Answer:
[tex]y(x)=-\frac{3}{4}-\frac{1}{2}x-\frac{1}{2}x^2+\frac{2}{3}e^{-2x}+\frac{1}{12}e^x[/tex]
Step-by-step explanation:
[tex]y''+y'-2y=x^2,\: y(0)=0,\: y'(0)=0\\\\\mathcal{L}\{y''\}+\mathcal{L}\{y'\}-2\mathcal{L}\{y\}=\mathcal{L}\{x^2\}\\\\s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0)-2Y(s)=\frac{2}{s^3}\\ \\s^2Y(s)+sY(s)-2Y(s)=\frac{2}{s^3}\\ \\(s^2+s-2)Y(s)=\frac{2}{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}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s+2}+\frac{E}{s-1}\\\\2=s^2(s+2)(s-1)A+s(s+2)(s-1)B+(s+2)(s-1)C+s^3(s-1)D+s^3(s+2)E\\\\2=s^{4} A + s^{4} D + s^{4} E + s^{3} A + s^{3} B + 2 s^{3} D - s^{3} E - 2 s^{2} A + s^{2} B + s^{2} C - 2 s B + s C - 2 C\\\\2=s^{4} \left(A + D + E\right) + s^{3} \left(A + B + 2 D - E\right) + s^{2} \left(- 2 A + B + C\right) + s \left(- 2 B + C\right) - 2 C[/tex]
Solve for each constant
[tex]\begin{cases} A + D + E = 0\\A + B + 2 D - E = 0\\- 2 A + B + C = 0\\- 2 B + C = 0\\- 2 C = 2 \end{cases}[/tex]
[tex]-2C=2\\C=-1[/tex]
[tex]-2B+C=0\\-2B+(-1)=0\\-2B-1=0\\-2B=1\\B=-\frac{1}{2}[/tex]
[tex]-2A+B+C=0\\-2A+(-\frac{1}{2})+(-1)=0\\-2A-\frac{1}{2}-1=0\\-2A-\frac{3}{2}=0\\-2A=\frac{3}{2}\\ A=-\frac{3}{4}[/tex]
[tex]A+D+E=0\\-\frac{3}{4}+D+E=0\\D+E=\frac{3}{4}\\D=\frac{3}{4}-E[/tex]
[tex]A+B+2D-E=0\\-\frac{3}{4}+(-\frac{1}{2})+2(\frac{3}{4}-E)-E=0\\-\frac{3}{4}-\frac{1}{2}+\frac{3}{2}-2E-E=0\\-\frac{3}{4}+1-3E=0\\\frac{1}{4}-3E=0\\\frac{1}{4}=3E\\\frac{1}{12}=E[/tex]
[tex]D=\frac{3}{4}-E\\D=\frac{3}{4}-\frac{1}{12}\\D=\frac{9}{12}-\frac{1}{12}\\D=\frac{8}{12}\\D=\frac{2}{3}[/tex]
Take the inverse transform and find the solution to the IVP
[tex]Y(s)=\frac{-\frac{3}{4}}{s}+\frac{-\frac{1}{2}}{s^2}+\frac{-1}{s^3}+\frac{\frac{2}{3}}{s+2}+\frac{\frac{1}{12}}{s-1}\\ \\y(x)=-\frac{3}{4}-\frac{1}{2}x-\frac{1}{2}x^2+\frac{2}{3}e^{-2x}+\frac{1}{12}e^x[/tex]
