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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - k^{2} y = 0

x^{4} - 8 x^{3} + 2

\int sin (π x) d x

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

- y d x + x d y = 0

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