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Show that the solution of the initial value problem

$$y''+p(t)y'+q(t)y=g(t),y(t_0)=y_0, y'(t_0)=y_0'$$

can be written as $y=u(t)+v(t)$, where $u$ and $v$ are solutions of the two initial value problems

$$u''+p(t)u'+q(t)u=0, u(t_0)=y_0, u'(t_0)=y_0'$$ $$v''+p(t)v'+q(t)v=g(t), v(t_0)=0, v'(t_0)=0$$

respectively. Be sure to check the initial values.

I was able to show that the solution to the IVP can be written as $y=u(t)+v(t)$ (I believe).

$$(u+v)''+p(t)(u+v)'+q(u+v)=[u''(t)+v''(t)]+[p(t)(u'(t)+v'(t))]+[q(t)(u(t)+v(t))]$$ $$=(u''+pu'+qu)+(v''+pv'+qv)$$ $$=0+g(t)$$ $$=g(t)$$

The problem I am having is checking the initial values. Generally, when it's a simple $$y''+y'-2y=2t, y(0)=0, y'(0)=1$$ type of problem, checking the initial values is very straightforward. For some reason or another, I can't quite seem to check the initial values on this one. I'm sure it is something silly that I'm having a mental block on because it looks different than normal.

Note: this is problem #21 in section 3.7 from Elementary Differential Equations and Boundary Value Problems, Eighth Edition. Boyce, DiPrima.

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2 Answers 2

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The question asks you to show that $y=u(t)+v(t)$ (with their respective initial conditions) is a solution to the given IVP. Thus, you need only show that $y=u(t)+v(t)$ satisfies (i.e. makes the equality true) the given IVP. You have already done the first part by plugging in $u(t)+v(t)$ into the IVP. The second part is much like the first, verify that the initial value of $u(t)+v(t)$ satisfy the initial values of the IVP.

Note: I am unfamiliar with writing in LaTeX so please excuse my poor notation.

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  • $\begingroup$ as to latex - basicaly you just put math symbols between dollar \$ symbols (double dollar if you want the equation to be centered) and if you need special symbols you just take a look here for example: artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols $\endgroup$
    – mm-aops
    May 15, 2014 at 17:35
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For the u part: y''+y'-2y=0, y(0)=0,y'(0)=1
Solve the characteristic polynomial, you can find the general solution to be $u=C_1e^{-2t}+C_2e^t$. Put the initial value in, that is $C_1+C_2=0$ and $-2C_1+C2=1$. Subtract the first one from the second one you can find $C_1=-\frac{1}{3}$ then $C_2=\frac{1}{3}$

For the v part: y''+y'-2y=2t, y(0)=0,y'(0)=0
The general solution is $v=C_3e^{-2t}+C_4e^t-t-\frac{1}{2}$. Put the initial value in, that is $C_3+C_4=\frac{1}{2}$ and $-2C_3+C_4=1$. Subtract the first one from the second one you can find $C_3=-\frac{1}{6}$ then $C_4=\frac{2}{3}$

So $u=-\frac{1}{3}e^{-2t}+\frac{1}{3}e^t$ and $v=-\frac{1}{6}e^{-2t}+\frac{2}{3}e^t-t-\frac{1}{2}$ Add them together we have $y=-\frac{1}{2}e^{-2t}+e^t-t-\frac{1}{2}$

For this case $y(0)=-\frac{1}{2}+1-0-\frac{1}{2}=0$ and $y'(0)=-\frac{1}{2}*(-2)+1-0-\frac{1}{2}=1$

Is this what you want?

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  • $\begingroup$ Not quite. You used the example to find the solution not the problem. $\endgroup$
    – Vincent
    May 15, 2014 at 16:10
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    $\begingroup$ For the initial value part, $y(t_0)=u(t_0)+v(t_0)=y_0+0=y_0$ and $y'(t_0)=u'(t_0)+v'(t_0)=y'_0+0=y'_0$ $\endgroup$
    – TooYoung
    May 16, 2014 at 0:39
  • $\begingroup$ I think that's what I was looking! So simple, thanks! $\endgroup$
    – Vincent
    May 16, 2014 at 2:33

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