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

y" - 4y' + 4y = 0 : y(1) = 1, y'(1) =1

Short Answer

Expert verified

The solution is y(t) = 2e2t-2- te3t+2.

Step by step solution

01

Find the solution of the differential equation.

The given differential equation is y" - 4y' + 4y = 0.

The auxiliary equation is r2- 4r + 4 = 0

Find the roots of the auxiliary equation;

r2-4r+4=0r=4±(-4)2-4(1)(4)2(1)r=4±16-162r=2,2

Therefore the solution is y(t) = c1e2t + c2te2t.

02

Apply initial conditions.

The initial conditions are y(1) = 1, y'(t) = 1.

Therefore,

y(1)=c1e2+c2(1)e2c1e2+c2(1)e2=1c1+c2=e-2

And

y'(t)=2c1e2t+c2(2te2t+e2t)y'(1)=2c1e2+c2(2(1)e2+e2)2c1e2+c2(3e2)=12c1+3c2=e-2

Solving for c1,c2 then;

c1 = 2e-2

c2 = -e2

Therefore, the solution is y(t)=2e-2e2t-e2te3t⇒y(t)=2e2t-2-te3t+2.

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