What is the solution for y^{''}-2y^'-3y=10e^{4x} ?

The solution for y^{''}-2y^'-3y=10e^{4x} is y=c_{1}e^{3x}+c_{2}e^{-x}+2e^{4x}

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y^{''}-2y^'-3y=10e^{4x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 10 e^{4 x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 10 e^{4 x}

Solve linear ODE:

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

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

f (x) = \frac{x}{x ^{2} + 7 x + 6}

3 + t

y = x^{2} + 4

\frac{d y}{d x} = (x - 3) e^{(- 2 y)}

\frac{d}{d x} (\frac{e ^{3 x} sin ( 2 x )}{5 ^{x}})

What is the solution for y^{''}-2y^'-3y=10e^{4x} ?
The solution for y^{''}-2y^'-3y=10e^{4x} is y=c_{1}e^{3x}+c_{2}e^{-x}+2e^{4x}