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This ODE came out in my exam

$$y''' -4y'' -12y'=x^2 $$

However, I was only taught till 2nd order linear ODEs. Can we reduce this to a 2nd order or even a 1st order ODE and then use substitution to find the general solutions at the end? How does it work?

Rodrigo de Azevedo
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user307640
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2 Answers2

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We are given

$$y''' -4y'' -12y'=x^2$$

We can set

$$\begin{align} y_1' &= y' = y_2 \\ y_2' &= y'' = y_3 \\ y_3' &= y''' = 4 y'' + 12y' + x^2 = 12 y_2 + 4 y_3 + x^2 \end{align}$$

In matrix form, we have a system $Y'(x) = AY(x) + g(x)$ as

$$Y' = \begin{bmatrix} y_1' \\ y_2' \\ y_3' \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 12 & 4 \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ x^2 \end{bmatrix}$$

Now you have a nonhomogeneous system and solve that using the standard methods or numerical methods.

Also, there are other methods to solve the original ODE, like Undetermined Coefficients, Laplace Transforms, Substitution/Integration and others.

You get the solution

$$y(x) = -\frac{1}{2} c_1 e^{-2 x}+\frac{1}{6} c_2 e^{6 x}+c_3-\frac{x^3}{36}+\frac{x^2}{36}-\frac{7 x}{216}$$

Moo
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  • is your matrix thing really better or something than the good ol' calculus 1 way of thinking $w=y'$ ? – BCLC Nov 19 '22 at 22:37
  • What does that matter? This is the answer to what is being asked. Period. I also provided hints for other approaches. – Moo Nov 19 '22 at 22:55
  • Moo, just wondering. If it's better, then why not do it right? – BCLC Nov 21 '22 at 01:02
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Part1. Yes because there's no original $y$ there. You can set $w=y'$ to get

$$w'' -4w' -12w=x^2 $$

Part2. You can't do this kind of substitution if there's an original $y$ eg

$$y''' -4y'' -12y'+7y=x^2 $$

because then you'd get something like

$$w'' -4w' -12w+7W=x^2 $$

where $W$ is some (any? Idk) antiderivative of $w$.

Part3. Anyway so go right ahead and apply what you know Gasai to

$$w'' -4w' -12w=x^2 $$

If I remember correctly, then this is

  1. Solve this $w'' -4w' -12w=0 $

  2. by solving this $d^2-4d-12=0$

  3. Solving $w'' -4w' -12w=x^2 $ given solution of $w'' -4w' -12w=0 $ or something. You know Gasai what to do.

BCLC
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