What is the solution for y^{''}-y^'-6y=36x,y(0)=4,y^'(0)=-17 ?

The solution for y^{''}-y^'-6y=36x,y(0)=4,y^'(0)=-17 is y=-e^{3x}+4e^{-2x}-6x+1

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y^{''}-y^'-6y=36x,y(0)=4,y^'(0)=-17

y^{^{''}} - y^{^{'}} - 6 y = 36 x, y (0) = 4, y^{^{'}} (0) = - 17

y = - e^{3 x} + 4 e^{- 2 x} - 6 x + 1

Solution steps

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

Solve linear ODE:

y = - e^{3 x} + 4 e^{- 2 x} - 6 x + 1

y = - e^{3 x} + 4 e^{- 2 x} - 6 x + 1

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

y^{^{''}} - y^{^{'}} = x^{2} + 2 x + 3

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 32 y = 6 + 3 x

y^{^{''}} + a y^{^{'}} - 2 a^{2} y = 1

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

What is the solution for y^{''}-y^'-6y=36x,y(0)=4,y^'(0)=-17 ?
The solution for y^{''}-y^'-6y=36x,y(0)=4,y^'(0)=-17 is y=-e^{3x}+4e^{-2x}-6x+1