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Synopsis:

Please check my work. I do not have a text "answers to odd problems" for reference as this is an "even" numbered problem. The following documents in good detail the steps taken to solve for this so that the root of any errors, if any occur, can easily be found. Your input is very graciously welcomed.

Given:

Solve the following initial value problem where $y(0)=1$ and $y'(0)=0$ by applying the convolution theorem...

$$y''+4y=-6\cos(t)$$

As a reference the convolution theorem formula is given and defined as...

$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$

My Solution:

Begin by finding the Laplace transform of $y''$...

$$\mathcal{L}\{y''\}=s^2\cdot Y-s\cdot y(0)-y'(0)=s^2Y-s(1)=s^2Y-s$$

Now we are ready to replace $y''$ with $\mathcal{L}\{y''\}$ while taking the Laplace transform of entire equation and solving for $Y$...

$$Y(s^2+4)=-6\cdot\frac{s}{s^2+1}+s$$

$$Y=-6\cdot\frac{s}{s^2+1}\cdot\frac{2}{2(s^2+4)}+\frac{s}{s^2+4}=-3\cdot\frac{s}{s^2+1}\cdot\frac{2}{s^2+4}+\frac{s}{s^2+4}$$

The problem shall be broken up into three parts for easier calculations...

$$Y_1(s)=\frac{2}{s^2+4}\cdot\frac{s}{s^2+1} \implies y_1(t)=\sin(2t)*\cos(t)$$

$$Y_2(s)=\frac{s}{s^2+4} \implies y_2(t)=\cos(2t)$$

$$Y(s)=-3Y_1(s)+Y_2(s) \implies y(t)=-3\cdot y_1(t)+y_2(t)$$

We already have $y_2(t)$ therefore we just need to find $y_1(t)$. We start by inserting the two associated functions of the transforms into the formula $h(t)$ and simplify by switching $t$ and $\tau$ in the cosine (no need to change the sign)...

$$h(t)= \sin(2t)*\cos(t)$$

$$=\int_0^t \sin(2\tau)\cdot \cos(t-\tau)d\tau$$

$$=\int_0^t \sin(2\tau)\cdot \cos(\tau-t)d\tau$$

We further simplify by applying the following trigonometric product formula identity given here as a reference...

$$sin(\alpha)\cdot \cos(\beta)=\frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]$$

This yields...

$$y_1(t)=\frac{1}{2}\int_0^t[\sin(3\tau-t)-\sin(\tau+t)]d\tau$$

$$=\frac{1}{2}\left[-\frac{\cos(3\tau-t)}{3}+\cos(\tau+t)\right]_0^t$$

$$=\frac{1}{2}\left[-\frac{\cos(2t)}{3}+\cos(2t)+\frac{\cos(t)}{3}-\cos(t)\right]$$

$$=\frac{1}{2}\left[\frac{2}{3}\cdot\cos(2t)-\frac{2}{3}\cdot\cos(t)\right]$$

$$=\frac{1}{3}\cdot\cos(2t)-\frac{1}{3}\cdot\cos(t)$$

Recall that...

$$y(t)=-3\cdot y_1(t)+y_2(t)$$

and that...

$$y_2(t)=\cos(2t)$$

Therefor we have...

$$y(t)=-3\left(\frac{1}{3}\cdot\cos(2t)-\frac{1}{3}\cdot\cos(t)\right)+\cos(2t)$$

$$=\cos(t)$$

Answer in Text:

No answer in text to compare this solution with.

Question:

I love solutions that are short, simple, and sweet as this but is it correct? Did you find any errors?

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    $\begingroup$ Not correct. You can easily check by plugging it into your original equation. $\endgroup$ – Euler....IS_ALIVE Feb 27 '15 at 5:57
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    $\begingroup$ Put slashes in front of all you trig latex like so \cos(t) or \cos t $\endgroup$ – dustin Feb 27 '15 at 5:57
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    $\begingroup$ @JulesManson: You should be getting $y(t) = -2 \cos(t) + 3 \cos(2t)$. It looks like you have an issue with the $Y(s)$ and that is causing the issues. $\endgroup$ – Amzoti Feb 27 '15 at 5:59
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    $\begingroup$ i hate foreign books by english authors with only odd number solutions. either give all or none or just hints. $\endgroup$ – RE60K Feb 27 '15 at 6:02
  • $\begingroup$ @Amzoti can you show me where my error occurred? $\endgroup$ – Jules Manson Feb 27 '15 at 6:02
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just checked this: \begin{align} \int_0^t \sin(2\tau)\cdot \cos(t-\tau)d\tau=\frac23(\cos t -\cos 2t) \end{align} which is different from your integral.

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