2.1.8 Problem 8
Internal
problem
ID
[4083]
Book
:
Differential
equations,
Shepley
L.
Ross,
1964
Section
:
2.4,
page
55
Problem
number
:
8
Date
solved
:
Saturday, December 06, 2025 at 04:16:51 PM
CAS
classification
:
[[_homogeneous, `class C`], _exact, _rational, [_Abel, `2nd type`, `class A`]]
2.1.8.1 Existence and uniqueness analysis
\begin{align*}
6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime }&=0 \\
y \left (\frac {1}{2}\right ) &= 3 \\
\end{align*}
This is non linear first order ODE. In canonical form it is written as \begin{align*} y^{\prime } &= f(x,y)\\ &= -\frac {6 x +4 y +1}{2 \left (2 x +y +1\right )} \end{align*}
The \(x\) domain of \(f(x,y)\) when \(y=3\) is
\[
\{x <-2\boldsymbol {\lor }-2<x\}
\]
And the point \(x_0 = {\frac {1}{2}}\) is inside this domain. The \(y\) domain of \(f(x,y)\) when \(x={\frac {1}{2}}\) is \[
\{y <-2\boldsymbol {\lor }-2<y\}
\]
And
the point \(y_0 = 3\) is inside this domain. Now we will look at the continuity of \begin{align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (-\frac {6 x +4 y +1}{2 \left (2 x +y +1\right )}\right ) \\ &= -\frac {2}{2 x +y +1}+\frac {6 x +4 y +1}{2 \left (2 x +y +1\right )^{2}} \end{align*}
The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=3\) is
\[
\{x <-2\boldsymbol {\lor }-2<x\}
\]
And the point \(x_0 = {\frac {1}{2}}\) is inside this domain. The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x={\frac {1}{2}}\) is \[
\{y <-2\boldsymbol {\lor }-2<y\}
\]
And
the point \(y_0 = 3\) is inside this domain. Therefore solution exists and is unique.
2.1.8.2 Solved using first_order_ode_exact
0.087 (sec)
Entering first order ode exact solver
\begin{align*}
6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime }&=0 \\
y \left (\frac {1}{2}\right ) &= 3 \\
\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 (4 x +2 y +2\right )\mathop {\mathrm {d}y} &= \left (-6 x -4 y -1\right )\mathop {\mathrm {d}x}\\ \left (6 x +4 y +1\right )\mathop {\mathrm {d}x} + \left (4 x +2 y +2\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(x,y) &= 6 x +4 y +1\\ N(x,y) &= 4 x +2 y +2 \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 (6 x +4 y +1\right )\\ &= 4 \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (4 x +2 y +2\right )\\ &= 4 \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 6 x +4 y +1\mathop {\mathrm {d}x} \\
\tag{3} \phi &= x \left (3 x +4 y +1\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} = 4 x+f'(y)
\end{equation}
But equation (2) says that \(\frac {\partial \phi }{\partial y} = 4 x +2 y +2\).
Therefore equation (4) becomes \begin{equation}
\tag{5} 4 x +2 y +2 = 4 x+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives \[
f'(y) = 2 y +2
\]
Integrating the above
w.r.t \(y\) gives \begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( 2 y +2\right ) \mathop {\mathrm {d}y} \\
f(y) &= y^{2}+2 y+ c_1 \\
\end{align*}
Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\)
into equation (3) gives \(\phi \) \[
\phi = x \left (3 x +4 y +1\right )+y^{2}+2 y+ c_1
\]
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 \[
c_1 = x \left (3 x +4 y +1\right )+y^{2}+2 y
\]
Simplifying the above gives \begin{align*}
y^{2}+\left (4 x +2\right ) y+3 x^{2}+x &= c_1 \\
\end{align*}
Solving for initial conditions the solution is \begin{align*}
y^{2}+\left (4 x +2\right ) y+3 x^{2}+x &= {\frac {89}{4}} \\
\end{align*}
Solving for \(y\) gives \begin{align*}
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
|
|
|
| Solution plot | Slope field \(6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime } = 0\) |
Summary of solutions found
\begin{align*}
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
2.1.8.3 Solved using first_order_ode_dAlembert
0.408 (sec)
Entering first order ode dAlembert solver
\begin{align*}
6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime }&=0 \\
y \left (\frac {1}{2}\right ) &= 3 \\
\end{align*}
Let \(p=y^{\prime }\) the ode becomes \begin{align*} 6 x +4 y +1+\left (4 x +2 y +2\right ) p = 0 \end{align*}
Solving for \(y\) from the above results in
\begin{align*}
\tag{1} y &= -\frac {\left (4 p +6\right ) x}{2 \left (2+p \right )}-\frac {2 p +1}{2 \left (2+p \right )} \\
\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 {-2 p -3}{2+p}\\ g &= \frac {-2 p -1}{4+2 p} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p -\frac {-2 p -3}{2+p} = \left (-\frac {2 x}{2+p}+\frac {2 x p}{\left (2+p \right )^{2}}+\frac {3 x}{\left (2+p \right )^{2}}-\frac {2}{4+2 p}+\frac {4 p}{\left (4+2 p \right )^{2}}+\frac {2}{\left (4+2 p \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 {-2 p -3}{2+p} = 0 \end{align*}
No valid singular solutions found.
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 {-2 p \left (x \right )-3}{2+p \left (x \right )}}{-\frac {2 x}{2+p \left (x \right )}+\frac {2 x p \left (x \right )}{\left (2+p \left (x \right )\right )^{2}}+\frac {3 x}{\left (2+p \left (x \right )\right )^{2}}-\frac {2}{4+2 p \left (x \right )}+\frac {4 p \left (x \right )}{\left (4+2 p \left (x \right )\right )^{2}}+\frac {2}{\left (4+2 p \left (x \right )\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+p \left (x \right )\right ) \left (p \left (x \right )+3\right ) \left (p \left (x \right )+1\right )}{2 x +3}
\end{equation}
is separable as it can be written as \begin{align*} p^{\prime }\left (x \right )&= -\frac {2 \left (2+p \left (x \right )\right ) \left (p \left (x \right )+3\right ) \left (p \left (x \right )+1\right )}{2 x +3}\\ &= f(x) g(p) \end{align*}
Where
\begin{align*} f(x) &= -\frac {2}{2 x +3}\\ g(p) &= \left (p +3\right ) \left (2+p \right ) \left (p +1\right ) \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\
\int { \frac {1}{\left (p +3\right ) \left (2+p \right ) \left (p +1\right )}\,dp} &= \int { -\frac {2}{2 x +3} \,dx} \\
\end{align*}
\[
-\ln \left (2+p \left (x \right )\right )+\frac {\ln \left (p \left (x \right )+1\right )}{2}+\frac {\ln \left (p \left (x \right )+3\right )}{2}=\ln \left (\frac {1}{2 x +3}\right )+c_1
\]
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 \[
\left (p +3\right ) \left (2+p \right ) \left (p +1\right )=0
\]
for \(p \left (x \right )\) gives
\begin{align*} p \left (x \right )&=-3\\ p \left (x \right )&=-2 \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*}
-\ln \left (2+p \left (x \right )\right )+\frac {\ln \left (p \left (x \right )+1\right )}{2}+\frac {\ln \left (p \left (x \right )+3\right )}{2} &= \ln \left (\frac {1}{2 x +3}\right )+c_1 \\
p \left (x \right ) &= -3 \\
p \left (x \right ) &= -2 \\
\end{align*}
Substituing the above solution for \(p\) in (2A) gives \begin{align*}
y &= \frac {x \left (\frac {2 \,{\mathrm e}^{2 c_1}-8 x^{2}+2 \sqrt {-4 \,{\mathrm e}^{2 c_1} x^{2}+16 x^{4}-12 \,{\mathrm e}^{2 c_1} x +96 x^{3}-9 \,{\mathrm e}^{2 c_1}+216 x^{2}+216 x +81}-24 x -18}{{\mathrm e}^{2 c_1}-4 x^{2}-12 x -9}-1\right )}{1-\frac {{\mathrm e}^{2 c_1}-4 x^{2}+\sqrt {-4 \,{\mathrm e}^{2 c_1} x^{2}+16 x^{4}-12 \,{\mathrm e}^{2 c_1} x +96 x^{3}-9 \,{\mathrm e}^{2 c_1}+216 x^{2}+216 x +81}-12 x -9}{{\mathrm e}^{2 c_1}-4 x^{2}-12 x -9}}+\frac {\frac {2 \,{\mathrm e}^{2 c_1}-8 x^{2}+2 \sqrt {-4 \,{\mathrm e}^{2 c_1} x^{2}+16 x^{4}-12 \,{\mathrm e}^{2 c_1} x +96 x^{3}-9 \,{\mathrm e}^{2 c_1}+216 x^{2}+216 x +81}-24 x -18}{{\mathrm e}^{2 c_1}-4 x^{2}-12 x -9}+1}{2-\frac {2 \left ({\mathrm e}^{2 c_1}-4 x^{2}+\sqrt {-4 \,{\mathrm e}^{2 c_1} x^{2}+16 x^{4}-12 \,{\mathrm e}^{2 c_1} x +96 x^{3}-9 \,{\mathrm e}^{2 c_1}+216 x^{2}+216 x +81}-12 x -9\right )}{{\mathrm e}^{2 c_1}-4 x^{2}-12 x -9}} \\
y &= -3 x -\frac {5}{2} \\
\end{align*}
Simplifying
the above gives \begin{align*}
y &= \frac {\left (-1-2 x \right ) \sqrt {-\left (2 x +3\right )^{2} {\mathrm e}^{2 c_1}+16 \left (x +\frac {3}{2}\right )^{4}}+4 \left (x +\frac {3}{2}\right ) \left (-\frac {{\mathrm e}^{2 c_1}}{4}+\left (x +\frac {3}{2}\right )^{2}\right )}{\sqrt {-\left (2 x +3\right )^{2} {\mathrm e}^{2 c_1}+16 \left (x +\frac {3}{2}\right )^{4}}} \\
y &= -3 x -\frac {5}{2} \\
\end{align*}
Solving for initial conditions the solution is \begin{align*}
y &= \frac {\left (-4 x -2\right ) \sqrt {\left (4 x^{2}+12 x +93\right ) \left (2 x +3\right )^{2}}+8 x^{3}+36 x^{2}+222 x +279}{2 \sqrt {\left (4 x^{2}+12 x +93\right ) \left (2 x +3\right )^{2}}} \\
y &= -3 x -\frac {5}{2} \\
\end{align*}
The solution \[
y = -3 x -\frac {5}{2}
\]
was found not to
satisfy the ode or the IC. Hence it is removed.
|
|
|
| Solution plot | Slope field \(6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime } = 0\) |
Summary of solutions found
\begin{align*}
y &= \frac {\left (-4 x -2\right ) \sqrt {\left (4 x^{2}+12 x +93\right ) \left (2 x +3\right )^{2}}+8 x^{3}+36 x^{2}+222 x +279}{2 \sqrt {\left (4 x^{2}+12 x +93\right ) \left (2 x +3\right )^{2}}} \\
\end{align*}
2.1.8.4 Solved using first_order_ode_homog_type_maple_C
0.546 (sec)
Entering first order ode homog type maple C solver
\begin{align*}
6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime }&=0 \\
y \left (\frac {1}{2}\right ) &= 3 \\
\end{align*}
Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to
new ode in \(Y(X)\) \[
\frac {d}{d X}Y \left (X \right ) = -\frac {6 X +6 x_{0} +4 Y \left (X \right )+4 y_{0} +1}{2 \left (2 X +2 x_{0} +Y \left (X \right )+y_{0} +1\right )}
\]
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 {3}{2}}\\ y_{0}&=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 {6 X +4 Y \left (X \right )}{2 \left (2 X +Y \left (X \right )\right )} \end{align*}
In canonical form, the ODE is
\begin{align*} Y' &= F(X,Y)\\ &= -\frac {3 X +2 Y}{2 X +Y}\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 \(M=-3 X -2 Y\) and \(N=2 X +Y\) 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 {-2 u -3}{u +2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {-2 u \left (X \right )-3}{u \left (X \right )+2}-u \left (X \right )}{X} \end{align*}
Or
\[ \frac {d}{d X}u \left (X \right )-\frac {\frac {-2 u \left (X \right )-3}{u \left (X \right )+2}-u \left (X \right )}{X} = 0 \]
Or \[ \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )+u \left (X \right )^{2}+2 \left (\frac {d}{d X}u \left (X \right )\right ) X +4 u \left (X \right )+3 = 0 \]
Or \[ X \left (u \left (X \right )+2\right ) \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}+4 u \left (X \right )+3 = 0 \]
Which is now solved as separable in \(u \left (X \right )\).
The ode
\begin{equation}
\frac {d}{d X}u \left (X \right ) = -\frac {\left (u \left (X \right )+3\right ) \left (u \left (X \right )+1\right )}{X \left (u \left (X \right )+2\right )}
\end{equation}
is separable as it can be written as \begin{align*} \frac {d}{d X}u \left (X \right )&= -\frac {\left (u \left (X \right )+3\right ) \left (u \left (X \right )+1\right )}{X \left (u \left (X \right )+2\right )}\\ &= f(X) g(u) \end{align*}
Where
\begin{align*} f(X) &= -\frac {1}{X}\\ g(u) &= \frac {\left (u +3\right ) \left (u +1\right )}{u +2} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\
\int { \frac {u +2}{\left (u +3\right ) \left (u +1\right )}\,du} &= \int { -\frac {1}{X} \,dX} \\
\end{align*}
\[
\frac {\ln \left (\left (u \left (X \right )+3\right ) \left (u \left (X \right )+1\right )\right )}{2}=\ln \left (\frac {1}{X}\right )+c_1
\]
Taking the exponential of both sides the solution becomes\[
\sqrt {\left (u \left (X \right )+3\right ) \left (u \left (X \right )+1\right )} = \frac {c_1}{X}
\]
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 \[
\frac {\left (u +3\right ) \left (u +1\right )}{u +2}=0
\]
for \(u \left (X \right )\) gives \begin{align*} u \left (X \right )&=-3\\ u \left (X \right )&=-1 \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*}
\sqrt {\left (u \left (X \right )+3\right ) \left (u \left (X \right )+1\right )} &= \frac {c_1}{X} \\
u \left (X \right ) &= -3 \\
u \left (X \right ) &= -1 \\
\end{align*}
Converting \(\sqrt {\left (u \left (X \right )+3\right ) \left (u \left (X \right )+1\right )} = \frac {c_1}{X}\) back to \(Y \left (X \right )\) gives \begin{align*} \sqrt {\frac {\left (Y \left (X \right )+3 X \right ) \left (Y \left (X \right )+X \right )}{X^{2}}} = \frac {c_1}{X} \end{align*}
Converting \(u \left (X \right ) = -3\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = -3 X \end{align*}
Converting \(u \left (X \right ) = -1\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = -X \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} \sqrt {\frac {\left (Y \left (X \right )+3 X \right ) \left (Y \left (X \right )+X \right )}{X^{2}}} = \frac {c_1}{X}\tag {A} \end{align*}
And replacing back terms in the above solution using
\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}
Or
\begin{align*} Y &= y +2\\ X &= x -\frac {3}{2} \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} \sqrt {\frac {\left (y+3 x +\frac {5}{2}\right ) \left (y+x -\frac {1}{2}\right )}{\left (x +\frac {3}{2}\right )^{2}}} = \frac {c_1}{x +\frac {3}{2}} \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = -3 X\tag {A} \end{align*}
And replacing back terms in the above solution using
\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}
Or
\begin{align*} Y &= y +2\\ X &= x -\frac {3}{2} \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} y-2 = -3 x -\frac {9}{2} \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = -X\tag {A} \end{align*}
And replacing back terms in the above solution using
\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}
Or
\begin{align*} Y &= y +2\\ X &= x -\frac {3}{2} \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} y-2 = -x -\frac {3}{2} \end{align*}
Simplifying the above gives
\begin{align*}
\sqrt {\frac {\left (2 y+6 x +5\right ) \left (2 y+2 x -1\right )}{\left (2 x +3\right )^{2}}} &= \frac {2 c_1}{2 x +3} \\
y-2 &= -3 x -\frac {9}{2} \\
y-2 &= -x -\frac {3}{2} \\
\end{align*}
Solving for initial conditions the solution is \begin{align*}
\sqrt {\frac {\left (2 y+6 x +5\right ) \left (2 y+2 x -1\right )}{\left (2 x +3\right )^{2}}} &= \frac {2 \sqrt {21}}{2 x +3} \\
y-2 &= -3 x -\frac {9}{2} \\
y-2 &= -x -\frac {3}{2} \\
\end{align*}
Solving for \(y\) gives \begin{align*}
y-2 &= -3 x -\frac {9}{2} \\
y-2 &= -x -\frac {3}{2} \\
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
The
solution \[
y-2 = -3 x -\frac {9}{2}
\]
was found not to satisfy the ode or the IC. Hence it is removed. The solution \[
y-2 = -x -\frac {3}{2}
\]
was found
not to satisfy the ode or the IC. Hence it is removed.
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|
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| Solution plot | Slope field \(6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime } = 0\) |
Summary of solutions found
\begin{align*}
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
Entering first order ode abel second kind solver\begin{align*}
6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime }&=0 \\
y \left (\frac {1}{2}\right ) &= 3 \\
\end{align*}
2.1.8.5 Solved using first_order_ode_abel_second_kind_case_5
0.076 (sec) Solving for initial conditions the solution is
\begin{align*}
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
|
|
|
| Solution plot | Slope field \(6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime } = 0\) |
|
|
|
| Solution plot | Slope field \(6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime } = 0\) |
Summary of solutions found
\begin{align*}
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
2.1.8.6 Solved using first_order_ode_LIE
0.939 (sec)
Entering first order ode LIE solver
\begin{align*}
6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime }&=0 \\
y \left (\frac {1}{2}\right ) &= 3 \\
\end{align*}
Writing the ode as \begin{align*} y^{\prime }&=-\frac {6 x +4 y +1}{2 \left (2 x +y +1\right )}\\ 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 (6 x +4 y +1\right ) \left (b_{3}-a_{2}\right )}{2 \left (2 x +y +1\right )}-\frac {\left (6 x +4 y +1\right )^{2} a_{3}}{4 \left (2 x +y +1\right )^{2}}-\left (-\frac {3}{2 x +y +1}+\frac {6 x +4 y +1}{\left (2 x +y +1\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {2}{2 x +y +1}+\frac {6 x +4 y +1}{2 \left (2 x +y +1\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives \[
\frac {24 x^{2} a_{2}-36 x^{2} a_{3}+20 x^{2} b_{2}-24 x^{2} b_{3}+24 x y a_{2}-48 x y a_{3}+16 x y b_{2}-24 x y b_{3}+8 y^{2} a_{2}-20 y^{2} a_{3}+4 y^{2} b_{2}-8 y^{2} b_{3}+24 x a_{2}-12 x a_{3}+4 x b_{1}+22 x b_{2}-16 x b_{3}-4 y a_{1}+10 y a_{2}+8 y b_{2}-4 y b_{3}+8 a_{1}+2 a_{2}-a_{3}+6 b_{1}+4 b_{2}-2 b_{3}}{4 \left (2 x +y +1\right )^{2}} = 0
\]
Setting the numerator to zero gives \begin{equation}
\tag{6E} 24 x^{2} a_{2}-36 x^{2} a_{3}+20 x^{2} b_{2}-24 x^{2} b_{3}+24 x y a_{2}-48 x y a_{3}+16 x y b_{2}-24 x y b_{3}+8 y^{2} a_{2}-20 y^{2} a_{3}+4 y^{2} b_{2}-8 y^{2} b_{3}+24 x a_{2}-12 x a_{3}+4 x b_{1}+22 x b_{2}-16 x b_{3}-4 y a_{1}+10 y a_{2}+8 y b_{2}-4 y b_{3}+8 a_{1}+2 a_{2}-a_{3}+6 b_{1}+4 b_{2}-2 b_{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} 24 a_{2} v_{1}^{2}+24 a_{2} v_{1} v_{2}+8 a_{2} v_{2}^{2}-36 a_{3} v_{1}^{2}-48 a_{3} v_{1} v_{2}-20 a_{3} v_{2}^{2}+20 b_{2} v_{1}^{2}+16 b_{2} v_{1} v_{2}+4 b_{2} v_{2}^{2}-24 b_{3} v_{1}^{2}-24 b_{3} v_{1} v_{2}-8 b_{3} v_{2}^{2}-4 a_{1} v_{2}+24 a_{2} v_{1}+10 a_{2} v_{2}-12 a_{3} v_{1}+4 b_{1} v_{1}+22 b_{2} v_{1}+8 b_{2} v_{2}-16 b_{3} v_{1}-4 b_{3} v_{2}+8 a_{1}+2 a_{2}-a_{3}+6 b_{1}+4 b_{2}-2 b_{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 (24 a_{2}-36 a_{3}+20 b_{2}-24 b_{3}\right ) v_{1}^{2}+\left (24 a_{2}-48 a_{3}+16 b_{2}-24 b_{3}\right ) v_{1} v_{2}+\left (24 a_{2}-12 a_{3}+4 b_{1}+22 b_{2}-16 b_{3}\right ) v_{1}+\left (8 a_{2}-20 a_{3}+4 b_{2}-8 b_{3}\right ) v_{2}^{2}+\left (-4 a_{1}+10 a_{2}+8 b_{2}-4 b_{3}\right ) v_{2}+8 a_{1}+2 a_{2}-a_{3}+6 b_{1}+4 b_{2}-2 b_{3} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} -4 a_{1}+10 a_{2}+8 b_{2}-4 b_{3}&=0\\ 8 a_{2}-20 a_{3}+4 b_{2}-8 b_{3}&=0\\ 24 a_{2}-48 a_{3}+16 b_{2}-24 b_{3}&=0\\ 24 a_{2}-36 a_{3}+20 b_{2}-24 b_{3}&=0\\ 24 a_{2}-12 a_{3}+4 b_{1}+22 b_{2}-16 b_{3}&=0\\ 8 a_{1}+2 a_{2}-a_{3}+6 b_{1}+4 b_{2}-2 b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=4 a_{3}+\frac {3 b_{3}}{2}\\ a_{2}&=4 a_{3}+b_{3}\\ a_{3}&=a_{3}\\ b_{1}&=-\frac {9 a_{3}}{2}-2 b_{3}\\ b_{2}&=-3 a_{3}\\ 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 &= x +\frac {3}{2} \\
\eta &= y -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 \\ &= y -2 - \left (-\frac {6 x +4 y +1}{2 \left (2 x +y +1\right )}\right ) \left (x +\frac {3}{2}\right ) \\ &= \frac {3 \left (x +\frac {y}{3}+\frac {5}{6}\right ) \left (x +y -\frac {1}{2}\right )}{2 x +y +1}\\ \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 {3 \left (x +\frac {y}{3}+\frac {5}{6}\right ) \left (x +y -\frac {1}{2}\right )}{2 x +y +1}}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (\left (6 x +2 y +5\right ) \left (2 x +2 y -1\right )\right )}{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 {6 x +4 y +1}{2 \left (2 x +y +1\right )} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {3}{6 x +2 y +5}+\frac {1}{2 x +2 y -1}\\ S_{y} &= \frac {1}{6 x +2 y +5}+\frac {1}{2 x +2 y -1} \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 (2 y+6 x +5\right )}{2}+\frac {\ln \left (2 y+2 x -1\right )}{2} = c_2 \end{align*}
The following diagram shows solution curves of the original ode and how they transform in the
canonical coordinates space using the mapping shown.
| | |
| Original ode in \(x,y\) coordinates |
Canonical coordinates
transformation |
ODE in canonical coordinates \((R,S)\) |
| | |
| \( \frac {dy}{dx} = -\frac {6 x +4 y +1}{2 \left (2 x +y +1\right )}\) |
|
\( \frac {d S}{d R} = 0\) |
|
|
\(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (6 x +2 y +5\right )}{2}+\frac {\ln \left (2 x +2 y -1\right )}{2} \end {aligned} \) |
|
| | |
Solving for initial conditions the solution is
\begin{align*}
\frac {\ln \left (2 y+6 x +5\right )}{2}+\frac {\ln \left (2 y+2 x -1\right )}{2} &= \ln \left (2\right )+\frac {\ln \left (7\right )}{2}+\frac {\ln \left (3\right )}{2} \\
\end{align*}
Solving for \(y\) gives \begin{align*}
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
|
|
|
| Solution plot | Slope field \(6 x +4 y+1+\left (4 x +2 y+2\right ) y^{\prime } = 0\) |
Summary of solutions found
\begin{align*}
y &= -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\
\end{align*}
2.1.8.7 ✓ Maple. Time used: 0.125 (sec). Leaf size: 23
ode:=6*x+4*y(x)+1+(4*x+2*y(x)+2)*diff(y(x),x) = 0;
ic:=[y(1/2) = 3];
dsolve([ode,op(ic)],y(x), singsol=all);
\[
y = -2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2}
\]
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} \\ {} & {} & \left [6 x +4 y \left (x \right )+1+\left (4 x +2 y \left (x \right )+2\right ) \left (\frac {d}{d x}y \left (x \right )\right )=0, y \left (\frac {1}{2}\right )=3\right ] \\ \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} \\ {} & {} & 4=4 \\ {} & \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 (6 x +4 y +1\right )d x +\textit {\_F1} \left (y \right ) \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & G \left (x , y\right )=3 x^{2}+4 y x +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} \\ {} & {} & 4 x +2 y +2=4 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 y +2 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_F1} \left (y \right ) \\ {} & {} & \textit {\_F1} \left (y \right )=y^{2}+2 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 )=3 x^{2}+4 y x +y^{2}+x +2 y \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} G \left (x , y\right )\hspace {3pt}\textrm {into the solution of the ODE}\hspace {3pt} \\ {} & {} & 3 x^{2}+4 y x +y^{2}+x +2 y =\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \left \{y \left (x \right )=-2 x -1-\sqrt {x^{2}+\mathit {C1} +3 x +1}, y \left (x \right )=-2 x -1+\sqrt {x^{2}+\mathit {C1} +3 x +1}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\frac {1}{2}\right )=3 \\ {} & {} & 3=-2-\sqrt {\mathit {C1} +\frac {11}{4}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \textrm {No solution}\hspace {3pt} \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (\frac {1}{2}\right )=3 \\ {} & {} & 3=-2+\sqrt {\mathit {C1} +\frac {11}{4}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\frac {89}{4} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =\frac {89}{4}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y \left (x \right )=-2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (x \right )=-2 x -1+\frac {\sqrt {4 x^{2}+12 x +93}}{2} \end {array} \]
2.1.8.8 ✓ Mathematica. Time used: 0.085 (sec). Leaf size: 28
ode=(6*x+4*y[x]+1)+(4*x+2*y[x]+2)*D[y[x],x]==0;
ic=y[1/2]==3;
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to \frac {1}{2} \left (\sqrt {4 x^2+12 x+93}-4 x-2\right ) \end{align*}
2.1.8.9 ✓ Sympy. Time used: 2.098 (sec). Leaf size: 22
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(6*x + (4*x + 2*y(x) + 2)*Derivative(y(x), x) + 4*y(x) + 1,0)
ics = {y(1/2): 3}
dsolve(ode,func=y(x),ics=ics)
\[
y{\left (x \right )} = - 2 x + \frac {\sqrt {4 x^{2} + 12 x + 93}}{2} - 1
\]