2.1.30 Internal
problem
ID
[8168]Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
31Date
solved
:
Tuesday, November 12, 2024 at 11:07:18 PMCAS
classification
:
[[_homogeneous, `class A`], _rational, _Riccati]
Solve
\begin{align*} y^{\prime }&=\frac {5 x^{2}-x y+y^{2}}{x^{2}} \end{align*}
Time used: 0.340 (sec)
In canonical form, the ODE is
\begin{align*} y' &= F(x,y)\\ &= \frac {5 x^{2}-x y +y^{2}}{x^{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 \(M=5 x^{2}-x y +y^{2}\) and \(N=x^{2}\) 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
\[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]
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 &= u^{2}-u +5\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x} \end{align*}
Or
\[ u^{\prime }\left (x \right )-\frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x} = 0 \]
Or
\[ u^{\prime }\left (x \right ) x -u \left (x \right )^{2}+2 u \left (x \right )-5 = 0 \]
Which is now solved as separable in \(u \left (x \right )\) .
The ode \(u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\) is separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= u^{2}-2 u +5 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{x} \,dx}\\ \frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2}&=\ln \left (x \right )+c_1 \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 \(u^{2}-2 u +5=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1-2 i\\ u \left (x \right )&=1+2 i \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 {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1\\ u \left (x \right ) = 1-2 i\\ u \left (x \right ) = 1+2 i \end{align*}
Solving for \(u \left (x \right )\) gives
\begin{align*}
u \left (x \right ) &= 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (x \right ) = 1-2 i\) back to \(y\) gives
\begin{align*} y = \left (1-2 i\right ) x \end{align*}
Converting \(u \left (x \right ) = 1+2 i\) back to \(y\) gives
\begin{align*} y = \left (1+2 i\right ) x \end{align*}
Converting \(u \left (x \right ) = 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\) back to \(y\) gives
\begin{align*} y = x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \end{align*}
Figure 2.71: Slope field plot\(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\) Summary of solutions found
\begin{align*}
y &= \left (1-2 i\right ) x \\
y &= \left (1+2 i\right ) x \\
y &= x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \\
\end{align*}
Time used: 0.189 (sec)
Applying change of variables \(y = u \left (x \right ) x\) , then the ode becomes
\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = \frac {5 x^{2}-x^{2} u \left (x \right )+u \left (x \right )^{2} x^{2}}{x^{2}} \end{align*}
Which is now solved The ode \(u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\) is separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= u^{2}-2 u +5 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{x} \,dx}\\ \frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2}&=\ln \left (x \right )+c_1 \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 \(u^{2}-2 u +5=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1-2 i\\ u \left (x \right )&=1+2 i \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 {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1\\ u \left (x \right ) = 1-2 i\\ u \left (x \right ) = 1+2 i \end{align*}
Solving for \(u \left (x \right )\) gives
\begin{align*}
u \left (x \right ) &= 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (x \right ) = 1-2 i\) back to \(y\) gives
\begin{align*} y = \left (1-2 i\right ) x \end{align*}
Converting \(u \left (x \right ) = 1+2 i\) back to \(y\) gives
\begin{align*} y = \left (1+2 i\right ) x \end{align*}
Converting \(u \left (x \right ) = 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\) back to \(y\) gives
\begin{align*} y = x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \end{align*}
Figure 2.72: Slope field plot\(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\) Summary of solutions found
\begin{align*}
y &= \left (1-2 i\right ) x \\
y &= \left (1+2 i\right ) x \\
y &= x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \left (1-2 i\right ) x \\
y &= \left (1+2 i\right ) x \\
y &= x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \\
\end{align*}
Time used: 0.382 (sec)
Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)
\[
\frac {d}{d X}Y \left (X \right ) = \frac {5 \left (x_{0} +X \right )^{2}-\left (x_{0} +X \right ) \left (Y \left (X \right )+y_{0} \right )+\left (Y \left (X \right )+y_{0} \right )^{2}}{\left (x_{0} +X \right )^{2}}
\]
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 {5 X^{2}-X Y \left (X \right )+Y \left (X \right )^{2}}{X^{2}} \end{align*}
In canonical form, the ODE is
\begin{align*} Y' &= F(X,Y)\\ &= \frac {5 X^{2}-X Y +Y^{2}}{X^{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 \(M=5 X^{2}-X Y +Y^{2}\) and \(N=X^{2}\) 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
\[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \]
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 &= u^{2}-u +5\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X} \end{align*}
Or
\[ \frac {d}{d X}u \left (X \right )-\frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X} = 0 \]
Or
\[ \left (\frac {d}{d X}u \left (X \right )\right ) X -u \left (X \right )^{2}+2 u \left (X \right )-5 = 0 \]
Which is now solved as separable in \(u \left (X \right )\) .
The ode \(\frac {d}{d X}u \left (X \right ) = \frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X}\) is separable as it can be written as
\begin{align*} \frac {d}{d X}u \left (X \right )&= \frac {u \left (X \right )^{2}-2 u \left (X \right )+5}{X}\\ &= f(X) g(u) \end{align*}
Where
\begin{align*} f(X) &= \frac {1}{X}\\ g(u) &= u^{2}-2 u +5 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX}\\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{X} \,dX}\\ \frac {\arctan \left (\frac {u \left (X \right )}{2}-\frac {1}{2}\right )}{2}&=\ln \left (X \right )+c_1 \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 \(u^{2}-2 u +5=0\) for \(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&=1-2 i\\ u \left (X \right )&=1+2 i \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 {\arctan \left (\frac {u \left (X \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (X \right )+c_1\\ u \left (X \right ) = 1-2 i\\ u \left (X \right ) = 1+2 i \end{align*}
Solving for \(u \left (X \right )\) gives
\begin{align*}
u \left (X \right ) &= 1+2 \tan \left (2 \ln \left (X \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (X \right ) = 1-2 i\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = \left (1-2 i\right ) X \end{align*}
Converting \(u \left (X \right ) = 1+2 i\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = \left (1+2 i\right ) X \end{align*}
Converting \(u \left (X \right ) = 1+2 \tan \left (2 \ln \left (X \right )+2 c_1 \right )\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = X \left (1+2 \tan \left (2 \ln \left (X \right )+2 c_1 \right )\right ) \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = \left (1-2 i\right ) X\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*}
Or
\begin{align*} Y &= y\\ X &= x \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} y = \left (1-2 i\right ) x \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = \left (1+2 i\right ) X\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*}
Or
\begin{align*} Y &= y\\ X &= x \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} y = \left (1+2 i\right ) x \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = X \left (1+2 \tan \left (2 \ln \left (X \right )+2 c_1 \right )\right )\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*}
Or
\begin{align*} Y &= y\\ X &= x \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} y = x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \end{align*}
Figure 2.73: Slope field plot\(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\) Time used: 0.237 (sec)
Solving for \(y'\) gives
\begin{align*}
\tag{1} y' &= \frac {5 x^{2}-x y+y^{2}}{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
\[ f(x,y) = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\tag {2} \]
\(m\)
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
\[ u \left (x \right )+x u^{\prime }\left (x \right ) = \frac {5 x^{2}-x^{2} u \left (x \right )+x^{2} u \left (x \right )^{2}}{x^{2}} \]
The ode \(u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\) is
separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right )^{2}-2 u \left (x \right )+5}{x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= u^{2}-2 u +5 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{u^{2}-2 u +5}\,du} &= \int { \frac {1}{x} \,dx}\\ \frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2}&=\ln \left (x \right )+c_1 \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 \(u^{2}-2 u +5=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1-2 i\\ u \left (x \right )&=1+2 i \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 {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} = \ln \left (x \right )+c_1\\ u \left (x \right ) = 1-2 i\\ u \left (x \right ) = 1+2 i \end{align*}
Solving for \(u \left (x \right )\) gives
\begin{align*}
u \left (x \right ) &= 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (x \right ) = 1-2 i\) back to \(y\) gives
\begin{align*} \frac {y}{x} = 1-2 i \end{align*}
Converting \(u \left (x \right ) = 1+2 i\) back to \(y\) gives
\begin{align*} \frac {y}{x} = 1+2 i \end{align*}
Converting \(u \left (x \right ) = 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\) back to \(y\) gives
\begin{align*} \frac {y}{x} = 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \\
\end{align*}
Figure 2.74: Slope field plot\(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\) Summary of solutions found
\begin{align*}
\frac {y}{x} &= 1-2 i \\
\frac {y}{x} &= 1+2 i \\
y &= x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right )\right ) \\
\end{align*}
Time used: 0.537 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {5 x^{2}-x y +y^{2}}{x^{2}}\\ 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 (5 x^{2}-x y +y^{2}\right ) \left (b_{3}-a_{2}\right )}{x^{2}}-\frac {\left (5 x^{2}-x y +y^{2}\right )^{2} a_{3}}{x^{4}}-\left (\frac {10 x -y}{x^{2}}-\frac {2 \left (5 x^{2}-x y +y^{2}\right )}{x^{3}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (2 y -x \right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{x^{2}} = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {5 x^{4} a_{2}+25 x^{4} a_{3}-2 b_{2} x^{4}-5 x^{4} b_{3}-10 x^{3} y a_{3}+2 x^{3} y b_{2}-x^{2} y^{2} a_{2}+12 x^{2} y^{2} a_{3}+x^{2} y^{2} b_{3}-4 x \,y^{3} a_{3}+y^{4} a_{3}-x^{3} b_{1}+x^{2} y a_{1}+2 x^{2} y b_{1}-2 x \,y^{2} a_{1}}{x^{4}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -5 x^{4} a_{2}-25 x^{4} a_{3}+2 b_{2} x^{4}+5 x^{4} b_{3}+10 x^{3} y a_{3}-2 x^{3} y b_{2}+x^{2} y^{2} a_{2}-12 x^{2} y^{2} a_{3}-x^{2} y^{2} b_{3}+4 x \,y^{3} a_{3}-y^{4} a_{3}+x^{3} b_{1}-x^{2} y a_{1}-2 x^{2} y b_{1}+2 x \,y^{2} 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
\[
\{x = v_{1}, y = v_{2}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} -5 a_{2} v_{1}^{4}+a_{2} v_{1}^{2} v_{2}^{2}-25 a_{3} v_{1}^{4}+10 a_{3} v_{1}^{3} v_{2}-12 a_{3} v_{1}^{2} v_{2}^{2}+4 a_{3} v_{1} v_{2}^{3}-a_{3} v_{2}^{4}+2 b_{2} v_{1}^{4}-2 b_{2} v_{1}^{3} v_{2}+5 b_{3} v_{1}^{4}-b_{3} v_{1}^{2} v_{2}^{2}-a_{1} v_{1}^{2} v_{2}+2 a_{1} v_{1} v_{2}^{2}+b_{1} v_{1}^{3}-2 b_{1} v_{1}^{2} v_{2} = 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 (-5 a_{2}-25 a_{3}+2 b_{2}+5 b_{3}\right ) v_{1}^{4}+\left (10 a_{3}-2 b_{2}\right ) v_{1}^{3} v_{2}+b_{1} v_{1}^{3}+\left (a_{2}-12 a_{3}-b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (-a_{1}-2 b_{1}\right ) v_{1}^{2} v_{2}+4 a_{3} v_{1} v_{2}^{3}+2 a_{1} v_{1} v_{2}^{2}-a_{3} v_{2}^{4} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} b_{1}&=0\\ 2 a_{1}&=0\\ -a_{3}&=0\\ 4 a_{3}&=0\\ -a_{1}-2 b_{1}&=0\\ 10 a_{3}-2 b_{2}&=0\\ a_{2}-12 a_{3}-b_{3}&=0\\ -5 a_{2}-25 a_{3}+2 b_{2}+5 b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ 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 \\
\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 {5 x^{2}-x y +y^{2}}{x^{2}}\right ) \left (x\right ) \\ &= \frac {-5 x^{2}+2 x y -y^{2}}{x}\\ \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 {-5 x^{2}+2 x y -y^{2}}{x}}} dy \end{align*}
Which results in
\begin{align*} S&= -\frac {\arctan \left (\frac {2 y -2 x}{4 x}\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 {5 x^{2}-x y +y^{2}}{x^{2}} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {y}{5 x^{2}-2 x y +y^{2}}\\ S_{y} &= -\frac {x}{5 x^{2}-2 x y +y^{2}} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= -\frac {1}{x}\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} &= -\frac {1}{R} \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 {-\frac {1}{R}\, dR}\\ S \left (R \right ) &= -\ln \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 {\arctan \left (\frac {x -y}{2 x}\right )}{2} = -\ln \left (x \right )+c_2 \end{align*}
Which gives
\begin{align*} y = -2 \tan \left (-2 \ln \left (x \right )+2 c_2 \right ) x +x \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 \((R,S)\)
\( \frac {dy}{dx} = \frac {5 x^{2}-x y +y^{2}}{x^{2}}\)
\( \frac {d S}{d R} = -\frac {1}{R}\)
\(\!\begin {aligned} R&= x\\ S&= \frac {\arctan \left (\frac {x -y}{2 x}\right )}{2} \end {aligned} \)
Figure 2.75: Slope field plot\(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\) Summary of solutions found
\begin{align*}
y &= -2 \tan \left (-2 \ln \left (x \right )+2 c_2 \right ) x +x \\
\end{align*}
Time used: 0.243 (sec)
In canonical form the ODE is
\begin{align*} y' &= F(x,y)\\ &= \frac {5 x^{2}-x y +y^{2}}{x^{2}} \end{align*}
This is a Riccati ODE. Comparing the ODE to solve
\[ y' = 5-\frac {y}{x}+\frac {y^{2}}{x^{2}} \]
With Riccati ODE standard form
\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows that \(f_0(x)=5\) , \(f_1(x)=-\frac {1}{x}\) and \(f_2(x)=\frac {1}{x^{2}}\) . Let
\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x^{2}}} \tag {1} \end{align*}
Using the above substitution in the given ODE results (after some simplification)in a second
order ODE to solve for \(u(x)\) which is
\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}
But
\begin{align*} f_2' &=-\frac {2}{x^{3}}\\ f_1 f_2 &=-\frac {1}{x^{3}}\\ f_2^2 f_0 &=\frac {5}{x^{4}} \end{align*}
Substituting the above terms back in equation (2) gives
\begin{align*} \frac {u^{\prime \prime }\left (x \right )}{x^{2}}+\frac {3 u^{\prime }\left (x \right )}{x^{3}}+\frac {5 u \left (x \right )}{x^{4}} = 0 \end{align*}
This is Euler second order ODE. Let the solution be \(u = x^r\) , then \(u'=r x^{r-1}\) and \(u''=r(r-1) x^{r-2}\) . Substituting these back
into the given ODE gives
\[ x^{2}(r(r-1))x^{r-2}+3 x r x^{r-1}+5 x^{r} = 0 \]
Simplifying gives
\[ r \left (r -1\right )x^{r}+3 r\,x^{r}+5 x^{r} = 0 \]
Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives
\[ r \left (r -1\right )+3 r+5 = 0 \]
Or
\[ r^{2}+2 r +5 = 0 \tag {1} \]
Equation (1) is the characteristic equation. Its roots determine the form of the general
solution. Using the quadratic equation the roots are
\begin{align*} r_1 &= -1-2 i\\ r_2 &= -1+2 i \end{align*}
The roots are complex conjugate of each others. Let the roots be
\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}
Where in this case \(\alpha =-1\) and \(\beta =-2\) . Hence the solution becomes
\begin{align*} u &= c_1 x^{r_1} + c_2 x^{r_2} \\ &= c_1 x^{\alpha + i \beta } + c_2 x^{\alpha - i \beta } \\ &= x^{\alpha } \left ( c_1 x^{i \beta } + c_2 x^{- i \beta }\right ) \\ &= x^{\alpha } \left ( c_1 e^{\ln \left (x^{i \beta }\right )} + c_2 e^{\ln \left (x^{-i \beta }\right )}\right ) \\ &= x^{\alpha } \left ( c_1 e^{i \left (\beta \ln {x}\right )} + c_2 e^{-i \left (\beta \ln {x}\right )}\right ) \end{align*}
Using the values for \(\alpha =-1, \beta =-2\) , the above becomes
\begin{align*} u&= x^{-1} \left ( c_1 e^{-2 i \ln \left (x \right )} + c_2 e^{2 i \ln \left (x \right )} \right ) \end{align*}
Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to
change. Applying this to the above result gives
\begin{align*} u&=\frac {1}{x}\left (c_1 \cos \left (2 \ln \left (x \right )\right )+c_2 \sin \left (2 \ln \left (x \right )\right )\right ) \end{align*}
Will add steps showing solving for IC soon.
Taking derivative gives
\[
u^{\prime }\left (x \right ) = -\frac {c_1 \cos \left (2 \ln \left (x \right )\right )+c_2 \sin \left (2 \ln \left (x \right )\right )}{x^{2}}+\frac {-\frac {2 c_1 \sin \left (2 \ln \left (x \right )\right )}{x}+\frac {2 c_2 \cos \left (2 \ln \left (x \right )\right )}{x}}{x}
\]
Doing change of constants, the solution becomes
\[
y = -\frac {\left (-\frac {c_3 \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right )}{x^{2}}+\frac {-\frac {2 c_3 \sin \left (2 \ln \left (x \right )\right )}{x}+\frac {2 \cos \left (2 \ln \left (x \right )\right )}{x}}{x}\right ) x^{3}}{c_3 \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right )}
\]
Figure 2.76: Slope field plot\(y^{\prime } = \frac {5 x^{2}-x y+y^{2}}{x^{2}}\) Summary of solutions found
\begin{align*}
y &= \frac {x \left (\left (c_3 -2\right ) \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right ) \left (2 c_3 +1\right )\right )}{c_3 \cos \left (2 \ln \left (x \right )\right )+\sin \left (2 \ln \left (x \right )\right )} \\
\end{align*}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {5 x^{2}-x y \left (x \right )+y \left (x \right )^{2}}{x^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {5 x^{2}-x y \left (x \right )+y \left (x \right )^{2}}{x^{2}} \end {array} \]
` 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 D
<- homogeneous successful `
Solving time : 0.010
dsolve ( diff ( y ( x ), x ) = (5*x^2-x*y(x)+y(x)^2)/x^2,
y(x),singsol=all)
\[
y = x \left (1+2 \tan \left (2 \ln \left (x \right )+2 c_{1} \right )\right )
\]
Solving time : 0.881
DSolve [{ D [ y [ x ], x ]==(5* x ^2- x * y [ x ]+ y [ x ]^2)/ x ^2,{}}, y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to x+2 x \tan (2 (\log (x)+c_1))
\]