What is the solution for y^{''}+4y=-12cos(2x)-4sin(2x) ?

The solution for y^{''}+4y=-12cos(2x)-4sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-3xsin(2x)+xcos(2x)

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y^{''}+4y=-12cos(2x)-4sin(2x)

y^{^{''}} + 4 y = - 12 cos (2 x) - 4 sin (2 x)

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

Solution steps

y^{''} (x) + 4 y = - 12 cos (2 x) - 4 sin (2 x)

Solve linear ODE:

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

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

\frac{\partial}{\partial y} (x y ln (x^{2} + y^{2}))

\int \frac{2}{4 - x ^{2}} d x

\int \frac{( x + 2 )}{( x + 1 )} d x

y^{^{'}} (t) = (y + 1) \cdot (y - 3), y (0) = 1

\frac{d}{d x} (lo g_{10} (sin (x)))

What is the solution for y^{''}+4y=-12cos(2x)-4sin(2x) ?
The solution for y^{''}+4y=-12cos(2x)-4sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-3xsin(2x)+xcos(2x)