What is the solution for y^{''}+y=8cos(2x)-4sin(x) ?

The solution for y^{''}+y=8cos(2x)-4sin(x) is y=c_{1}cos(x)+c_{2}sin(x)-8/3 cos(2x)+2xcos(x)

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y^{''}+y=8cos(2x)-4sin(x)

y^{^{''}} + y = 8 cos (2 x) - 4 sin (x)

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

Solution steps

y^{''} (x) + y = 8 cos (2 x) - 4 sin (x)

Solve linear ODE:

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

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

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

y^{^{''}} - y = e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

4 y^{^{''}} + y = cos (t)

y^{^{''}} + y = 7 x sin (x)

y^{^{''}} - 2 y^{^{'}} = x + 1

What is the solution for y^{''}+y=8cos(2x)-4sin(x) ?
The solution for y^{''}+y=8cos(2x)-4sin(x) is y=c_{1}cos(x)+c_{2}sin(x)-8/3 cos(2x)+2xcos(x)