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

The solution for y^{''}-2y^'-3y=e^4x is y=c_{1}e^{3x}+c_{2}e^{-x}-(e^4x)/3+(2e^4)/9

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

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

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{e ^{4} x}{3} + \frac{2 e ^{4}}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{e ^{4} x}{3} + \frac{2 e ^{4}}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{e ^{4} x}{3} + \frac{2 e ^{4}}{9}

What is the solution for y^{''}-2y^'-3y=e^4x ?
The solution for y^{''}-2y^'-3y=e^4x is y=c_{1}e^{3x}+c_{2}e^{-x}-(e^4x)/3+(2e^4)/9