Step 1

The differential equation is \(y'-3y=5e^{3x}\)

Using the method of undetermined coefficient, the solution of the equation will be of the form \(y=y_{h}+y_{p}\) where \(y_{h}\) is the general solution to the corresponding homogeneous equation and \(y_{p}\) is the particular solution.

Find the general solution to \(y'−3y=0\).

The characteristic equation is \(r−3=0\).

\(r-3=0\)

\(r=3\)

Thus, the general solution is, \(y_{h}=c_{1}e^{3x}\).

Step 2

To find the particular solution, suppose \(y_{p}=Axe^{3x}\).

\(y_{p}'=3Axe^{3x}-3Axe^{3x}=5e^{3x}\)

\(Ae^{3x}=5e^{3x}\)

\(A=5\)

Thus, the particular solution is \(y_{p}=5xe^{3x}\).

Therefore, the solution of the equation is,

\(y=y_{h}+y_{p}\)

\(=c_{1}e^{3x}+5xe^{3x}\)

That is, \(y = c_{1}e^{3x}+5xe^{3x}\).