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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} ((2 x^{2} - 1) + (5 x - 6))

f (x) = 12 x^{2}

\frac{\partial}{\partial x} (10 x^{\frac{1}{4}} \cdot 2 y^{\frac{3}{4}})

x \to 1 lim (4 x^{2} - 2 x - 6)

\int (1 - x)^{\frac{3}{2}} d x

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