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

The solution for y^{''}+4y=-cos(3x)-2sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2/5 sin(3x)+1/5 cos(3x)

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y^{''}+4y=-cos(3x)-2sin(3x)

y^{^{''}} + 4 y = - cos (3 x) - 2 sin (3 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} sin (3 x) + \frac{1}{5} cos (3 x)

Solution steps

y^{''} (x) + 4 y = - cos (3 x) - 2 sin (3 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} sin (3 x) + \frac{1}{5} cos (3 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} sin (3 x) + \frac{1}{5} cos (3 x)

y^{^{''}} + 4 y^{^{'}} = t^{2} + 6 e^{t}

y^{^{''}} - 9 y^{^{'}} = 5 e^{- 2 t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 4 y = 4 x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (3 x) cos (2 x)

3 y^{^{''}} + 3 y^{^{'}} - 1 = 0

What is the solution for y^{''}+4y=-cos(3x)-2sin(3x) ?
The solution for y^{''}+4y=-cos(3x)-2sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2/5 sin(3x)+1/5 cos(3x)