Answer to Question #205569 in Differential Equations for adi

Question

Use Laplace transform to solve the differential equationy

′′−2y′−3y=0y″−2y′−3y=0

with the initial conditions y(0)=2y(0)=2 and y′(0)=−1y′(0)=−1 and y is a function of time t.

Answer 1

Solution

Let L[y(t)] = Y(s)

Then Laplace transform of the equation is

s²Y(s)-sy(0)-y’(0)-2[sY(s)-y(0)]-3Y(s)=0

(s²-2s-3)Y(s) = sy(0)+y’(0)-2y(0)

Using the initial conditions (s²-2s-3)Y(s) = 2s-5 => Y(s) = (2s-5)/(s²-2s-3) = (2s-5)/[(s-3)(s+1)] 

(2s-5)/[(s-3)(s+1)] = A/(s-3)+B/(s+1), where A,B are arbitrary constants

From this equality

2s-5 = A(s+1)+B(s-3)  => 2s-5 = (A+B)s+A-3B => A+B=2, A-3B=-5 => 4B=7, A+B=2  =>  B=7/4, A=1/4

So Y(s) = (1/4)/(s-3)+(7/4)/(s+1)

From the Table of Laplace Transforms   1/(s-a)<->e^-at  

Therefore y(t) = e^3t /4+7 e^-t/4

Answer

y(t) = e^3t /4+7 e^-t/4