# Please solve the 2nd order differential equation by (PLEASE FOLLOW GIVEN METHOD) LAPLACE TRANSFORMATION ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE L(y

Laplace transform

ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right] \cdot b$$
Problem 2 Solve the differential equation
$$\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)$$ where $$y(0)=0$$
and $$y'(0)=0$$

2021-01-28
Step 1
Given a differential equation, $$y"+a^2y=b\sin(\omega t)$$, where $$y(0)=0, y'(0)=0$$
Taking the Laplace transform of both sides of the given differential equation,
$$L(y")+a^2L(y)=bL(\sin(\omega t))$$
$$s^2L(y)−sy(0)−y'(0)+a^2L(y)=\frac{b\omega}{s^2+\omega^2}$$
$$(s^2+a^2)L(y)=\frac{b\omega}{s^2+\omega^2}$$
$$\therefore, L(y)=\frac{b\omega}{(s^2+\omega^2)}(s^2+\omega^2)$$
Step 2
Applying partial fraction,
$$\frac{b\omega}{(s^2+\omega^2)(s^2+a^2)}=\frac{A}{s^2+\omega^2}+\frac{B}{s^2+\omega^2}\dots (1)$$
$$A(s^2+a^2)+B(s^2+\omega^2)=b\omega$$
$$As^2+Aa^2+Bs^2+B\omega^2=b\omega$$
$$(A+B)s^2+Aa^2+B\omega^2=b\omega$$
Comparing coefficients,
$$A+B=0 \Rightarrow A=-B$$
$$Aa^2+B\omega^2=b\omega$$
$$\text{Substitute A in the above equation, }$$
$$-Ba^2+B\omega^2=b\omega$$
$$B=\frac{b\omega}{\omega^2-a^2}$$
Therefore, $$A=(\frac{b\omega}{a})^2-\omega^2$$
Step 3
Substitute A and B in equation (1),
$$\frac{b\omega}{(s^2+\omega^2)(s^2+a^2)}=\frac{b\omega}{(a^2-\omega^2)(s^2+\omega^2)}-\frac{b\omega}{(a^2-w^2)(s^2+a^2)}$$
$$=\frac{b\omega}{(a^2-w^2)\frac{1}{s^2+\omega^2}}-\frac{1}{(s^2+a^2)}$$
Substitute in L(y),
$$L(y)=\frac{b\omega}{a^2-\omega^2} \left(\frac{1}{s^2+\omega^2}-\frac{1}{s^2+a^2}\right)$$
$$\text{Taking inverse Laplace transform, }$$
$$y(t)=\frac{b\omega}{a^2-\omega^2}L^{-1}\left(\frac{1}{s^2+\omega^2}-\frac{1}{s^2+a^2}\right)$$
$$y(t)=\frac{b\omega}{a^2-\omega^2}\left[L^{-1}\left(\frac{1}{s^2+\omega^2}\right)-L^{-1}\left(\frac{1}{s^2+a^2}\right)\right]$$
$$=\frac{b\omega}{a^2-\omega^2}\left[\frac{1}{\omega} \sin(\omega t)-\frac{1}{a} \sin(at)\right]$$
Therefore, the required solution is,
$$y(t)=\frac{b\omega}{a^2-\omega^2}\left[\frac{1}{\omega} \sin(\omega t)-\frac{1}{a} \sin(at)\right]$$