What is the solution for y^{''}+y=2cos(3x) ?

The solution for y^{''}+y=2cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)-1/4 cos(3x)

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y^{''}+y=2cos(3x)

y^{^{''}} + y = 2 cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{4} cos (3 x)

Solution steps

y^{''} (x) + y = 2 cos (3 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{4} cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{4} cos (3 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = e^{4 x}

y^{^{''}} - y^{^{'}} + 49 y = 7 sin (7 t)

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 3

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 4

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- t}

What is the solution for y^{''}+y=2cos(3x) ?
The solution for y^{''}+y=2cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)-1/4 cos(3x)