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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} - 1) y = \frac{2}{1 + e ^{x}}

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 x^{2} + x + 9

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{9 + e ^{x}}

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 θ (t) = csc^{2} (2 t)

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