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

The solution for y^{''}+y=sin(2x)+cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)-1/3 sin(2x)-1/8 cos(3x)

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

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

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} sin (2 x) - \frac{1}{8} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} sin (2 x) - \frac{1}{8} cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} sin (2 x) - \frac{1}{8} cos (3 x)

\frac{1}{2} y^{^{''}} + \frac{1}{2} y^{^{'}} + 6 y = 10 cos (3 t)

y^{^{'''}} - 6 y^{^{''}} - y^{^{'}} + 6 y = e^{5 x}

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

y^{^{''}} - 4 \cdot y = e^{- α \cdot x}

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = (x^{2} + 4 x + 8) e^{2} x

What is the solution for y^{''}+y=sin(2x)+cos(3x) ?
The solution for y^{''}+y=sin(2x)+cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)-1/3 sin(2x)-1/8 cos(3x)