Please solve the 2nd order differential equation by (PLEASE FOLLOW GIVEN METHOD) LAPLACE TRANSFORMATION ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE L(y) = left[frac{w}{(s^2 + a^2)(s^2+w^2)}right]*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

Question
Laplace transform
asked 2021-01-27
Please solve the 2nd order differential equation by (PLEASE FOLLOW GIVEN METHOD) LAPLACE TRANSFORMATION
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
\(L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right]*b\)
Problem 2 Solve the differential equation
\(\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)\) where \(y(0)=0\)
and \(y'(0)=0\)

Answers (1)

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,
\(y(t)=\frac{b\omega}{a^2-\omega^2}\left[\frac{1}{\omega} \sin(\omega t)-\frac{1}{a} \sin(at)\right]\)
0

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