Question

Solve the initial value problem
y"(t)+ 2y'(t)+y(t)=3e^(-t) ; y'(0)=2 y(0)=4

Answer 1

$y''+2y'+y=3e^{-t}\\ y(0)=4, y'(0)=2\\$

Solution of equation find in form

$y=y_1+y_2$

Consider equation

$y''+2y'+y=0$

Write characteristic equation

$\lambda^2+2\lambda+1=0 \to \lambda_1=\lambda_2=-1$

then the solution of equation is

$y_1=c_1e^{-t}+c_2te^{-t}$

The solution $y_2$ find in form

$y_2=at^2e^{-t}$

input $y_2$ in the first equation

$2ae^{-t}-2tae^{-t} -2tae^{-t} +t^2ae^{-t}+\\ +4tae^{-t}-2t^2ae^{-t}+t^2ae^{-t}=3e^{-t}\\ 2a=3 , a=1.5$

so

$y_2=1.5t^2e^{-t}$

and

$y=c_1e^{-t}+c_2te^{-t}+1.5t^2e^{-t}$

Find

$y'=-c_1e^{-t}+c_2e^{-t}-c_2te^{-t}+\\ +2t\cdot 1.5e^{-t}-1.5t^2e^{-t}\\ y(0)=c_1 \to c_1=4\\ y'(0)=-c_1+c_2 \to -c_1+c_2=2 \to\\ -4+c_2=2 \to c_2=6$

So

$y=4e^{-t}+6te^{-t}+1.5t^2e^{-t}$