What is the solution for y^{''}-2y^'=8sin(x)+2xe^x ?

The solution for y^{''}-2y^'=8sin(x)+2xe^x is y=c_{1}e^{2x}+c_{2}-8/5 sin(x)+16/5 cos(x)-2e^xx

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y^{''}-2y^'=8sin(x)+2xe^x

y^{^{''}} - 2 y^{^{'}} = 8 sin (x) + 2 x e^{x}

y = c_{1} e^{2 x} + c_{2} - \frac{8}{5} sin (x) + \frac{16}{5} cos (x) - 2 e^{x} x

Solution steps

y^{''} (x) - 2 y^{'} (x) = 8 sin (x) + 2 x e^{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - \frac{8}{5} sin (x) + \frac{16}{5} cos (x) - 2 e^{x} x

y = c_{1} e^{2 x} + c_{2} - \frac{8}{5} sin (x) + \frac{16}{5} cos (x) - 2 e^{x} x

\int (x + x^{2}) d x

\frac{d}{d x} (4 - 5 sin (2 x + \frac{π}{4}))

f (x) = 2 x 1 - 3 x, at x = - 2

\int \frac{1}{x ^{2} - 3 x - 4} d x

\int (x + 1)^{\frac{3}{2}} d x

What is the solution for y^{''}-2y^'=8sin(x)+2xe^x ?
The solution for y^{''}-2y^'=8sin(x)+2xe^x is y=c_{1}e^{2x}+c_{2}-8/5 sin(x)+16/5 cos(x)-2e^xx