What is the solution for y^{''}+4y=7sin(2x) ?

The solution for y^{''}+4y=7sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-(7xcos(2x))/4

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y^{''}+4y=7sin(2x)

y^{^{''}} + 4 y = 7 sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{7 x cos ( 2 x )}{4}

Solution steps

y^{''} (x) + 4 y = 7 sin (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{7 x cos ( 2 x )}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{7 x cos ( 2 x )}{4}

5 x y (\frac{d y}{d x}) = 2

\frac{3}{t ^{4}}

\frac{\partial}{\partial u} (uv - 1)

\frac{2 s}{( s - 1 ) ( s ^{2} + 25 )}

\frac{\partial}{\partial y} (e^{- y} (x^{2} + y^{2}))

What is the solution for y^{''}+4y=7sin(2x) ?
The solution for y^{''}+4y=7sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-(7xcos(2x))/4