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I have troubles finding the solution for the Cauchy problem $$u''(t) + u(t) = \mid t\mid$$ with conditions $$u(0) = 1, \, u'(0) = -1$$ for $t \in \mathbb{R}$

I tried to solve it with the formula for the solution of a linear system, reducing the order of the equation and ending up with a two dimensional system, but I get stuck. Is there another way to do it? Thanks in advance for any help!

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Here is the standard procedure: the general solution of the homogeneous equation is $a\sin\, x + b\cos \, x$ There is particular solution $u_0$ of the form $u_o(x)=a(x)\sin\, x + b(x)\cos \, x$ where $a(x)$ and $b(x)$ are now functions. Plug this into the equation and determine the functions $a(x)$ and $b(x)$. The general solution to the given non-homogeneous equation is $c\sin\, x + d\cos \, x+u_0$ where $c$ and $d$ are constants. Find the values of these constants using the initial conditions.

This is called the method of variation of parameters and you can find more details in Wikipedia.

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You should get, for instance by the method of unknown parameters for the two simpler equations $u''+u=\pm t$, $$ u(t)=\begin{cases} A\cos(t)+B\sin(t)+t&\text{ for }t\ge 0\\ C\cos(t)+D\sin(t)-t&\text{ for }t<0 \end{cases} $$ and now have to determine the integration constants to match the initial conditions from both sides of $t=0$, $$ A=1,~ B+1=-1\text{ and } C=1,~ D-1=-1 $$ so that $$ u(t)=\cos(t)-\sin(t)+[|t|-\sin(|t|)]. $$

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  • $\begingroup$ Thanks! We can do this in general for a linear equation of order two, or just in case $u'$ doesn't appear? $\endgroup$ – astrobarrel Aug 22 '19 at 13:21
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    $\begingroup$ It depends on "what" means. You can in general find a particular solution for $L[u]=|t|$ where $L$ is any linear differential operator by finding a solution $L[u_0]=t$ with initial conditions all zero, and then set $u_p(t)={\rm sign}(t)\,u_0(t)$. Here obviously $u_0(t)=t-\sin(t)=O(t^3)$ satisfies these conditions. $\endgroup$ – Lutz Lehmann Aug 22 '19 at 13:27
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You could use Laplace transform to solve a Cauchy problem like this.

Start considering $t\ge0$, so $|t|=t$.

First apply the Laplace transform to the equation:

$$u''(t)+u(t)=t\Rightarrow \mathcal{L}\{u''(t)+u(t)\}(s)=\mathcal{L}\{t\}(s)$$

You will get $$s^2\mathcal{L}\{u(t)\}(s)-su(0)-u'(0)+\mathcal{L}\{u(t)\}(s)=\frac{1}{s^2}$$

Then solve the equation to $\mathcal{L}\{u(t)\}(s)$: $$\mathcal{L}\{u(t)\}(s)=\frac{1}{s^2}-\frac{2}{1+s^2}+\frac{s}{1+s^2}$$

Apply the Laplace transform inverse:

$$u(t)=\mathcal{L}^{-1} \left(\frac{1}{s^2}-\frac{2}{1+s^2}+\frac{s}{1+s^2}\right)$$

And you will get the solution:

$$u(t)=t-2\sin{t}+\cos{t}$$

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  • $\begingroup$ Thanks, nice solution! $\endgroup$ – astrobarrel Aug 22 '19 at 13:15

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