Find a second-order lincar equation for which

Find a second-order lincar equation for which
$y\left(x\right)={c}_{1}{e}^{3x}+{c}_{2}{e}^{5x}+\mathrm{sin}\left(2x\right)$
is the general solution
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The homogenous part of the linear equation is, ${c}_{1}{e}^{3x}+{c}_{2}{e}^{5x}$
The particular part of the linear equation is $\mathrm{sin}\left(2x\right)$
From the homogenous part, it can be concluded that the roots of the characteristic equation will be 3 and 5.
Find the characteristic equation.
$\left(m-3\right)\left(m-5\right)=0$
${m}^{2}-8m+15=0$
Assume f(x) to be the right-hand side of the differential equation. The differential equation will be, $y{}^{″}-8{y}^{\prime }+15y=f\left(x\right)$
Substitute $y=\mathrm{sin}2x$ which is the particular solution into the differential equation obtained.
$\frac{{d}^{2}\mathrm{sin}\left(2x\right)}{{dx}^{2}}-8\frac{d\mathrm{sin}\left(2x\right)}{dx}+15\mathrm{sin}\left(2x\right)=f\left(x\right)$
$-4\mathrm{sin}\left(2x\right)-16\mathrm{cos}\left(2x\right)+15\mathrm{sin}\left(2x\right)=f\left(x\right)$
$-16\mathrm{cos}\left(2x\right)+11\mathrm{sin}\left(2x\right)=f\left(x\right)$
Thus, the differential equation will be,
$y{}^{″}-8{y}^{\prime }+15y=11\mathrm{sin}\left(2x\right)-16\mathrm{cos}\left(2x\right)$
Answer: $y{}^{″}-8{y}^{\prime }+15y=11\mathrm{sin}\left(2x\right)-16\mathrm{cos}\left(2x\right)$
Jeffrey Jordon