What is the solution for y^{''}-4y^'=1-2e^{3x} ?

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

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

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

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

Solution steps

y^{''} (x) - 4 y^{'} (x) = 1 - 2 e^{3 x}

Solve linear ODE:

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

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

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

y^{^{''}} + y^{^{'}} + y = x + cos (x)

2 y^{^{''}} - 5 y^{^{'}} + 2 y = 5 x^{4} cosh (x)

y^{^{''''}} - y = 14 t + cos (t)

y^{^{''}} + y = - x^{- 2} sin (x) + 2 x^{- 1} cos (x)

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