What is the solution for y^{''}-y^'-2y=15x+8 ?

The solution for y^{''}-y^'-2y=15x+8 is y=c_{1}e^{2x}+c_{2}e^{-x}-(15x)/2-1/4

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y^{''}-y^'-2y=15x+8

y^{^{''}} - y^{^{'}} - 2 y = 15 x + 8

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{15 x}{2} - \frac{1}{4}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 15 x + 8

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{15 x}{2} - \frac{1}{4}

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{15 x}{2} - \frac{1}{4}

y^{^{''}} + 6 y^{^{'}} + 25 y = 15 cos (5 t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{6} e^{2 t}

y^{^{''}} - 2 y^{^{'}} + y = t^{- 7} e^{t}

y^{^{''}} - y - 2 = 0, y (0) = 3

y^{^{''}} - 6 y^{^{'}} + 8 y = 2

What is the solution for y^{''}-y^'-2y=15x+8 ?
The solution for y^{''}-y^'-2y=15x+8 is y=c_{1}e^{2x}+c_{2}e^{-x}-(15x)/2-1/4