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How is $\ y'' + 2y' +2y = e^{-x} \cdot \sin(x)$, $y(0)=0$, $y'(0)=0$ converted into a system of 1st order ODEs?

I worked the following: $$\ y{_1}=y,\; y_1'=y'=y_2,\; y_2=y',\; y_2'=y'' $$ such that I have: $$\ y_1'=y_2,\; y_2'=e^{-x} \cdot \sin(x)-2y_2-2y_1$$. Separately, I have the exact solution of the original 2nd order ODE to be: $$\ y(x) = \frac{1}{2}e^{-x}(\sin(x)-x\cos(x)) $$

However, I don't know how to express that in terms of the solutions for $\ y_1 $ and $\ y_2 $ which I need to then test some numerical approximations (e.g.: Euler, RK4, Midpt) for some pre-written MATLAB code. Is there a quick way for me to use the exact solution I already have to get $\ y_1 $ and $\ y_2 $ or do I need to solve the 1st order ODE system on its own? If so, how?

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    $\begingroup$ The solutions are $y_1(x) = \dfrac{1}{2} e^{-x} ( \sin x - x \cos x), y_2(x) = \dfrac{1}{2} e^{-x} ( (x - 1) \sin x + x \cos x)$. Compare $y_1(x)$ to your $y(x)$. What do you notice? $\endgroup$
    – Moo
    Nov 13 '16 at 17:59
  • $\begingroup$ Thank you both very much. Clearly I wasn't connecting the relation between my own variable substitutions and the exact solution. $\endgroup$
    – xq1515426
    Nov 13 '16 at 18:17
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You wrote that in the first line of definitions of $y_1,y_2$, relative to a solution $y$ of the second order equation, the first order components are $$ y_1=y,\, y_2=y' $$ so that the only thing you have to do is to compute the derivative of the exact solution. See also the comment of Moo.

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