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The problem im given is:

Use Laplace transforms to solve the initial value problem.

$$\ddot x +x=\sin(2t)$$

$$x(0)=0=\dot x(0)$$

I first do the following Laplace transforms:

$$\mathcal{L}\{\ \ddot x(t)\}=s^2X(s),$$

$$\mathcal{L}\{ \sin(2t)\}=\frac{2}{s^2+4}$$

Taking into account the initial values, rewriting the equation as:

$$s^2X(s)+X(s)=\frac{2}{s^2+4}$$

Then solving for $X(s)$:

$$X(s)=\frac{2}{(s^2+1)(s^2+4)}$$

Then here is where i get stuck, in that i dont understand how to correctly use the method of partial fractions on this problem, to get to a point where i could do an inverse Laplace transform or if that's the right approach?

$$\frac{2}{(s^2+1)(s^2+4)}=\frac{A}{s^2+1}+\frac{B}{s^2+4}$$

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You're definitely on the right track here. Let's solve your system. If we multiply both sides by $(s^2+1)(s^2+4)$, we see that

$$2 = A(s^2+4) + B(s^2+1).$$

Or equivalently

$$2 = (A+B)s^2 + (4A+B).$$

By equating coefficients (which can be made rigorous by differentiating both sides sufficiently many times), we see that $A+B = 0$ since there is no $s^2$ term on the left hand side and $4A+B = 2$. Can you take it form here?

As a note for future posts, if you put two sets of two dollar signs around your LaTeX code, it will render it as an honest-to-goodness equation (not just inline math) which helps readability a lot.

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  • $\begingroup$ What do you mean by equating coefficients? The book used that term as well, but didnt illustrate. $\endgroup$ May 17, 2014 at 3:46
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    $\begingroup$ I'll illustrate by example, suppose we had the following equation: $$s^2 + 5 = A(s^2+2) + B(s^2+5).$$ Rearranging the left and right hand sides we have $$s^2 + 5 = (A+B)s^2 + (2A+5B).$$ Equating coefficients means that we equate the coefficients of the powers of $s$ on each side. So in this case, we have $1s^2$ on the left hand side and $(A+B)s^2$ on the right hand side, so we would say that $1 = A+B$. Similarly, if we equate the $s^0$ power (constant terms), we see that $5 = 2A+5B$ as well. Does this make sense? $\endgroup$ May 17, 2014 at 3:49
  • $\begingroup$ Yes, now i get the right answer of $A=-B=\frac{2}{3}$! $\endgroup$ May 17, 2014 at 5:08

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