Solving a third order Euler-Cauchy ODE I have been given the following ODE:
$$(2x+3)^3 y''' + 3 (2x+3) y' - 6 y=0$$
and I have to solve it using Euler's method, which I am fairly familiar with.
Now, I let $ 2x+3 = e^t$ and $y=e^{λt}$
After differentiating $y$, I get that $$y''' = \frac{y_t'''-3y_t''+2y_t'}{e^{3t}}$$
and $y'$ is $$\frac{y_t'}{e^t}$$
Now after substituting in the given equation I get
$$e^{3t} \frac{y_t'''-3y_t''+2y_t'}{e^{3t}} + 3e^t \frac{y_t'}{e^t} -6y=0 $$
After which I am left with the following homogeneous equation:
$$y''' - 3y'' + 5y' -6y =0$$
Which can be easily solved and the solutions are (I checked in wolframalpha):
$$C_1 e^{2t} + e^{\frac{t}{2}}(C_2 \cos(\frac{\sqrt {11}}{2} t) + C_3 \sin(\frac{\sqrt {11}}{2} t))$$
When I plug $2x+3=e^t$ back in, I get: $$y(x) = C_1(2x+3)^2 + C_2 \sqrt{2x+3}   \cos(\frac{\sqrt {11}}{2}\ln(2x+3)) +  C_3 \sqrt{2x+3}   \sin(\frac{\sqrt {11}}{2}\ln(2x+3))$$
But the wolframalpha solution for the whole eqauation is
$$C_2(2x+3)^{\frac{3}{2}} + C_3(2x+3) + C_1\sqrt{2x+3}$$
Now, I am new to ODES so I can't rule out that I made a silly mistake. What I did when substituting back is essentially $e^t = 2x+3$ and $t=\ln(2x+3)$
Can anyone point out my mistakes?
 A: If you set $2x+3=e^t$, then in $u(t)=y(x)$ you get $u(t)=y(\frac{e^t-3}2)$. Thus computing the derivatives gives
$$
u'(t)=y'(x)\frac{e^t}2\\
u''(t)=y''(x)\frac{e^{2t}}4+y'(x)\frac{e^t}2\\
u'''(t)=y'''(x)\frac{e^{3t}}8+y''(x)\frac{3e^{2t}}4+y'(x)\frac{e^t}2
$$
This can also be solved for the derivatives of $y$ to get
$$
y'(x)=2e^{-t}u(t)\\
y''(x)=4e^{-2t}(u''(t)-u'(t))\\
y'''(x)=8e^{-3t}(u'''(t)-3u''(t)+2u'(t))
$$
This means in your initial calculations you did not consider the inner derivative/linear coefficient $2$ in $e^t=2x+3$. You could have chosen to set $e^t=x+\frac32$, then the powers of $2$ originate in the polynomial coefficients.
A: $$(2x+3)^3 y'''+3(2x+3)y'-6y=0$$
Let $2x+3=z$, the the ODE converts to
$$8z^3 \frac{d^3 y}{dz^3}+6z\frac{dy}{dz}-6y=0$$
THis is Eulars Eq. which is solves by taking $y=z^m$, then
$$8m(m-1)(m-2)+6m-6=0 \implies m=1/2,1,3/2$$
So the solution of the ODE is
$$y=C_1 z^{1/2}+ C_2 z +C_3 z^{3/2},~~z=(2x+3).$$
A: $$(2x+3)^3 y''' + 3(2x+3)y' -6y=0$$
Substitute $u=2x+3$
$$8u^3 y''' + 6uy' -6y=0$$
$$4u^3 y''' + 3uy' -3y=0$$
Then $u=e^t \implies t =\ln u$
$$y'=\dfrac {dy}{du}=\dfrac {dy}{dt}\dfrac {dt}{du}=\dfrac 1 u\dfrac {dy}{dt}$$
$$6uy'_u=6y'_t$$
The DE becomes:
$$4y'''-12y''+11y'-3y=0$$
And the solution is:
$$y(t)=c_1e^{t}+c_2e^{3/2t}+c_3e^{1/2t}$$

So here you have a mistake for $y'$
$$e^{3t} \frac{y_t'''-3y_t'''+2y_t'}{e^{3t}} + \color{red}{3e^t \frac{y_t'}{e^t}} -6y=0$$
it should be $6y'_t$
