What is the solution for y^{''}+16y=sin(4x) ?

The solution for y^{''}+16y=sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)-(xcos(4x))/8

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y^{''}+16y=sin(4x)

y^{^{''}} + 16 y = sin (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) - \frac{x cos ( 4 x )}{8}

Solution steps

y^{''} (x) + 16 y = sin (4 x)

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) - \frac{x cos ( 4 x )}{8}

y = c_{1} cos (4 x) + c_{2} sin (4 x) - \frac{x cos ( 4 x )}{8}

\int \frac{cosh ( x )}{1 + sinh ( x )} d x

\int \frac{1}{( e ^{x} + 1 )} d x

\frac{d}{d x} (\frac{1 + sin ( x )}{cos ( x )})

y^{^{'}} - \frac{4 y}{x} = - \frac{2}{x ^{2}}

\frac{- 6 x}{x ^{2} + 1}

What is the solution for y^{''}+16y=sin(4x) ?
The solution for y^{''}+16y=sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)-(xcos(4x))/8