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

The solution for y^{''}+4y=xsin(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(xsin(x))/3-2/9 cos(x)

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( x )}{3} - \frac{2}{9} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( x )}{3} - \frac{2}{9} cos (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( x )}{3} - \frac{2}{9} cos (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{- x}

4 y^{^{''}} - 4 y^{^{'}} + y = 40 e^{\frac{t}{2}}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = sec (x)

y^{^{''}} - y^{^{'}} + 225 y = 15 sin (15 t)

y^{^{''}} + y^{^{'}} - 6 y = x^{2} + 3

What is the solution for y^{''}+4y=xsin(x) ?
The solution for y^{''}+4y=xsin(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(xsin(x))/3-2/9 cos(x)