What is the solution for f^{''}+16f=sin(2x) ?

The solution for f^{''}+16f=sin(2x) is f(x)=c_{1}cos(4x)+c_{2}sin(4x)+1/12 sin(2x)

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f^{''}+16f=sin(2x)

f^{^{''}} + 16 f = sin (2 x)

f (x) = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{12} sin (2 x)

Solution steps

f^{''} (x) + 16 f (x) = sin (2 x)

Solve linear ODE:

f (x) = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{12} sin (2 x)

f (x) = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{12} sin (2 x)

\int_{6}^{12} 144 - x^{2} d x

x^{2} y^{^{''}} - 6 y = 0, y (1) = 6, y^{^{'}} (1) = - 2

\frac{\partial}{\partial x} (e^{x y} y^{2} + \frac{x}{y})

\int e^{- \frac{x ^{2}}{2}} d x

\int_{0}^{1} \frac{x ^{4} + 3}{x} d x

What is the solution for f^{''}+16f=sin(2x) ?
The solution for f^{''}+16f=sin(2x) is f(x)=c_{1}cos(4x)+c_{2}sin(4x)+1/12 sin(2x)