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}+1/3 e^{2x}

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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{1}{3} e^{2 x}

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{1}{3} e^{2 x}

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

y^{^{''}} - 12 y^{^{'}} + 35 y = - 40 e^{- 3 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} ln (1 + x^{2})

y^{^{''}} + y^{^{'}} - 6 y = e^{- 3 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e^{x} - 9

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin (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}+1/3 e^{2x}