What is the solution for y^{''}+2y=-18xe^{2x} ?

The solution for y^{''}+2y=-18xe^{2x} is y=c_{1}cos(sqrt(2)x)+c_{2}sin(sqrt(2)x)-3xe^{2x}+2e^{2x}

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y^{''}+2y=-18xe^{2x}

y^{^{''}} + 2 y = - 18 x e^{2 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 3 x e^{2 x} + 2 e^{2 x}

Solution steps

y^{''} (x) + 2 y = - 18 x e^{2 x}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 3 x e^{2 x} + 2 e^{2 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 3 x e^{2 x} + 2 e^{2 x}

y^{^{''}} - y^{^{'}} - 6 y = 2 cos (x) - 2 e^{3 x} + 10

y^{^{''}} + y = sin (4 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = x e^{- 2 x}

\frac{d ^{2} x}{d t ^{2}} + 7 \frac{d x}{d t} + 6 x = 5 e^{- t}

y^{^{''}} + 2 y^{^{'}} + 5 = 2 cos (t + 0.523599)

What is the solution for y^{''}+2y=-18xe^{2x} ?
The solution for y^{''}+2y=-18xe^{2x} is y=c_{1}cos(sqrt(2)x)+c_{2}sin(sqrt(2)x)-3xe^{2x}+2e^{2x}