Question

Solve the initial value problem (IVP):

(D^2+4D+4)=9e^x

y(0)=2

y'(0)=1

Answer 1

$D^2+4D+4=9e^x$

Where $D$ is the linear differential operator $\frac{d}{dx}$

Thus we can write the equation as:

$\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=9e^x$

The associated auxilliary equation is;

$m^2+4m+4=0\\ (m+2)(m+2)\\ m=-2 \text{ twice}$

Thus the solution of the homogeneous part is

$y_c=(A+Bx)e^{-2x}$

For the particular solution

Consider the solution $y=Ae^x$

$\implies y'=Ae^x,y''=Ae^x$

Substitute this into the equation.

$Ae^x+4Ae^x+4Ae^x=9e^x\\ 9Ae^x=9e^x\\ A=1$

Thus the particular solution is;

$y_p=e^x$

Hence, the general solution is

$y=(A+Bx)e^{-2x}+e^x$

Now, consider the initial value given. $y(0)=2,y'(0)=1$

$y(0)=A+1=2\\A=1$

$y'(x)=Be^{-2x}-2(A+Bx)e^{-2x}+e^x\\ y'(0)=B-2A+1=1\\ B=2A=2$

Hence the solution to the IVP is;

$y=(1+2x)e^{-2x}+e^x$