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)

\frac{3 s - 1}{s ^{2} + 2 s + 5}

sin^{2} (x) + cos^{2} (x)

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

(- 5, - 1), (5, 2)

\int cos^{2} (\frac{3 π x}{L}) d x

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)