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

The solution for y^{''}-y^'-2y=xe^{4x} is y=c_{1}e^{2x}+c_{2}e^{-x}+(e^{4x}x)/(10)-7/100 e^{4x}

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{e ^{4 x} x}{10} - \frac{7}{100} e^{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{e ^{4 x} x}{10} - \frac{7}{100} e^{4 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{e ^{4 x} x}{10} - \frac{7}{100} e^{4 x}

q^{^{''}} + 100 q^{^{'}} + 2500 q = 40, q (0) = 0

(D^{2} + 5) y = cos (5) x

(D^{2} + 4 D + 4) y = 5 e^{3 x}

3 y^{^{'}} + 2 y^{^{''}} = - 1000 cos (3.14 \cdot \frac{t}{0.2})

y^{^{''}} - \frac{1}{2} y^{^{'}} - 1 = 0, y (0) = 100, y (3) = 10

What is the solution for y^{''}-y^'-2y=xe^{4x} ?
The solution for y^{''}-y^'-2y=xe^{4x} is y=c_{1}e^{2x}+c_{2}e^{-x}+(e^{4x}x)/(10)-7/100 e^{4x}