What is the solution for y^{''}-2y^'-3y=-5xe^{4x} ?

The solution for y^{''}-2y^'-3y=-5xe^{4x} is y=c_{1}e^{3x}+c_{2}e^{-x}-e^{4x}x+6/5 e^{4x}

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y^{''}-2y^'-3y=-5xe^{4x}

y^{^{''}} - 2 y^{^{'}} - 3 y = - 5 x e^{4 x}

y = c_{1} e^{3 x} + c_{2} e^{- x} - e^{4 x} x + \frac{6}{5} e^{4 x}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = - 5 x e^{4 x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - e^{4 x} x + \frac{6}{5} e^{4 x}

y = c_{1} e^{3 x} + c_{2} e^{- x} - e^{4 x} x + \frac{6}{5} e^{4 x}

f (x) = \frac{3}{x + 5}

\int (\frac{x}{x ^{2} + y ^{2}}) d y

x^{- 2}, 8

What is the solution for y^{''}-2y^'-3y=-5xe^{4x} ?
The solution for y^{''}-2y^'-3y=-5xe^{4x} is y=c_{1}e^{3x}+c_{2}e^{-x}-e^{4x}x+6/5 e^{4x}