Assuming the two numbers $r_1$ and $r_2$ are different, solve the IVP

$L[y]$:$=y^{\text{'}\text{'}}-(r_1+r_2)y^{\text{'}}+r_1{r}_{2}y=g(t),y(0)=0=y^{\text{'}}(0)$

In particular make sure and represent the solution ultimately as the quadrature

$y(t)={\displaystyle \underset{0}{\overset{t}{}}}dsG(t,s)g(s)={\displaystyle \underset{0}{\overset{t}{}}}dsG(t-s)g(s)$

where

$G(t,s)=G(t-s)$

is the famed "Green's Function" for the linear operator

$L[y]$:$=y^{\text{'}\text{'}}-(r_1+r_2)y^{\text{'}}+r_1{r}_{2}y$

Hints: Variation of parameters will work, whence using some of the results from the previous problem

will be useful. But to get the final result in the required form $(17)$ you will have to carry through the

usual variation of parameters recipe further than, say, the book did, in which they represented a

particular solution of the nonhomogeneous equation as sort of sum of two integrals. Instead, in order to

get your solution in the form $(17),$ you will evidently have to combine any linear combination of integrals

into just one integral.

Another hint: In the case that the ODE in $(16)$ is actually

$L[y]$:$=y^{\text{'}\text{'}}-(r_1+r_2)y^{\text{'}}+r_1{r}_{2}y=y^{\text{'}\text{'}}-(i-i)y^{\text{'}}+i(-i)y$

$=y^{\text{'}\text{'}}+{}^{2}y=g(t)$

then the solution to the IVP with the indicated quiescent data is $(17)$ with

$G(t)$:$=\frac{2isin(t)}{2i}=\frac{sin(t)}{}=G_{}(t)$

Importantly$-$way too much information$-$this Green's function for the linear operator

$L[y]$:$=y^{\text{'}\text{'}}+{}^{2}y=L_{}[y]$

satisfies itself the IVP

$L_{}[G_{}]=0,G_{}(0)=0,G_{}^{\text{'}}(0)=1$