Differential Equation, Solve for function

[GreenPrint]

Homework Statement

Given

$\frac{dx}{dt} + ax = Asin(ωt), x(0) = b$

Solve for $x(t)$

Homework Equations

The Attempt at a Solution

I take the Laplace transform of both sides and get

$sX(s) - x(0) + aX(s) = \frac{Aω}{s^{2} + ω^{2}}$
$X(s) = \frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})(s+1)}$

The solution then state that the expression above is equal to

$(b + \frac{ωA}{a^{2}+ω^{2}})\frac{1}{s + a} + \frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} - \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}$

I don't know how the two expressions are equal to each other or were $\frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} - \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}$ came from. I believe that $\frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})(s+a)} = (b + \frac{ωA}{a^{2}+ω^{2}})\frac{1}{s + a}$ as it's just factoring. So then if this is true than $\frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} = \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}$, which I don't see how that's true.

Thanks for any help.

 

[LCKurtz]

Homework Statement

Given

$\frac{dx}{dt} + ax = Asin(ωt), x(0) = b$

Solve for $x(t)$

Homework Equations

The Attempt at a Solution

I take the Laplace transform of both sides and get

$sX(s) - x(0) + aX(s) = \frac{Aω}{s^{2} + ω^{2}}$
$X(s) = \frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})\color{red}{(s+a)}}$

The solution then state that the expression above is equal to...

I think you meant $s+a$ where I have corrected in red. I didn't bother to work through your next calculations. This is going to be messy even without any mistakes. What the author surely did was use the standard procedure of partial fractions to break up that last fraction:$$
\frac{Aω}{(s^2 + ω^2)(s+a)}= \frac C {s+a} + \frac{Ds+E}{s^2+\omega^2}$$Use your calculus partial fraction methods to solve for $C,D,E$. They will be a bit messy and if neither you nor the author made any mistakes it should work. That last fraction is easy enough to take the inverse transform. I would inverse just as it stands first, then substitute the values for $C,D$, and $E$ in your answer.

I should comment that LaPlace transforms is an terrible way to work this problem. I'm assuming you were asked to do it that way or I would have a rather different hint.

 

[GreenPrint]

How would you recommend solving this problem then, if there's a much easier way to solve this than I would like to know how that can be done. Resolving that into partial fractions is very difficult to do.

 

[Chestermiller]

Another way of inverting the second term is to use convolution, especially with the s+a term there in the denominator.

 

[pasmith]

How would you recommend solving this problem then, if there's a much easier way to solve this than I would like to know how that can be done. Resolving that into partial fractions is very difficult to do.

The method of undertermined coefficients suggests a solution of the form
$$x(t) = Be^{-at} + C\cos \omega t + D \sin \omega t$$
for some constants $B$, $C$ and $D$.

 

[LCKurtz]

How would you recommend solving this problem then, if there's a much easier way to solve this than I would like to know how that can be done. Resolving that into partial fractions is very difficult to do.

It's a constant coefficient equation. I would have recommended the same as pasmith did above.