1.7 problem 2(a)
Internal
problem
ID
[3683]
Book
:
Elementary
Differential
equations,
Chaundy,
1969
Section
:
Exercises
3,
page
60
Problem
number
:
2(a)
Date
solved
:
Sunday, October 20, 2024 at 06:52:05 PM
CAS
classification
:
[_linear]
Solve
\begin{align*} y^{\prime } x +y&=x \end{align*}
1.7.1 Solved as first order linear ode
Time used: 0.112 (sec)
In canonical form a linear first order is
\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {1}{x}\\ p(x) &=1 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{x}d x}\\ &= x \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (x y\right ) &= x\\ \mathrm {d} \left (x y\right ) &= x\mathrm {d} x \end{align*}
Integrating gives
\begin{align*} x y&= \int {x \,dx} \\ &=\frac {x^{2}}{2} + c_{1} \end{align*}
Dividing throughout by the integrating factor \(x\) gives the final solution
\[ y = \frac {x^{2}+2 c_{1}}{2 x} \]
Figure 19: Slope field plot
\(y^{\prime } x +y = x\)
1.7.2 Solved as first order homogeneous class A ode
Time used: 1.239 (sec)
In canonical form, the ODE is
\begin{align*} y' &= F(x,y)\\ &= -\frac {-x +y}{x}\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 \(M=x -y\) and \(N=x\) 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 &= 1-u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {1-2 u \left (x \right )}{x} \end{align*}
Or
\[ u^{\prime }\left (x \right )-\frac {1-2 u \left (x \right )}{x} = 0 \]
Or
\[ u^{\prime }\left (x \right ) x +2 u \left (x \right )-1 = 0 \]
Which is now solved as separable in \(u \left (x \right )\) .
The ode \(u^{\prime }\left (x \right ) = -\frac {2 u \left (x \right )-1}{x}\) is separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= -\frac {2 u \left (x \right )-1}{x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= -2 u +1 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{-2 u +1}\,du} &= \int { \frac {1}{x} \,dx}\\ -\frac {\ln \left (2 u \left (x \right )-1\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 \(-2 u +1=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&={\frac {1}{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*} -\frac {\ln \left (2 u \left (x \right )-1\right )}{2} = \ln \left (x \right )+c_{1}\\ u \left (x \right ) = {\frac {1}{2}} \end{align*}
Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (x \right )&={\frac {1}{2}}\\ u \left (x \right )&=\frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x^{2}} \end{align*}
Converting \(u \left (x \right ) = {\frac {1}{2}}\) back to \(y\) gives
\begin{align*} y = \frac {x}{2} \end{align*}
Converting \(u \left (x \right ) = \frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x^{2}}\) back to \(y\) gives
\begin{align*} y = \frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x} \end{align*}
Figure 20: Slope field plot
\(y^{\prime } x +y = x\)
1.7.3 Solved as first order homogeneous class D2 ode
Time used: 0.201 (sec)
Applying change of variables \(y = u \left (x \right ) x\) , then the ode becomes
\begin{align*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x +u \left (x \right ) x = x \end{align*}
Which is now solved The ode \(u^{\prime }\left (x \right ) = -\frac {2 u \left (x \right )-1}{x}\) is separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= -\frac {2 u \left (x \right )-1}{x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= -2 u +1 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{-2 u +1}\,du} &= \int { \frac {1}{x} \,dx}\\ -\frac {\ln \left (2 u \left (x \right )-1\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 \(-2 u +1=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&={\frac {1}{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*} -\frac {\ln \left (2 u \left (x \right )-1\right )}{2} = \ln \left (x \right )+c_{1}\\ u \left (x \right ) = {\frac {1}{2}} \end{align*}
Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (x \right )&={\frac {1}{2}}\\ u \left (x \right )&=\frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x^{2}} \end{align*}
Converting \(u \left (x \right ) = {\frac {1}{2}}\) back to \(y\) gives
\begin{align*} y = \frac {x}{2} \end{align*}
Converting \(u \left (x \right ) = \frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x^{2}}\) back to \(y\) gives
\begin{align*} y = \frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x} \end{align*}
Figure 21: Slope field plot
\(y^{\prime } x +y = x\)
1.7.4 Solved as first order homogeneous class Maple C ode
Time used: 1.691 (sec)
Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)
\[
\frac {d}{d X}Y \left (X \right ) = -\frac {-x_{0} -X +Y \left (X \right )+y_{0}}{x_{0} +X}
\]
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 {-X +Y \left (X \right )}{X} \end{align*}
In canonical form, the ODE is
\begin{align*} Y' &= F(X,Y)\\ &= -\frac {-X +Y}{X}\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 \(M=X -Y\) and \(N=X\) 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 &= 1-u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {1-2 u \left (X \right )}{X} \end{align*}
Or
\[ \frac {d}{d X}u \left (X \right )-\frac {1-2 u \left (X \right )}{X} = 0 \]
Or
\[ \left (\frac {d}{d X}u \left (X \right )\right ) X +2 u \left (X \right )-1 = 0 \]
Which is now solved as separable in \(u \left (X \right )\) .
The ode \(\frac {d}{d X}u \left (X \right ) = -\frac {2 u \left (X \right )-1}{X}\) is separable as it can be written as
\begin{align*} \frac {d}{d X}u \left (X \right )&= -\frac {2 u \left (X \right )-1}{X}\\ &= f(X) g(u) \end{align*}
Where
\begin{align*} f(X) &= \frac {1}{X}\\ g(u) &= -2 u +1 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX}\\ \int { \frac {1}{-2 u +1}\,du} &= \int { \frac {1}{X} \,dX}\\ -\frac {\ln \left (2 u \left (X \right )-1\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 \(-2 u +1=0\) for \(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&={\frac {1}{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*} -\frac {\ln \left (2 u \left (X \right )-1\right )}{2} = \ln \left (X \right )+c_{1}\\ u \left (X \right ) = {\frac {1}{2}} \end{align*}
Solving for \(u \left (X \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (X \right )&={\frac {1}{2}}\\ u \left (X \right )&=\frac {X^{2}+{\mathrm e}^{-2 c_{1}}}{2 X^{2}} \end{align*}
Converting \(u \left (X \right ) = {\frac {1}{2}}\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = \frac {X}{2} \end{align*}
Converting \(u \left (X \right ) = \frac {X^{2}+{\mathrm e}^{-2 c_{1}}}{2 X^{2}}\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = \frac {X^{2}+{\mathrm e}^{-2 c_{1}}}{2 X} \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = \frac {X}{2}\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 = \frac {x}{2} \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = \frac {X^{2}+{\mathrm e}^{-2 c_{1}}}{2 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 = \frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x} \end{align*}
Figure 22: Slope field plot
\(y^{\prime } x +y = x\)
1.7.5 Solved as first order Exact ode
Time used: 0.069 (sec)
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 (x\right )\mathop {\mathrm {d}y} &= \left (x -y\right )\mathop {\mathrm {d}x}\\ \left (-x +y\right )\mathop {\mathrm {d}x} + \left (x\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(x,y) &= -x +y\\ N(x,y) &= x \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 (-x +y\right )\\ &= 1 \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x\right )\\ &= 1 \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 (2) w.r.t. \(y\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int N\mathop {\mathrm {d}y} \\
\int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int x\mathop {\mathrm {d}y} \\
\tag{3} \phi &= x y+ f(x) \\
\end{align*}
Where \(f(x)\) 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 \(x\) gives
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial x} = y+f'(x)
\end{equation}
But equation (1) says that \(\frac {\partial \phi }{\partial x} = -x +y\) .
Therefore equation (4) becomes
\begin{equation}
\tag{5} -x +y = y+f'(x)
\end{equation}
Solving equation (5) for \( f'(x)\) gives
\[
f'(x) = -x
\]
Integrating the above w.r.t \(x\)
gives
\begin{align*}
\int f'(x) \mathop {\mathrm {d}x} &= \int \left ( -x\right ) \mathop {\mathrm {d}x} \\
f(x) &= -\frac {x^{2}}{2}+ c_{1} \\
\end{align*}
Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(x)\) into equation
(3) gives \(\phi \)
\[
\phi = x y -\frac {1}{2} x^{2}+ c_{1}
\]
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
\[
c_{1} = x y -\frac {1}{2} x^{2}
\]
Solving for \(y\) from the
above solution(s) gives (after possible removing of solutions that do not verify)
\begin{align*} y = \frac {x^{2}+2 c_{1}}{2 x} \end{align*}
Figure 23: Slope field plot
\(y^{\prime } x +y = x\)
1.7.6 Solved as first order isobaric ode
Time used: 5.635 (sec)
Solving for \(y'\) gives
\begin{align*}
\tag{1} y' &= -\frac {-x +y}{x} \\
\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
\[ f(x,y) = -\frac {-x +y}{x}\tag {2} \]
\(m\)
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
\[ u \left (x \right )+x u^{\prime }\left (x \right ) = -\frac {-x +x u \left (x \right )}{x} \]
The ode \(u^{\prime }\left (x \right ) = -\frac {2 u \left (x \right )-1}{x}\) is
separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= -\frac {2 u \left (x \right )-1}{x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= -2 u +1 \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{-2 u +1}\,du} &= \int { \frac {1}{x} \,dx}\\ -\frac {\ln \left (2 u \left (x \right )-1\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 \(-2 u +1=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&={\frac {1}{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*} -\frac {\ln \left (2 u \left (x \right )-1\right )}{2} = \ln \left (x \right )+c_{1}\\ u \left (x \right ) = {\frac {1}{2}} \end{align*}
Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (x \right )&={\frac {1}{2}}\\ u \left (x \right )&=\frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x^{2}} \end{align*}
Converting \(u \left (x \right ) = {\frac {1}{2}}\) back to \(y\) gives
\begin{align*} \frac {y}{x} = {\frac {1}{2}} \end{align*}
Converting \(u \left (x \right ) = \frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x^{2}}\) back to \(y\) gives
\begin{align*} \frac {y}{x} = \frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x^{2}} \end{align*}
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} \frac {y}{x}&={\frac {1}{2}}\\ y&=\frac {x^{2}+{\mathrm e}^{-2 c_{1}}}{2 x} \end{align*}
Figure 24: Slope field plot
\(y^{\prime } x +y = x\)
1.7.7 Solved using Lie symmetry for first order ode
Time used: 0.813 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=-\frac {-x +y}{x}\\ 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 (-x +y \right ) \left (b_{3}-a_{2}\right )}{x}-\frac {\left (-x +y \right )^{2} a_{3}}{x^{2}}-\left (\frac {1}{x}+\frac {-x +y}{x^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )+\frac {x b_{2}+y b_{3}+b_{1}}{x} = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {x^{2} a_{2}+x^{2} a_{3}-2 b_{2} x^{2}-x^{2} b_{3}-2 x y a_{3}+2 y^{2} a_{3}-x b_{1}+y a_{1}}{x^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -x^{2} a_{2}-x^{2} a_{3}+2 b_{2} x^{2}+x^{2} b_{3}+2 x y a_{3}-2 y^{2} a_{3}+x b_{1}-y 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} -a_{2} v_{1}^{2}-a_{3} v_{1}^{2}+2 a_{3} v_{1} v_{2}-2 a_{3} v_{2}^{2}+2 b_{2} v_{1}^{2}+b_{3} v_{1}^{2}-a_{1} v_{2}+b_{1} v_{1} = 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 (-a_{2}-a_{3}+2 b_{2}+b_{3}\right ) v_{1}^{2}+2 a_{3} v_{1} v_{2}+b_{1} v_{1}-2 a_{3} v_{2}^{2}-a_{1} v_{2} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} b_{1}&=0\\ -a_{1}&=0\\ -2 a_{3}&=0\\ 2 a_{3}&=0\\ -a_{2}-a_{3}+2 b_{2}+b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=2 b_{2}+b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=b_{2}\\ 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 {-x +y}{x}\right ) \left (x\right ) \\ &= 2 y -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}{2 y -x}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (2 y -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 {-x +y}{x} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {1}{-4 y +2 x}\\ S_{y} &= \frac {1}{2 y -x} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= -\frac {1}{2 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}{2 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}{2 R}\, dR}\\ S \left (R \right ) &= -\frac {\ln \left (R \right )}{2} + 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-x \right )}{2} = -\frac {\ln \left (x \right )}{2}+c_2 \end{align*}
Which gives
\begin{align*} y = \frac {x^{2}+{\mathrm e}^{2 c_2}}{2 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 \((R,S)\)
\( \frac {dy}{dx} = -\frac {-x +y}{x}\)
\( \frac {d S}{d R} = -\frac {1}{2 R}\)
\(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (2 y -x \right )}{2} \end {aligned} \)
Figure 25: Slope field plot
\(y^{\prime } x +y = x\)
1.7.8 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x +y=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Isolate the derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=1-\frac {y}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+\frac {y}{x}=1 \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+\frac {y}{x}\right )=\mu \left (x \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+\frac {y}{x}\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=\frac {\mu \left (x \right )}{x} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int \mu \left (x \right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \mu \left (x \right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (x \right )d x +\mathit {C1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=x \\ {} & {} & y=\frac {\int x d x +\mathit {C1}}{x} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\frac {x^{2}}{2}+\mathit {C1}}{x} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}+2 \mathit {C1}}{2 x} \end {array} \]
1.7.9 Maple trace
` Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful `
1.7.10 Maple dsolve solution
Solving time : 0.003
(sec)
Leaf size : 13
dsolve ( diff ( y ( x ), x )* x + y ( x ) = x,
y(x),singsol=all)
\[
y = \frac {x}{2}+\frac {c_{1}}{x}
\]
1.7.11 Mathematica DSolve solution
Solving time : 0.039
(sec)
Leaf size : 17
DSolve [{ x * D [ y [ x ], x ]+ y [ x ]== x ,{}},
y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to \frac {x}{2}+\frac {c_1}{x}
\]