What is the solution for y^{''}+y^'-6y=36x ?

The solution for y^{''}+y^'-6y=36x is y=c_{1}e^{2x}+c_{2}e^{-3x}-6x-1

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y^{''}+y^'-6y=36x

y^{^{''}} + y^{^{'}} - 6 y = 36 x

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - 6 x - 1

Solution steps

y^{''} (x) + y^{'} (x) - 6 y = 36 x

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - 6 x - 1

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - 6 x - 1

y^{^{''}} + 2 y^{^{'}} + 5 y = 20 e^{- t} cos (2 t)

\frac{d ^{2} y}{d x ^{2}} + y = 3 cos (x) - 2 sin (x)

y^{^{''}} - 4 y^{^{'}} + 5 y = x e^{x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{- 2 t}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x}, y^{^{'}} (0) = - 1, y (0) = 1

What is the solution for y^{''}+y^'-6y=36x ?
The solution for y^{''}+y^'-6y=36x is y=c_{1}e^{2x}+c_{2}e^{-3x}-6x-1