What is the solution for y^{''}+4y=4598cos(2x) ?

The solution for y^{''}+4y=4598cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(2299xsin(2x))/2

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y^{''}+4y=4598cos(2x)

y^{^{''}} + 4 y = 4598 cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2299 x sin ( 2 x )}{2}

Solution steps

y^{''} (x) + 4 y = 4598 cos (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2299 x sin ( 2 x )}{2}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2299 x sin ( 2 x )}{2}

x^{^{''}} - 2 x^{^{'}} + 2 x = e^{- t} cos (2 t)

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + y = x

y^{^{''}} - y^{^{'}} - 6 y = 32 sin (3 x)

y^{^{''}} - 2 y^{^{'}} + y = 576 e^{9 t}

What is the solution for y^{''}+4y=4598cos(2x) ?
The solution for y^{''}+4y=4598cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(2299xsin(2x))/2