Answer to Question #343385 in Differential Equations for jweoij

Question

Use the Laplace transform to solve the given initial-value problem.

y′ + 2y = sin 4t, y(0) = 1

Answer 1

Solution

Let L(y) = Y(s)

Then L(y’) = sY(s) – y(0) = sY – 1; L(sin(4t) = 4/(s² + 4²)

So from the given initial-value problem

sY + 2Y – 1 = 4/(s² + 4²)

Y = 4/[(s² + 4²)(s + 2)] + 1/(s + 2)

4/[(s² + 4²)(s + 2)] = (As+B)/(s² + 4²) + C/(s + 2) => (As+B)(s+2) + C(s² + 4²) = 4 => As² +(2A+B)s + 2B +Cs² +16C = 4 => A + C = 0, 2A + B = 0, 2B + 16C = 4 => A = -C, B – 2C = 0, B + 8C = 2 => C = 0.2, B = 0.4, A = -0.2 =>

Y(s) = 0.2(2 – s)/(s² + 4²) + 1.2/(s + 2) = 0.4/(s² + 4²) – 0.2s/(s² + 4²) + 1.2/(s + 2)

Therefore using tables of inverse Laplace transform

y(t) = L^-1[Y(s)] = 0.1sin(4t) – 0.2 cos(4t) + 1.2e^-2t  

Answer

y(t) = 0.1sin(4t) – 0.2 cos(4t) + 1.2e^-2t