What is the solution for y^{''}+y=16cos(3x) ?

The solution for y^{''}+y=16cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)-2cos(3x)

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y^{''}+y=16cos(3x)

y^{^{''}} + y = 16 cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) - 2 cos (3 x)

Solution steps

y^{''} (x) + y = 16 cos (3 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - 2 cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) - 2 cos (3 x)

y^{^{''}} + 7 y^{^{'}} + 12 y = e^{t}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = e^{- x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{(- x)} cos (2 x)

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{8 e ^{- 3 x}}{x ^{2} + 1}

y^{^{'''}} + 2 y^{^{''}} = 32 e^{2 x} + 24 x

What is the solution for y^{''}+y=16cos(3x) ?
The solution for y^{''}+y=16cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)-2cos(3x)