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

y"-6y'+9y=0:y(0)=2,y'(0)=253

Short Answer

Expert verified

The solution is y(t)=2e3t+73te3t.

Step by step solution

01

Find the solution of the differential equation.

The given differential equation is y" -6y' + 9y = 0.

The auxiliary equation is;

r2-6r1+9r0=0r2-6r+9=0r2-3r-3r+9=0r(r-3)-3(r-3)=0(r-3)2=0 r=3,3

Therefore, the solution is y(t) = c1e3t + c2te3t.

02

Apply initial conditions.

The initial conditions are y(0)=2,y'(0)=253.

Therefore,

y(0)=c1e0+c2(0)e0c1=2

And

y'(t)=3c1e3t+c2(3te3t+e3t)y'(0)=3c1e0+c2(0+e0)3c1+c2=253

Solving for c1,c2 then;

c1=2c2=73

Therefore, the solution is y(t)=2e3t+73te3t.

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