What is the solution for y^{''}+16y=1+2sin(4x) ?

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

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y^{''}+16y=1+2sin(4x)

y^{^{''}} + 16 y = 1 + 2 sin (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{16} - \frac{x cos ( 4 x )}{4}

Solution steps

y^{''} (x) + 16 y = 1 + 2 sin (4 x)

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{16} - \frac{x cos ( 4 x )}{4}

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{16} - \frac{x cos ( 4 x )}{4}

\frac{3 x}{4 - x ^{2}}, at x = 1

\int_{b}^{1} 2 ln (x) d x

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

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{5}}})

\int \frac{x ^{3}}{( 4 - x ^{2} ) ^{3}} d x

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