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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.