What is the solution for y^{''}+4y=16xcos(2x) ?

The solution for y^{''}+4y=16xcos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2x^2sin(2x)+xcos(2x)

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y^{''}+4y=16xcos(2x)

y^{^{''}} + 4 y = 16 x cos (2 x)

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

Solution steps

y^{''} (x) + 4 y = 16 x cos (2 x)

Solve linear ODE:

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

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

y^{^{''}} + 36 y = sin (9 x)

0.5 x^{^{''}} + 3 x^{^{'}} + 4 x = 3 cos (t)

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

- 2 y^{^{''}} + 18 y + 2 (3 x^{2} + x + \frac{10}{3}) = 0

D C = 1.5226 D U S

What is the solution for y^{''}+4y=16xcos(2x) ?
The solution for y^{''}+4y=16xcos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2x^2sin(2x)+xcos(2x)