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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} - y = 8 t + cos (t)

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

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

\frac{d ^{2} y}{d x ^{2}} + 100 y = csc (10 x)

y^{^{''}} + y^{^{'}} + y = x^{4} + 4 x^{3} + 12 x^{2}

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