What is the solution for y^{''}-2y^'+2y=e^2x(cos(x)-9sin(x)) ?

The solution for y^{''}-2y^'+2y=e^2x(cos(x)-9sin(x)) is

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y^{''}-2y^'+2y=e^2x(cos(x)-9sin(x))

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 9 sin (x))

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{11 e ^{2} x sin ( x )}{5} - \frac{17 e ^{2} x cos ( x )}{5} + (- 18 e^{2} + \frac{418 e ^{2}}{25}) sin (x) - \frac{124 e ^{2}}{25} cos (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = e^{2} x (cos (x) - 9 sin (x))

Solve linear ODE:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{11 e ^{2} x sin ( x )}{5} - \frac{17 e ^{2} x cos ( x )}{5} + (- 18 e^{2} + \frac{418 e ^{2}}{25}) sin (x) - \frac{124 e ^{2}}{25} cos (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{11 e ^{2} x sin ( x )}{5} - \frac{17 e ^{2} x cos ( x )}{5} + (- 18 e^{2} + \frac{418 e ^{2}}{25}) sin (x) - \frac{124 e ^{2}}{25} cos (x)

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} + y = 5 e^{t}

y^{^{''}} - x e^{- x} = 2 y^{^{'}}

y^{^{''}} + 4 y = 3, 2 sin (4 t) y (0) = 1

y^{^{''}} - y^{^{'}} - 6 y = x e^{x}

x^{^{''}} + 4 x = - t^{2}

What is the solution for y^{''}-2y^'+2y=e^2x(cos(x)-9sin(x)) ?
The solution for y^{''}-2y^'+2y=e^2x(cos(x)-9sin(x)) is