Find the solution to the initial value problem y 4y t2 3e

Find the solution to the initial value problem y\" + 4y = t^2 + 3e^t, y(0) = 0, y\'(0) = 2.

Solution

First we solve the homogeneous ode

y\'\'+4y=0

y\'\'=-2^2y

y=A sin(2t)+B cos(2t)

Now we look for a particular solutoin using method of undetermined coefficients.

We guess the particular solution based on the inhomogenous part

Guess is

yp=Ct^2+Dt+E+Fe^t

Substituting gives

2C+Fe^t+4(Ct^2+Dt+E+Fe^t)=t^2+3e^t

2C+4E+5Fe^t+4Dt+4Ct^2=t^2+3e^t

Comparing coefficients gives

D=0

2C+4E=0 ie C=-2E

4C=1 ie C=1/4

E=-1/8

5F=3

F=3/5

yp=t^2/4-1/8+3e^t/5

Hence,

y=A sin(2t)+B cos(2t)+t^2/4-1/8+3e^t/5

y(0)=0=B-1/8+3 ie B=-23/8

y\'(t)=2A cos(2t)-2B sin(2t)+t/2+3e^t/5

y\'(0)=2A+3=0 ie A=-3/2

y=-3 sin(2t)/2-23 cos(2t)/8+t^2/4-1/8+3e^t/5