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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?

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3 Answers 3

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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.

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  • $\begingroup$ Perfectly understood! Thanks for your answer! $\endgroup$
    – john doe
    Jul 13, 2020 at 10:03
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$$(2x+3)^3 y'''+3(2x+3)y'-6y=0$$ 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 $$y=C_1 z^{1/2}+ C_2 z +C_3 z^{3/2},~~z=(2x+3).$$

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

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

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