# 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?

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.

• Perfectly understood! Thanks for your answer! Jul 13, 2020 at 10:03

$$(2x+3)^3 y'''+3(2x+3)y'-6y=0$$ 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 $$y=C_1 z^{1/2}+ C_2 z +C_3 z^{3/2},~~z=(2x+3).$$

$$(2x+3)^3 y''' + 3(2x+3)y' -6y=0$$ 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 $$6y'_t$$

• $8u^3y′′′+6uy′−6y=0$ how did you get the coefficients 8, 6 and -6? When you substitute $u=2x+3$ I simply get $u^3y''' +3uy' -6y=0$ Jul 12, 2020 at 16:36
• I'm not sure why 6 instead of 3? Jul 12, 2020 at 16:37
• You simplify by $2$ not $3$ @johndoe Jul 12, 2020 at 16:38