# ODE $x'' + 2x' +5x = \sin3t$, $x(0)=1,\ x'(0)=-1$, Solve using Laplace Transform

While solving the differential equation $$x'' + 2 x' + 5 x = \sin3t, \quad x(0) = 1, \quad x'(0) = -1$$ by use of Laplace transform I got to

$$X(s^2 +2s+5)=\frac{(3)}{s^2 +9} +s +1$$ $$X=\frac{\left(\:s^3+s^2+9s+12\right)}{\left(s^2+9\right)\left(s^2-2s+5^{\:}\right)}$$

Is it correct, if yes how do i find the inverse of it?

X=f(s)

• Partial fraction decomposition, may be? – Michael Galuza Aug 3 '15 at 17:29
• @Michael Galuza have you verified whether my answer is correct? The partial fraction is pretty complex... – problematic Aug 3 '15 at 17:32
• @Michael Galuza Please just verify my f(s), i will do the rest myself – problematic Aug 3 '15 at 18:14

Notes:

1. The first part of this solution is for an equation that contained an error. It remains here as a demonstration. The corrcted equation is the second half of this solution.
2. The process of the proposer is correct in process but does need some guidance in finding a nice result.

Consider the differential equation $$x'' + 2 x' + 5 x = 0, \quad x(0) = 1, \quad x'(0) = -1$$ and make use of \begin{align} \mathcal{L}\{x''(t)\} &= s^{2} f(s) - s f(0) - f'(0) = s^{2} f(s) - s + 1 \\ \mathcal{L}\{x'(t)\} &= s f(s) - f(0) = s f(s) - 1 \end{align} then \begin{align} (s^2 + 2s + 5) f(s) = s + 1 \end{align} such that \begin{align} f(s) = \frac{s+1}{(s+1)^{2} + 2^{2}}. \end{align} Using the known transform \begin{align} \mathcal{L}\{ e^{-at} \, \cos(b t)\} = \frac{s+a}{(s+a)^{2} + b^{2}} \end{align} then the solution to the differential equation is $x(t) = e^{-t} \, \cos(2 t)$.

Amended equation

$$x'' + 2 x' + 5 x = \sin(3 t)$$ becomes, after the Laplace transform, \begin{align} ((s+1)^{2} + 4) \, f(s) = s+1 + \frac{3}{s^{2} + 3^{2}} \end{align} for which $$f(s) = \frac{s+1}{(s+1)^{2} + 2^{2}} + \frac{1}{2} \, \frac{3}{s^{2} + 3^{2}} \cdot \frac{2}{(s+1)^{2} + 2^{2}}.$$ By using \begin{align} \mathcal{L}\{ e^{-at} \, \sin(bt)\} &= \frac{b}{(s+a)^{2} + b^{2}} \\ \mathcal{L}\left\{\int_{0}^{t} F(t-u) \, G(u) \, du \right\} &= f(s) \cdot g(s) \end{align} then \begin{align} x(t) &= e^{-t} \, \cos(2t) + \frac{1}{2} \, \int_{0}^{t} e^{-u} \, \sin(2u) \, \sin(3t-3u) \, du \\ &= \frac{1}{52} \, e^{-t} \, \left( 9 \, \sin(2t) + 58 \, \cos(2t) \right) - \frac{1}{26} \, \left( 2 \sin(3t) + 3 \cos(3t) \right). \end{align}

• i'm sorry... i made a mistake, please check the question again if you want to – problematic Aug 3 '15 at 17:49
• @problematic The corrected equation solution has also been presented. – Leucippus Aug 3 '15 at 18:17