Internal
problem
ID
[9266] Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948 Section
:
Chapter
1,
linear
first
order Problem
number
:
285 Date
solved
:
Thursday, October 17, 2024 at 01:36:24 PM CAS
classification
:
[[_homogeneous, `class A`], _exact, _rational, _dAlembert]
Solve
\begin{align*} \left (4 y^{2}+2 x y+3 x^{2}\right ) y^{\prime }+y^{2}+6 x y+2 x^{2}&=0 \end{align*}
1.284.1 Solved as first order homogeneous class A ode
Time used: 0.588 (sec)
In canonical form, the ODE is
\begin{align*} y' &= F(x,y)\\ &= -\frac {2 x^{2}+6 x y +y^{2}}{3 x^{2}+2 x y +4 y^{2}}\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 \(M=-2 x^{2}-6 x y -y^{2}\) and \(N=3 x^{2}+2 x y +4 y^{2}\) are both homogeneous and of the same order \(n=2\). 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
\[ 4 u^{\prime }\left (x \right ) u \left (x \right )^{2} x +2 u^{\prime }\left (x \right ) u \left (x \right ) x +4 u \left (x \right )^{3}+3 u^{\prime }\left (x \right ) x +3 u \left (x \right )^{2}+9 u \left (x \right )+2 = 0 \]
Or
\[ 2+x \left (4 u \left (x \right )^{2}+2 u \left (x \right )+3\right ) u^{\prime }\left (x \right )+4 u \left (x \right )^{3}+3 u \left (x \right )^{2}+9 u \left (x \right ) = 0 \]
Which is now solved as separable in \(u \left (x \right )\).
The ode \(u^{\prime }\left (x \right ) = -\frac {4 u \left (x \right )^{3}+3 u \left (x \right )^{2}+9 u \left (x \right )+2}{x \left (4 u \left (x \right )^{2}+2 u \left (x \right )+3\right )}\) is separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= -\frac {4 u \left (x \right )^{3}+3 u \left (x \right )^{2}+9 u \left (x \right )+2}{x \left (4 u \left (x \right )^{2}+2 u \left (x \right )+3\right )}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= -\frac {1}{x}\\ g(u) &= \frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3} \end{align*}
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 \(\frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3}=0\) for \(u \left (x \right )\) gives
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.
Which is now solved The ode \(u^{\prime }\left (x \right ) = -\frac {4 u \left (x \right )^{3}+3 u \left (x \right )^{2}+9 u \left (x \right )+2}{x \left (4 u \left (x \right )^{2}+2 u \left (x \right )+3\right )}\) is separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= -\frac {4 u \left (x \right )^{3}+3 u \left (x \right )^{2}+9 u \left (x \right )+2}{x \left (4 u \left (x \right )^{2}+2 u \left (x \right )+3\right )}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= -\frac {1}{x}\\ g(u) &= \frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3} \end{align*}
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 \(\frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3}=0\) for \(u \left (x \right )\) gives
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.
Solving for possible values of \(x_{0}\) and \(y_{0}\)
which makes the above ode a homogeneous ode results in
\begin{align*} x_{0}&=0\\ y_{0}&=0 \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 X^{2}+6 X Y \left (X \right )+Y \left (X \right )^{2}}{3 X^{2}+2 X Y \left (X \right )+4 Y \left (X \right )^{2}} \end{align*}
In canonical form, the ODE is
\begin{align*} Y' &= F(X,Y)\\ &= -\frac {2 X^{2}+6 X Y +Y^{2}}{3 X^{2}+2 X Y +4 Y^{2}}\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 \(M=-2 X^{2}-6 X Y -Y^{2}\) and \(N=3 X^{2}+2 X Y +4 Y^{2}\) are both homogeneous and of the same order \(n=2\). 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
\[ 4 \left (\frac {d}{d X}u \left (X \right )\right ) u \left (X \right )^{2} X +2 \left (\frac {d}{d X}u \left (X \right )\right ) u \left (X \right ) X +4 u \left (X \right )^{3}+3 \left (\frac {d}{d X}u \left (X \right )\right ) X +3 u \left (X \right )^{2}+9 u \left (X \right )+2 = 0 \]
Or
\[ 2+X \left (4 u \left (X \right )^{2}+2 u \left (X \right )+3\right ) \left (\frac {d}{d X}u \left (X \right )\right )+4 u \left (X \right )^{3}+3 u \left (X \right )^{2}+9 u \left (X \right ) = 0 \]
Which is now solved as separable in \(u \left (X \right )\).
The ode \(\frac {d}{d X}u \left (X \right ) = -\frac {4 u \left (X \right )^{3}+3 u \left (X \right )^{2}+9 u \left (X \right )+2}{X \left (4 u \left (X \right )^{2}+2 u \left (X \right )+3\right )}\) is separable as it can be written as
\begin{align*} \frac {d}{d X}u \left (X \right )&= -\frac {4 u \left (X \right )^{3}+3 u \left (X \right )^{2}+9 u \left (X \right )+2}{X \left (4 u \left (X \right )^{2}+2 u \left (X \right )+3\right )}\\ &= f(X) g(u) \end{align*}
Where
\begin{align*} f(X) &= -\frac {1}{X}\\ g(u) &= \frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3} \end{align*}
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 \(\frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3}=0\) for \(u \left (X \right )\) gives
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.
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
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
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
\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (2 x^{2}+6 x y +y^{2}\right )\\ &= 6 x +2 y \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (3 x^{2}+2 x y +4 y^{2}\right )\\ &= 6 x +2 y \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 \(\phi \left (x,y\right )\)
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{align*}
\tag{4} \frac {\partial \phi }{\partial y} &= 3 x^{2}+2 x y+f'(y) \\
&=x \left (3 x +2 y \right )+f'(y) \\
\end{align*}
But equation (2) says that \(\frac {\partial \phi }{\partial y} = 3 x^{2}+2 x y +4 y^{2}\).
Therefore equation (4) becomes
\begin{equation}
\tag{5} 3 x^{2}+2 x y +4 y^{2} = x \left (3 x +2 y \right )+f'(y)
\end{equation}
Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into
equation (3) gives \(\phi \)
\[
\phi = \frac {2}{3} x^{3}+3 y \,x^{2}+y^{2} x +\frac {4}{3} y^{3}+ c_1
\]
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
\[
c_1 = \frac {2}{3} x^{3}+3 y \,x^{2}+y^{2} x +\frac {4}{3} y^{3}
\]
Figure 393: Slope field plot \(\left (4 y^{2}+2 x y+3 x^{2}\right ) y^{\prime }+y^{2}+6 x y+2 x^{2} = 0\)
1.284.5 Solved as first order isobaric ode
Time used: 0.321 (sec)
Solving for \(y'\) gives
\begin{align*}
\tag{1} y' &= -\frac {y^{2}+6 x y+2 x^{2}}{4 y^{2}+2 x y+3 x^{2}} \\
\end{align*}
Each of the above ode’s is now solved An ode \(y^{\prime }=f(x,y)\) is isobaric if
\[ f(t x, t^m y) = t^{m-1} f(x,y)\tag {1} \]
Where here
\[ f(x,y) = -\frac {y^{2}+6 x y+2 x^{2}}{4 y^{2}+2 x y+3 x^{2}}\tag {2} \]
\(m\)
is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives
\[ m = 1 \]
Since the ode is
isobaric of order \(m=1\), then the substitution
\begin{align*} y&=u x^m \\ &=u x \end{align*}
Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives
\[ u \left (x \right )+x u^{\prime }\left (x \right ) = -\frac {x^{2} u \left (x \right )^{2}+6 x^{2} u \left (x \right )+2 x^{2}}{4 x^{2} u \left (x \right )^{2}+2 x^{2} u \left (x \right )+3 x^{2}} \]
The ode \(u^{\prime }\left (x \right ) = -\frac {4 u \left (x \right )^{3}+3 u \left (x \right )^{2}+9 u \left (x \right )+2}{\left (4 u \left (x \right )^{2}+2 u \left (x \right )+3\right ) x}\) is
separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= -\frac {4 u \left (x \right )^{3}+3 u \left (x \right )^{2}+9 u \left (x \right )+2}{\left (4 u \left (x \right )^{2}+2 u \left (x \right )+3\right ) x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= -\frac {1}{x}\\ g(u) &= \frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3} \end{align*}
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 \(\frac {4 u^{3}+3 u^{2}+9 u +2}{4 u^{2}+2 u +3}=0\) for \(u \left (x \right )\) gives
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.
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}-\frac {\left (2 x^{2}+6 x y +y^{2}\right ) \left (b_{3}-a_{2}\right )}{3 x^{2}+2 x y +4 y^{2}}-\frac {\left (2 x^{2}+6 x y +y^{2}\right )^{2} a_{3}}{\left (3 x^{2}+2 x y +4 y^{2}\right )^{2}}-\left (-\frac {4 x +6 y}{3 x^{2}+2 x y +4 y^{2}}+\frac {\left (2 x^{2}+6 x y +y^{2}\right ) \left (6 x +2 y \right )}{\left (3 x^{2}+2 x y +4 y^{2}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {6 x +2 y}{3 x^{2}+2 x y +4 y^{2}}+\frac {\left (2 x^{2}+6 x y +y^{2}\right ) \left (2 x +8 y \right )}{\left (3 x^{2}+2 x y +4 y^{2}\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
\frac {6 x^{4} a_{2}-4 x^{4} a_{3}+23 x^{4} b_{2}-6 x^{4} b_{3}+8 x^{3} y a_{2}-24 x^{3} y a_{3}+2 x^{3} y b_{2}-8 x^{3} y b_{3}+33 x^{2} y^{2} a_{2}-54 x^{2} y^{2} a_{3}+6 x^{2} y^{2} b_{2}-33 x^{2} y^{2} b_{3}+48 x \,y^{3} a_{2}-2 x \,y^{3} a_{3}+16 x \,y^{3} b_{2}-48 x \,y^{3} b_{3}+4 y^{4} a_{2}+21 y^{4} a_{3}+16 y^{4} b_{2}-4 y^{4} b_{3}+14 x^{3} b_{1}-14 x^{2} y a_{1}-10 x^{2} y b_{1}+10 x \,y^{2} a_{1}-22 x \,y^{2} b_{1}+22 y^{3} a_{1}}{\left (3 x^{2}+2 x y +4 y^{2}\right )^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} 6 x^{4} a_{2}-4 x^{4} a_{3}+23 x^{4} b_{2}-6 x^{4} b_{3}+8 x^{3} y a_{2}-24 x^{3} y a_{3}+2 x^{3} y b_{2}-8 x^{3} y b_{3}+33 x^{2} y^{2} a_{2}-54 x^{2} y^{2} a_{3}+6 x^{2} y^{2} b_{2}-33 x^{2} y^{2} b_{3}+48 x \,y^{3} a_{2}-2 x \,y^{3} a_{3}+16 x \,y^{3} b_{2}-48 x \,y^{3} b_{3}+4 y^{4} a_{2}+21 y^{4} a_{3}+16 y^{4} b_{2}-4 y^{4} b_{3}+14 x^{3} b_{1}-14 x^{2} y a_{1}-10 x^{2} y b_{1}+10 x \,y^{2} a_{1}-22 x \,y^{2} b_{1}+22 y^{3} a_{1} = 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
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 \\
\eta &= y \\
\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 - \left (-\frac {2 x^{2}+6 x y +y^{2}}{3 x^{2}+2 x y +4 y^{2}}\right ) \left (x\right ) \\ &= \frac {2 x^{3}+9 y \,x^{2}+3 y^{2} x +4 y^{3}}{3 x^{2}+2 x y +4 y^{2}}\\ \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
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 x^{3}+9 y \,x^{2}+3 y^{2} x +4 y^{3}}{3 x^{2}+2 x y +4 y^{2}}}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (2 x^{3}+9 y \,x^{2}+3 y^{2} x +4 y^{3}\right )}{3} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
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 x^{2}+6 x y +y^{2}}{3 x^{2}+2 x y +4 y^{2}} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {2 x^{2}+6 x y +y^{2}}{2 x^{3}+9 y \,x^{2}+3 y^{2} x +4 y^{3}}\\ S_{y} &= \frac {3 x^{2}+2 x y +4 y^{2}}{2 x^{3}+9 y \,x^{2}+3 y^{2} x +4 y^{3}} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
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 \(f(R)\).
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= \frac {-p -3+\sqrt {-11 p^{2}-5 p +7}}{4 p +1}\\ g &= 0 \end{align*}
Hence (2) becomes
\begin{align*} p -\frac {-p -3+\sqrt {-11 p^{2}-5 p +7}}{4 p +1} = \left (-\frac {x}{4 p +1}-\frac {11 x p}{\left (4 p +1\right ) \sqrt {-11 p^{2}-5 p +7}}-\frac {5 x}{2 \left (4 p +1\right ) \sqrt {-11 p^{2}-5 p +7}}+\frac {4 x p}{\left (4 p +1\right )^{2}}+\frac {12 x}{\left (4 p +1\right )^{2}}-\frac {4 x \sqrt {-11 p^{2}-5 p +7}}{\left (4 p +1\right )^{2}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p -\frac {-p -3+\sqrt {-11 p^{2}-5 p +7}}{4 p +1} = 0 \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{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {-p \left (x \right )-3+\sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}{4 p \left (x \right )+1}}{-\frac {x}{4 p \left (x \right )+1}-\frac {11 x p \left (x \right )}{\left (4 p \left (x \right )+1\right ) \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}-\frac {5 x}{2 \left (4 p \left (x \right )+1\right ) \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}+\frac {4 x p \left (x \right )}{\left (4 p \left (x \right )+1\right )^{2}}+\frac {12 x}{\left (4 p \left (x \right )+1\right )^{2}}-\frac {4 x \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}{\left (4 p \left (x \right )+1\right )^{2}}}\tag {3} \end{align*}
This ODE is now solved for \(p \left (x \right )\). No inversion is needed. The ode \(p^{\prime }\left (x \right ) = -\frac {2 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\, \left (4 p \left (x \right )+1\right ) \left (-4 p \left (x \right )^{2}+\sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}-2 p \left (x \right )-3\right )}{x \left (-2 p \left (x \right )-61+22 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\right )}\) is separable as it can be
written as
\begin{align*} p^{\prime }\left (x \right )&= -\frac {2 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\, \left (4 p \left (x \right )+1\right ) \left (-4 p \left (x \right )^{2}+\sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}-2 p \left (x \right )-3\right )}{x \left (-2 p \left (x \right )-61+22 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\right )}\\ &= f(x) g(p) \end{align*}
Where
\begin{align*} f(x) &= -\frac {2}{x}\\ g(p) &= \frac {\sqrt {-11 p^{2}-5 p +7}\, \left (4 p +1\right ) \left (-4 p^{2}+\sqrt {-11 p^{2}-5 p +7}-2 p -3\right )}{-2 p -61+22 \sqrt {-11 p^{2}-5 p +7}} \end{align*}
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 \(\frac {\sqrt {-11 p^{2}-5 p +7}\, \left (4 p +1\right ) \left (-4 p^{2}+\sqrt {-11 p^{2}-5 p +7}-2 p -3\right )}{-2 p -61+22 \sqrt {-11 p^{2}-5 p +7}}=0\) for \(p \left (x \right )\) gives
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.
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= \frac {-p -3-\sqrt {-11 p^{2}-5 p +7}}{4 p +1}\\ g &= 0 \end{align*}
Hence (2) becomes
\begin{align*} p -\frac {-p -3-\sqrt {-11 p^{2}-5 p +7}}{4 p +1} = \left (-\frac {x}{4 p +1}+\frac {11 x p}{\left (4 p +1\right ) \sqrt {-11 p^{2}-5 p +7}}+\frac {5 x}{2 \left (4 p +1\right ) \sqrt {-11 p^{2}-5 p +7}}+\frac {4 x p}{\left (4 p +1\right )^{2}}+\frac {12 x}{\left (4 p +1\right )^{2}}+\frac {4 x \sqrt {-11 p^{2}-5 p +7}}{\left (4 p +1\right )^{2}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p -\frac {-p -3-\sqrt {-11 p^{2}-5 p +7}}{4 p +1} = 0 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = -\frac {x \left (\left (i \sqrt {3}-1\right ) \left (1+6 \sqrt {37}\right )^{{2}/{3}}+\sqrt {443+11 \left (1+2 i \sqrt {111}+i \sqrt {3}+2 \sqrt {37}\right ) \left (1+6 \sqrt {37}\right )^{{1}/{3}}-443 i \sqrt {3}+4 i \sqrt {111}-4 \sqrt {37}+242 \left (1+6 \sqrt {37}\right )^{{2}/{3}}}\, \sqrt {6}+11 i \sqrt {3}+22 \left (1+6 \sqrt {37}\right )^{{1}/{3}}+11\right )}{4 i \sqrt {3}\, \left (1+6 \sqrt {37}\right )^{{2}/{3}}-4 \left (1+6 \sqrt {37}\right )^{{2}/{3}}+44 i \sqrt {3}+44}\\ y = \frac {\left (\left (-i \sqrt {3}-1\right ) \left (1+6 \sqrt {37}\right )^{{2}/{3}}+\sqrt {\left (-22 i \sqrt {111}-11 i \sqrt {3}+22 \sqrt {37}+11\right ) \left (1+6 \sqrt {37}\right )^{{1}/{3}}+443 i \sqrt {3}-4 i \sqrt {111}-4 \sqrt {37}+242 \left (1+6 \sqrt {37}\right )^{{2}/{3}}+443}\, \sqrt {6}-11 i \sqrt {3}+22 \left (1+6 \sqrt {37}\right )^{{1}/{3}}+11\right ) x}{4 i \sqrt {3}\, \left (1+6 \sqrt {37}\right )^{{2}/{3}}+4 \left (1+6 \sqrt {37}\right )^{{2}/{3}}+44 i \sqrt {3}-44} \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{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {-p \left (x \right )-3-\sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}{4 p \left (x \right )+1}}{-\frac {x}{4 p \left (x \right )+1}+\frac {11 x p \left (x \right )}{\left (4 p \left (x \right )+1\right ) \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}+\frac {5 x}{2 \left (4 p \left (x \right )+1\right ) \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}+\frac {4 x p \left (x \right )}{\left (4 p \left (x \right )+1\right )^{2}}+\frac {12 x}{\left (4 p \left (x \right )+1\right )^{2}}+\frac {4 x \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}}{\left (4 p \left (x \right )+1\right )^{2}}}\tag {3} \end{align*}
This ODE is now solved for \(p \left (x \right )\). No inversion is needed. The ode \(p^{\prime }\left (x \right ) = \frac {2 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\, \left (4 p \left (x \right )+1\right ) \left (4 p \left (x \right )^{2}+2 p \left (x \right )+\sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}+3\right )}{x \left (2 p \left (x \right )+61+22 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\right )}\) is separable as it can be
written as
\begin{align*} p^{\prime }\left (x \right )&= \frac {2 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\, \left (4 p \left (x \right )+1\right ) \left (4 p \left (x \right )^{2}+2 p \left (x \right )+\sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}+3\right )}{x \left (2 p \left (x \right )+61+22 \sqrt {-11 p \left (x \right )^{2}-5 p \left (x \right )+7}\right )}\\ &= f(x) g(p) \end{align*}
Where
\begin{align*} f(x) &= \frac {2}{x}\\ g(p) &= \frac {\sqrt {-11 p^{2}-5 p +7}\, \left (4 p +1\right ) \left (4 p^{2}+\sqrt {-11 p^{2}-5 p +7}+2 p +3\right )}{2 p +61+22 \sqrt {-11 p^{2}-5 p +7}} \end{align*}
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 \(\frac {\sqrt {-11 p^{2}-5 p +7}\, \left (4 p +1\right ) \left (4 p^{2}+\sqrt {-11 p^{2}-5 p +7}+2 p +3\right )}{2 p +61+22 \sqrt {-11 p^{2}-5 p +7}}=0\) for \(p \left (x \right )\) gives
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.
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryinghomogeneous D<-homogeneous successful`