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

The solution for y^{''}+4y^'-5y=e^{3x} is y=c_{1}e^x+c_{2}e^{-5x}+1/16 e^{3x}

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

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

y = c_{1} e^{x} + c_{2} e^{- 5 x} + \frac{1}{16} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 5 x} + \frac{1}{16} e^{3 x}

y = c_{1} e^{x} + c_{2} e^{- 5 x} + \frac{1}{16} e^{3 x}

\frac{d}{d x} (\frac{x ^{9}}{45})

y = 6 x, y = 6 x^{2}

\frac{( - 6 x )}{8 - 2 x}

y^{^{''}} + y = x, y (1) = 0, y^{^{'}} (1) = 1

\int 3 x^{2} d x

What is the solution for y^{''}+4y^'-5y=e^{3x} ?
The solution for y^{''}+4y^'-5y=e^{3x} is y=c_{1}e^x+c_{2}e^{-5x}+1/16 e^{3x}