Answer:
[tex]y(x)=-2x-\frac{11}{6}e^{-x}+\frac{3}{2}e^x+\frac{1}{3}e^{-4x}[/tex]
Step-by-step explanation:
[tex]y''-y=5e^{-4x}+2x,\: y(0)=y'(0)=0\\\\\mathcal{L}\{y''\}-\mathcal{L}\{y\}=\mathcal{L}\{5e^{-4x}\}+\mathcal{L}\{2x\}\\\\s^2Y(s)-sy(0)-y'(0)-Y(s)=\frac{5}{s+4}+\frac{2}{s^2}\\ \\s^2Y(s)-Y(s)=\frac{5}{s+4}+\frac{2}{s^2}\\ \\(s^2-1)Y(s)=\frac{5}{s+4}+\frac{2}{s^2}\\ \\Y(s)=\frac{5}{(s+4)(s^2-1)}+\frac{2}{s^2(s^2-1)}\\ \\Y(s)=\frac{5s^2+2(s+4)}{s^2(s+4)(s^2-1)}\\ \\Y(s)=\frac{5s^2+2s+8}{s^2(s-1)(s+1)(s+4)}[/tex]
Perform the partial fraction decomposition
[tex]\frac{5 s^{2} + 2 s + 8}{s^{2} \left(s - 1\right) \left(s + 1\right) \left(s + 4\right)}=\frac{A}{s}+\frac{B}{s^{2}}+\frac{C}{s + 1}+\frac{D}{s - 1}+\frac{E}{s + 4}\\\\5 s^{2} + 2 s + 8=s^{2} \left(s - 1\right) \left(s + 1\right) E + s^{2} \left(s - 1\right) \left(s + 4\right) C + s^{2} \left(s + 1\right) \left(s + 4\right) D + s \left(s - 1\right) \left(s + 1\right) \left(s + 4\right) A + \left(s - 1\right) \left(s + 1\right) \left(s + 4\right) B[/tex]
[tex]5 s^{2} + 2 s + 8=s^{4} A + s^{4} C + s^{4} D + s^{4} E + 4 s^{3} A + s^{3} B + 3 s^{3} C + 5 s^{3} D - s^{2} A + 4 s^{2} B - 4 s^{2} C + 4 s^{2} D - s^{2} E - 4 s A - s B - 4 B\\\\5 s^{2} + 2 s + 8=s^{4} \left(A + C + D + E\right) + s^{3} \left(4 A + B + 3 C + 5 D\right) + s^{2} \left(- A + 4 B - 4 C + 4 D - E\right) + s \left(- 4 A - B\right) - 4 B[/tex]
Solve for each constant
[tex]\begin{cases} A + C + D + E = 0\\4 A + B + 3 C + 5 D = 0\\- A + 4 B - 4 C + 4 D - E = 5\\- 4 A - B = 2\\- 4 B = 8 \end{cases}[/tex]
[tex]-4B=8\\B=-2[/tex]
[tex]-4A-B=2\\-4A-(-2)=2\\-4A+2=2\\-4A=0\\A=0[/tex]
[tex]A+C+D+E=0\\C+D+E=0\\E=-C-D[/tex]
[tex]-A+4B-4C+4D-E=5\\4(-2)-4C+4D-(-C-D)=5\\-8-4C+4D+C+D=5\\-3C+5D=13\\5D=13+3C[/tex]
[tex]4A+B+3C+5D=0\\4(0)+(-2)+3C+13+3C=0\\-2+6C+13=0\\11+6C=0\\6C=-11\\C=-\frac{11}{6}[/tex]
[tex]5D=13+3C\\5D=13+3(-\frac{11}{6})\\5D=13-\frac{33}{6}\\5D=\frac{15}{2}\\D=\frac{15}{10}\\D=\frac{3}{2}[/tex]
[tex]E=-C-D\\E=-(-\frac{11}{6})-(\frac{3}{2})\\E=\frac{11}{6}-\frac{3}{2}\\E=\frac{2}{6}\\E=\frac{1}{3}[/tex]
Take the inverse transform and solve for the IVP
[tex]Y(s)=\frac{0}{s}+\frac{-2}{s^2}+\frac{-\frac{11}{6}}{s+1}+\frac{\frac{3}{2}}{s-1}+\frac{\frac{1}{3}}{s+4}\\ \\y(x)=-2x-\frac{11}{6}e^{-x}+\frac{3}{2}e^x+\frac{1}{3}e^{-4x}[/tex]
