What is the solution for y^{''}+y^'=16e^{3x} ?

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

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y^{''}+y^'=16e^{3x}

y^{^{''}} + y^{^{'}} = 16 e^{3 x}

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

Solution steps

y^{''} (x) + y^{'} (x) = 16 e^{3 x}

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} + 169 y = 13 sin (13 t)

y^{^{''}} + 4 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y = 8 x e^{x} + 2 e^{x}

y^{^{''}} + 5 y^{^{'}} = 4 x + 8

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