; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

\begin{align*} y(x)&\to -\frac {\frac {\sqrt {\frac {4 c x (g-a x)+b^2 x^2+2 b e x+4 c^2 c_1+e^2}{c}}}{\sqrt {\frac {1}{c}}}+b x+e}{2 c}\\ y(x)&\to -\frac {-\frac {\sqrt {\frac {4 c x (g-a x)+b^2 x^2+2 b e x+4 c^2 c_1+e^2}{c}}}{\sqrt {\frac {1}{c}}}+b x+e}{2 c} \end{align*}

2.3.3.10 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
e = symbols("e") 
g = symbols("g") 
y = Function("y") 
ode = Eq(2*a*x + b*y(x) - g + (b*x + 2*c*y(x) + e)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)

NotImplementedError : The given ODE -(-2*a*x - b*y(x) + g)/(b*x + 2*c*y(x) + e) + Derivative(y(x), x

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0

classify_ode(ode,func=y(x)) 
 
('factorable', '1st_exact', '1st_power_series', 'lie_group', '1st_exact_Integral')

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