2.10 problem 35
Internal
problem
ID
[3949]
Book
:
Ordinary
differential
equations
and
their
solutions.
By
George
Moseley
Murphy.
1960
Section
:
Various
2
Problem
number
:
35
Date
solved
:
Thursday, October 17, 2024 at 05:05:32 AM
CAS
classification
:
[_separable]
Solve
\begin{align*} y^{\prime }&=\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \end{align*}
2.10.1 Solved as first order linear ode
Time used: 0.161 (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) &=-a -\cos \left (\ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right )\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \left (-a -\cos \left (\ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right )\right )d x}\\ &= {\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\) gives the final solution
\[ y = {\mathrm e}^{x \left (a +\sin \left (\ln \left (x \right )\right )\right )} c_1 \]
2.10.2 Solved as first order separable ode
Time used: 0.124 (sec)
The ode \(y^{\prime } = \left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\) is separable as it can be written as
\begin{align*} y^{\prime }&= \left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\\ &= f(x) g(y) \end{align*}
Where
\begin{align*} f(x) &= a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\\ g(y) &= y \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{y}\,dy} &= \int { a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right ) \,dx}\\ \ln \left (y\right )&=x \left (a +\sin \left (\ln \left (x \right )\right )\right )+c_1 \end{align*}
We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is
zero, since we had to divide by this above. Solving \(g(y)=0\) or \(y=0\) for \(y\) gives
\begin{align*} y&=0 \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.
Therefore the solutions found are
\begin{align*} \ln \left (y\right ) = x \left (a +\sin \left (\ln \left (x \right )\right )\right )+c_1\\ y = 0 \end{align*}
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} y&=0\\ y&={\mathrm e}^{a x +\sin \left (\ln \left (x \right )\right ) x +c_1} \end{align*}
2.10.3 Solved as first order homogeneous class D2 ode
Time used: 0.136 (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 ) = \left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) u \left (x \right ) x \end{align*}
Which is now solved In canonical form a linear first order is
\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=-\frac {\cos \left (\ln \left (x \right )\right ) x +\sin \left (\ln \left (x \right )\right ) x +a x -1}{x}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {\cos \left (\ln \left (x \right )\right ) x +\sin \left (\ln \left (x \right )\right ) x +a x -1}{x}d x}\\ &= x \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u x \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} u x \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}
Dividing throughout by the integrating factor \(x \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\) gives the final solution
\[ u \left (x \right ) = \frac {{\mathrm e}^{x \left (a +\sin \left (\ln \left (x \right )\right )\right )} c_1}{x} \]
Converting \(u \left (x \right ) = \frac {{\mathrm e}^{x \left (a +\sin \left (\ln \left (x \right )\right )\right )} c_1}{x}\) back to \(y\)
gives
\begin{align*} y = {\mathrm e}^{x \left (a +\sin \left (\ln \left (x \right )\right )\right )} c_1 \end{align*}
2.10.4 Solved as first order Exact ode
Time used: 0.179 (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*} \mathop {\mathrm {d}y} &= \left (\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\right )\mathop {\mathrm {d}x}\\ \left (-\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(x,y) &= -\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\\ N(x,y) &= 1 \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 (-\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\right )\\ &= -a -\cos \left (\ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right ) \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=1\left ( \left ( -a -\cos \left (\ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right )\right ) - \left (0 \right ) \right ) \\ &=-a -\cos \left (\ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right ) \end{align*}
Since \(A\) does not depend on \(y\), then it can be used to find an integrating factor. The integrating
factor \(\mu \) is
\begin{align*} \mu &= e^{ \int A \mathop {\mathrm {d}x} } \\ &= e^{\int -a -\cos \left (\ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right )\mathop {\mathrm {d}x} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{-a x -\sin \left (\ln \left (x \right )\right ) x } \\ &= {\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )} \end{align*}
\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\)
for now so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= {\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\left (-\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\right ) \\ &= -\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= {\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\left (1\right ) \\ &= {\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )} \end{align*}
Now a modified ODE is ontained from the original ODE, which is exact and can be solved.
The modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (-\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\right ) + \left ({\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)
\begin{align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {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 \overline {N}\mathop {\mathrm {d}y} \\
\int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int {\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\mathop {\mathrm {d}y} \\
\tag{3} \phi &= y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}+ 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{align*}
\tag{4} \frac {\partial \phi }{\partial x} &= \left (-a -\cos \left (\ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right )\right ) y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}+f'(x) \\
&=-\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}+f'(x) \\
\end{align*}
But equation (1) says that \(\frac {\partial \phi }{\partial x} = -\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}\).
Therefore equation (4) becomes
\begin{equation}
\tag{5} -\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )} = -\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}+f'(x)
\end{equation}
Solving equation (5) for \( f'(x)\) gives
\[ f'(x) = 0 \]
Therefore
\[ f(x) = c_1 \]
Where \(c_1\) is
constant of integration. Substituting this result for \(f(x)\) into equation (3) gives \(\phi \)
\[
\phi = y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}+ c_1
\]
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
\[
c_1 = y \,{\mathrm e}^{-x \left (a +\sin \left (\ln \left (x \right )\right )\right )}
\]
Solving for \(y\) from the
above solution(s) gives (after possible removing of solutions that do not verify)
\begin{align*} y = {\mathrm e}^{x \left (a +\sin \left (\ln \left (x \right )\right )\right )} c_1 \end{align*}
2.10.5 Solved using Lie symmetry for first order ode
Time used: 0.439 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y\\ 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}+\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \left (b_{3}-a_{2}\right )-\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right )^{2} y^{2} a_{3}-\left (-\frac {\sin \left (\ln \left (x \right )\right )}{x}+\frac {\cos \left (\ln \left (x \right )\right )}{x}\right ) y \left (x a_{2}+y a_{3}+a_{1}\right )-\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {\cos \left (\ln \left (x \right )\right )^{2} x \,y^{2} a_{3}+2 \cos \left (\ln \left (x \right )\right ) \sin \left (\ln \left (x \right )\right ) x \,y^{2} a_{3}+2 \cos \left (\ln \left (x \right )\right ) a x \,y^{2} a_{3}+\sin \left (\ln \left (x \right )\right )^{2} x \,y^{2} a_{3}+2 \sin \left (\ln \left (x \right )\right ) a x \,y^{2} a_{3}+a^{2} x \,y^{2} a_{3}+\cos \left (\ln \left (x \right )\right ) x^{2} b_{2}+2 \cos \left (\ln \left (x \right )\right ) x y a_{2}+\cos \left (\ln \left (x \right )\right ) y^{2} a_{3}+\sin \left (\ln \left (x \right )\right ) x^{2} b_{2}-\sin \left (\ln \left (x \right )\right ) y^{2} a_{3}+a \,x^{2} b_{2}+a x y a_{2}+\cos \left (\ln \left (x \right )\right ) x b_{1}+\cos \left (\ln \left (x \right )\right ) y a_{1}+\sin \left (\ln \left (x \right )\right ) x b_{1}-\sin \left (\ln \left (x \right )\right ) y a_{1}+a x b_{1}-x b_{2}}{x} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} -\cos \left (\ln \left (x \right )\right )^{2} x \,y^{2} a_{3}-2 \cos \left (\ln \left (x \right )\right ) \sin \left (\ln \left (x \right )\right ) x \,y^{2} a_{3}-2 \cos \left (\ln \left (x \right )\right ) a x \,y^{2} a_{3}-\sin \left (\ln \left (x \right )\right )^{2} x \,y^{2} a_{3}-2 \sin \left (\ln \left (x \right )\right ) a x \,y^{2} a_{3}-a^{2} x \,y^{2} a_{3}-\cos \left (\ln \left (x \right )\right ) x^{2} b_{2}-2 \cos \left (\ln \left (x \right )\right ) x y a_{2}-\cos \left (\ln \left (x \right )\right ) y^{2} a_{3}-\sin \left (\ln \left (x \right )\right ) x^{2} b_{2}+\sin \left (\ln \left (x \right )\right ) y^{2} a_{3}-a \,x^{2} b_{2}-a x y a_{2}-\cos \left (\ln \left (x \right )\right ) x b_{1}-\cos \left (\ln \left (x \right )\right ) y a_{1}-\sin \left (\ln \left (x \right )\right ) x b_{1}+\sin \left (\ln \left (x \right )\right ) y a_{1}-a x b_{1}+x b_{2} = 0
\end{equation}
Simplifying the above gives
\begin{equation}
\tag{6E} -a^{2} x \,y^{2} a_{3}-a \,x^{2} b_{2}-a x y a_{2}-a x b_{1}+x b_{2}-y^{2} a_{3} x -2 \sin \left (\ln \left (x \right )\right ) a x \,y^{2} a_{3}-2 \cos \left (\ln \left (x \right )\right ) a x \,y^{2} a_{3}-x \,y^{2} a_{3} \sin \left (2 \ln \left (x \right )\right )-\sin \left (\ln \left (x \right )\right ) x^{2} b_{2}+\sin \left (\ln \left (x \right )\right ) y^{2} a_{3}-\cos \left (\ln \left (x \right )\right ) x^{2} b_{2}-2 \cos \left (\ln \left (x \right )\right ) x y a_{2}-\cos \left (\ln \left (x \right )\right ) y^{2} a_{3}-\sin \left (\ln \left (x \right )\right ) x b_{1}+\sin \left (\ln \left (x \right )\right ) y a_{1}-\cos \left (\ln \left (x \right )\right ) x b_{1}-\cos \left (\ln \left (x \right )\right ) y a_{1} = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y, \cos \left (\ln \left (x \right )\right ), \ln \left (x \right ), \sin \left (2 \ln \left (x \right )\right ), \sin \left (\ln \left (x \right )\right )\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\{x = v_{1}, y = v_{2}, \cos \left (\ln \left (x \right )\right ) = v_{3}, \ln \left (x \right ) = v_{4}, \sin \left (2 \ln \left (x \right )\right ) = v_{5}, \sin \left (\ln \left (x \right )\right ) = v_{6}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} -a^{2} v_{1} v_{2}^{2} a_{3}-2 v_{3} a v_{1} v_{2}^{2} a_{3}-2 v_{6} a v_{1} v_{2}^{2} a_{3}-v_{1} v_{2}^{2} a_{3} v_{5}-a v_{1} v_{2} a_{2}-a v_{1}^{2} b_{2}-2 v_{3} v_{1} v_{2} a_{2}-v_{2}^{2} a_{3} v_{1}-v_{3} v_{2}^{2} a_{3}+v_{6} v_{2}^{2} a_{3}-v_{3} v_{1}^{2} b_{2}-v_{6} v_{1}^{2} b_{2}-a v_{1} b_{1}-v_{3} v_{2} a_{1}+v_{6} v_{2} a_{1}-v_{3} v_{1} b_{1}-v_{6} v_{1} b_{1}+v_{1} b_{2} = 0
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} -v_{3} v_{1}^{2} b_{2}-v_{6} v_{1}^{2} b_{2}-a v_{1}^{2} b_{2}-2 v_{3} a v_{1} v_{2}^{2} a_{3}-v_{1} v_{2}^{2} a_{3} v_{5}-2 v_{6} a v_{1} v_{2}^{2} a_{3}+\left (-a^{2} a_{3}-a_{3}\right ) v_{1} v_{2}^{2}-2 v_{3} v_{1} v_{2} a_{2}-a v_{1} v_{2} a_{2}-v_{3} v_{1} b_{1}-v_{6} v_{1} b_{1}+\left (-a b_{1}+b_{2}\right ) v_{1}-v_{3} v_{2}^{2} a_{3}+v_{6} v_{2}^{2} a_{3}-v_{3} v_{2} a_{1}+v_{6} v_{2} a_{1} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} a_{1}&=0\\ a_{3}&=0\\ -a_{1}&=0\\ -2 a_{2}&=0\\ -a_{3}&=0\\ -b_{1}&=0\\ -b_{2}&=0\\ -a a_{2}&=0\\ -2 a a_{3}&=0\\ -a b_{2}&=0\\ -a^{2} a_{3}-a_{3}&=0\\ -a b_{1}+b_{2}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=0\\ 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 &= 0 \\
\eta &= y \\
\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}{y}} dy \end{align*}
Which results in
\begin{align*} S&= \ln \left (y \right ) \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) &= \left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= 0\\ S_{y} &= \frac {1}{y} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\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} &= a +\cos \left (\ln \left (R \right )\right )+\sin \left (\ln \left (R \right )\right ) \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 {a +\cos \left (\ln \left (R \right )\right )+\sin \left (\ln \left (R \right )\right )\, dR}\\ S \left (R \right ) &= a R +\sin \left (\ln \left (R \right )\right ) R + 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*} \ln \left (y\right ) = a x +\sin \left (\ln \left (x \right )\right ) x +c_2 \end{align*}
Which gives
\begin{align*} y = {\mathrm e}^{a x +\sin \left (\ln \left (x \right )\right ) x +c_2} \end{align*}
2.10.6 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right ) y \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}=a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}d x =\int \left (a +\cos \left (\ln \left (x \right )\right )+\sin \left (\ln \left (x \right )\right )\right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (x \right )\right )=x a +\sin \left (\ln \left (x \right )\right ) x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )={\mathrm e}^{x a +\sin \left (\ln \left (x \right )\right ) x +\mathit {C1}} \end {array} \]
2.10.7 Maple trace
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
2.10.8 Maple dsolve solution
Solving time : 0.003 (sec)
Leaf size : 14
dsolve(diff(y(x),x) = (a+cos(ln(x))+sin(ln(x)))*y(x),
y(x),singsol=all)
\[
y \left (x \right ) = c_{1} {\mathrm e}^{x \left (\sin \left (\ln \left (x \right )\right )+a \right )}
\]
2.10.9 Mathematica DSolve solution
Solving time : 0.048 (sec)
Leaf size : 22
DSolve[{D[y[x],x]==(a+Cos[Log[x]]+Sin[Log[x]]) y[x],{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to c_1 e^{x (a+\sin (\log (x)))} \\
y(x)\to 0 \\
\end{align*}