What is the solution for y^{''}+4y=32sin(2x)+16cos(2x) ?

The solution for y^{''}+4y=32sin(2x)+16cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+4xsin(2x)-8xcos(2x)

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y^{''}+4y=32sin(2x)+16cos(2x)

y^{^{''}} + 4 y = 32 sin (2 x) + 16 cos (2 x)

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

Solution steps

y^{''} (x) + 4 y = 32 sin (2 x) + 16 cos (2 x)

Solve linear ODE:

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

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

5 y^{^{''}} + y^{^{'}} = - 6 x, y (0) = 0, y^{^{'}} (0) = - 10

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

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 x

y^{^{''}} + 2 y^{^{'}} + y = \frac{ln ( x )}{e ^{x}}

y^{^{''}} - y^{^{'}} - 2 y = \frac{( e ^{2 t} + e ^{- 2 t} )}{2}

What is the solution for y^{''}+4y=32sin(2x)+16cos(2x) ?
The solution for y^{''}+4y=32sin(2x)+16cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+4xsin(2x)-8xcos(2x)