Differential equation with resonance

Question

Problem:

$$y''+20y=100\sin(wt)$$

conditions:

$$y(0)=2, y'(0)=0$$

What would the solution for $y(t)$ be when $w=2\sqrt 5$

So far I have GS(H)=Acos($\sqrt 20$)+Bsin($\sqrt 20$).

Answer 1

So the Differential equation can be rewritten as:

$$(D^2+20)y=100\sin(t\sqrt{20})$$ where $D=\dfrac{d}{dx}$.

Complementary Function, $C.F.=y_c=A\cos(t\sqrt{20})+B\sin(t\sqrt{20})$, and

The Particular Integral, \begin{align} y_p&=\dfrac{1}{D^2+20}100\sin(t\sqrt{20})\\ &=100\cdot\dfrac{1}{D^2+20}\sin(t\sqrt{20})\\ &=100t\cdot\dfrac{1}{2D}\sin(t\sqrt{20})\\ &=50t\cdot\dfrac{1}{D}\sin(t\sqrt{20})\\ &=-5\sqrt5t\cos(t\sqrt{20})\\ \end{align} Therefore General Solution: $y=y_c+y_p=A\cos(t\sqrt{20})+B\sin(t\sqrt{20})-5\sqrt5t\cos(t\sqrt{20})$.

Now, $y(0)=2\implies A=2$ and, $y'(0)=0\implies B=\dfrac{5\sqrt5}{\sqrt{20}}=\dfrac52$.

So, $y=2\cos(t\sqrt{20})+\dfrac52\sin(t\sqrt{20})-5\sqrt5t\cos(t\sqrt{20})\\ \implies \boxed{y=\dfrac52\sin(2\sqrt5t)+(2-5\sqrt5t)\cos(2\sqrt5t)}$