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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = x^{2} + 3, (1, 4)

\frac{d}{d x} (\frac{x ^{p (x)}}{x ^{m} - a ^{m}})

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

\frac{d}{d x} (\frac{1 + csc ( x ^{7} )}{1 - cot ( x ^{7} )})

\frac{9}{x - 6}

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