What is the solution for y^{''}-y^'-2y=3x+4 ?

The solution for y^{''}-y^'-2y=3x+4 is y=c_{1}e^{2x}+c_{2}e^{-x}-(3x)/2-5/4

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y^{''}-y^'-2y=3x+4

y^{^{''}} - y^{^{'}} - 2 y = 3 x + 4

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3 x}{2} - \frac{5}{4}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 3 x + 4

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3 x}{2} - \frac{5}{4}

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3 x}{2} - \frac{5}{4}

y^{^{''}} + y = g (t), y (0) = 0, y^{^{'}} (0) = 4

y^{^{''}} + 4 y = \frac{e ^{2 x}}{x}

y^{^{''}} - y^{^{'}} - 2 y = e^{3 t}

y^{^{''}} - y = 3 x^{2} - 1

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

What is the solution for y^{''}-y^'-2y=3x+4 ?
The solution for y^{''}-y^'-2y=3x+4 is y=c_{1}e^{2x}+c_{2}e^{-x}-(3x)/2-5/4