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

The solution for y^{''}+y=3e^{-4x} is y=c_{1}cos(x)+c_{2}sin(x)+3/17 e^{-4x}

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

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

y = c_{1} cos (x) + c_{2} sin (x) + \frac{3}{17} e^{- 4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + \frac{3}{17} e^{- 4 x}

y = c_{1} cos (x) + c_{2} sin (x) + \frac{3}{17} e^{- 4 x}

\int \frac{3 ln ^{2} ( x )}{x ^{2}} d x

\frac{d}{d x} (\frac{x sec ( - ( \frac{y}{x} ) + y )}{x})

\int (x^{2} - 1)^{3} d x

\int sin (t^{4}) d t

f (t) = 1 + t

What is the solution for y^{''}+y=3e^{-4x} ?
The solution for y^{''}+y=3e^{-4x} is y=c_{1}cos(x)+c_{2}sin(x)+3/17 e^{-4x}