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

The solution for y^{''}+4y^'+3y=5e^2x is y=c_{1}e^{-x}+c_{2}e^{-3x}+(5e^2x)/3-(20e^2)/9

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

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

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{5 e ^{2} x}{3} - \frac{20 e ^{2}}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{5 e ^{2} x}{3} - \frac{20 e ^{2}}{9}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{5 e ^{2} x}{3} - \frac{20 e ^{2}}{9}

y^{^{''}} + 13 y = 7 e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = - 19 t e^{4 t}

2 y^{^{''}} - y^{^{'}} = 7

y^{^{''}} + y^{^{'}} - 6 y = x^{2} - x + e^{2 x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 e^{- 5 x}

What is the solution for y^{''}+4y^'+3y=5e^2x ?
The solution for y^{''}+4y^'+3y=5e^2x is y=c_{1}e^{-x}+c_{2}e^{-3x}+(5e^2x)/3-(20e^2)/9