Find a second-order lincar equation for which

The homogenous part of the linear equation is, ${\displaystyle c_1{e}^{3x}+c_2{e}^{5x}}$
The particular part of the linear equation is ${\displaystyle \sin \left(2x\right)}$
From the homogenous part, it can be concluded that the roots of the characteristic equation will be 3 and 5.
Find the characteristic equation.
${\displaystyle (m-3)(m-5)=0}$
${\displaystyle m^2-8m+15=0}$
Assume f(x) to be the right-hand side of the differential equation. The differential equation will be, ${\displaystyle y{}^{″}-8y^{\prime }+15y=f\left(x\right)}$
Substitute ${\displaystyle y=\sin 2x}$ which is the particular solution into the differential equation obtained.
${\displaystyle \frac{d^2\sin \left(2x\right)}{{dx}^2}-8\frac{d\sin \left(2x\right)}{dx}+15\sin \left(2x\right)=f\left(x\right)}$
${\displaystyle -4\sin \left(2x\right)-16\cos \left(2x\right)+15\sin \left(2x\right)=f\left(x\right)}$
${\displaystyle -16\cos \left(2x\right)+11\sin \left(2x\right)=f\left(x\right)}$
Thus, the differential equation will be,
${\displaystyle y{}^{″}-8y^{\prime }+15y=11\sin \left(2x\right)-16\cos \left(2x\right)}$
Answer: ${\displaystyle y{}^{″}-8y^{\prime }+15y=11\sin \left(2x\right)-16\cos \left(2x\right)}$