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

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 $$y'(0)=0$$

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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,
$$y(t)=\frac{b\omega}{a^2-\omega^2}\left[\frac{1}{\omega} \sin(\omega t)-\frac{1}{a} \sin(at)\right]$$