Laplace transform of $\ddot{x} +4x = f(t)$

Question

I am stuck in an excercise that on first sight didn't look that strange: Given the initial problem: $$\ddot{x} +4x = f(t), x(t=0)=3, \dot{x}(t=0)=-1$$ So I started:

$$s^2(X(s) -sx(0)-\dot{x}(0) +4X(s) = \mathcal{L}(f(t))$$ Now substitute the given values: $$s^2X(s) -3s-(-1) +4X(s) = \mathcal{L}(f(t))$$ Rearranging: $$X(s)(s^2+4) -3s+1 = \mathcal{L}(f(t))$$

The answer give: The Laplace transform is of the form: $$X(t)= A \cos2t+B \sin2t +\frac{1}{2}\int_{0}^{t}f(\tau)\sin2(t-\tau)d\tau $$

Is there anybody that can help me to get the given form?

Answer 1

I think I found the solution already:

$$X(s)= \frac{\mathcal{L(f(t))}}{s^2+4} + \frac{3s-1}{s^2+4}$$

$$= \frac{\mathcal{L(f(t))}}{s^2+4} + \frac{3s}{s^2+4} + \frac{-1}{s^2+4}$$

Since the 'standard results of: $\frac{s}{s^2+4} = A\cos2t$ and $\frac{1}{s^2+4}= B\sin2t$ The first term can be arranged by the convolution theorem: $$\frac{\mathcal{L(f(t))}}{s^2+4} = \frac{1}{2}\int_{0}^{t}\sin2(t-\tau)\cdot f(\tau)dt $$

combining this all: $$X(t)= A \cos2t+B \sin2t +\frac{1}{2}\int_{0}^{t}f(\tau)\sin2(t-\tau)d\tau$$

Answer 2

Hint

You were almost done put all the s terme at the right side then use the convolution formula $$X(s)(s^2+4) -3s+1 = \mathcal{L}(f(t))$$ $$X(s)(s^2+4) =3s-1+ \mathcal{L}(f(t))$$ For convenience I substitute $h(s)=\mathcal{L}(f(t))$ $$X(s) =\frac {3s-1}{s^2+4}+ h(s) * \, \frac 1 {s^2+4}$$ $$X(s) =3\frac {s}{s^2+4}-\frac 12\frac {2}{s^2+4}+ \frac 12h(s) * \, \frac 2 {s^2+4}$$ Because $\mathcal{L^{-1}}(h(s))=\mathcal{L^{-1}}\mathcal{L}(f(t))=f(t)$ Using the convolution formula, we get that : $$\boxed {x(t) =3\cos(2t)-\frac 12\sin(2t)+ \frac 12\int_0^t f(\tau) \, \sin(2(t-\tau))d\tau}$$