Answer:
[tex]c_1e^{3x}+c_2e^{-x}+\frac{1}{2} xe^{3x}[/tex]
Step-by-step explanation:
For homogeneous solution:
[tex]y''-2y'-3=0\\r^2-2r-3=0\\(r-3)(r+1)=0[/tex]
So since roots are r = 3 and r = -1, [tex]y_h=c_1e^{3x}+c_2e^{-x}[/tex]
Since we are given [tex]2e^{3x}[/tex], we will use undetermined coefficients. However, here the trick is we have [tex]c_1e^{3x}[/tex] in homogeneous solution. So in particular solution, as undetermined coefficients, we will use [tex]Axe^{3x}[/tex] instead of [tex]Ae^{3x}[/tex].
Hence,
[tex]Y = Axe^{3x}\\Y' = Ae^{3x} + 3Axe^{3x}\\Y'' = 3Ae^{3x} + 3Ae^{3x} + 9Axe^{3x} = 6Ae^{3x} + 9Axe^{3x}[/tex]
So,
[tex]6Ae^{3x}+9Axe^{3x}-2Ae^{3x}-6Axe^{3x}-3Axe^{3x}=2e^{3x}\\4Ae^{3x}=2e^{3x}\\A=\frac{1}{2}[/tex]
Hence, [tex]y_p = \frac{1}{2} xe^{3x}[/tex]
General solution is:
[tex]y=y_h+y_p=c_1e^{3x}+c_2e^{-x}+\frac{1}{2} xe^{3x}[/tex]
