What is the solution for y^{''}+4y=8sin(2x) ?

The solution for y^{''}+4y=8sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-2xcos(2x)

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y^{''}+4y=8sin(2x)

y^{^{''}} + 4 y = 8 sin (2 x)

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

Solution steps

y^{''} (x) + 4 y = 8 sin (2 x)

Solve linear ODE:

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

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

x \to 0.5 lim (\frac{sin ( x )}{x})

\int cos^{2} (\frac{3 π x}{L}) d x

n = 0 \sum \infty (- \frac{3}{10})^{n}

s l o p e (4, - 3), (7, - 2)

\frac{d}{d x} (\frac{x}{12})

What is the solution for y^{''}+4y=8sin(2x) ?
The solution for y^{''}+4y=8sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-2xcos(2x)