Answer:
[tex]y(x)=\frac{1}{2}e^{-3x}-\frac{2}{3}e^{-2x}+\frac{1}{6}e^{x}[/tex]
Step-by-step explanation:
[tex]y''+y'-2y=e^{-3x},\: y(0)=0,\: y'(0)=0\\\\\mathcal{L}\{y''\}+\mathcal{L}\{y'\}-2\mathcal{L}\{y\}=\mathcal{L}\{e^{-3x}\}\\\\s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0)-2Y(s)=\frac{1}{s+3} \\ \\s^2Y(s)+sY(s)-2Y(s)=\frac{1}{s+3}\\ \\(s^2+s-2)Y(s)=\frac{1}{s+3}\\ \\Y(s)=\frac{2}{(s+3)(s^2+s-2)}\\ \\Y(s)=\frac{2}{(s+3)(s+2)(s-1)}[/tex]
Perform the partial fraction decomposition
[tex]\frac{2}{(s+3)(s+2)(s-1)}=\frac{A}{s+3}+\frac{B}{s+2}+\frac{C}{s-1}\\ \\2=(s+2)(s-1)A+(s+3)(s-1)B+(s+2)(s+3)C[/tex]
Solve for each constant
[tex]2=((-2)+2)((-2)-1)A+((-2)+3)((-2)-1)B+((-2)+2)((-2)+3)C\\\\2=-3B\\\\-\frac{2}{3}=B[/tex]
[tex]2=(1+2)(1-1)A+(1+3)(1-1)B+(1+2)(1+3)C\\ \\2=12C\\\\\frac{2}{12}=C\\ \\\frac{1}{6}=C[/tex]
[tex]2=((-3)+2)((-3)-1)A+((-3)+3)((-3)-1)B+((-3)+2)((-3)+3)C\\ \\2=4A\\\\\frac{2}{4}=A\\ \\\frac{1}{2}=A[/tex]
Take the inverse transform and solve for the IVP
[tex]Y(s)=\frac{\frac{1}{2}}{s+3}+\frac{-\frac{2}{3}}{s+2}+\frac{\frac{1}{6}}{s-1}\\\\y(x)=\frac{1}{2}e^{-3x}-\frac{2}{3}e^{-2x}+\frac{1}{6}e^{x}[/tex]
