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Chapter 4: Q19E (page 164)

In Problems 13鈥�20, solve the given initial value problem.

y" + 2y' + y = 0 : y(0) = 1, y'(0) = -3

Short Answer

Expert verified

The solution is y(t) = e(-t) -2e(-t).

Step by step solution

01

Find the solution of the differential equation.

The given differential equation is y" + 2y' + y = 0.

The auxiliary equation is r2 + 2r + 1 = 0

r2+2r+1=0r=-2卤22-4(1)(1)2(1)r=-2卤4-42r=-1,-1

Therefore the solution is y(t) = c1e-t + c2te-t.

02

Apply initial conditions.

The initial conditions are y(0) = 1, y'(0) =-3.

Therefore,

y(0)=ce0_c2(0)e0c1=1

And

y'(t)=-c1e(-t)+c2-te(-1)+e-1y'(0)=-c1e0+c2(0)e0+e0-c1+c2=-3

Solving for c1,c2 then;

c1=1c2=-2

Therefore, the solution is y(t) = e(-t)- 2e(-1).

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