What is the solution for y^{''}+4y=12sin(2x) ?

The solution for y^{''}+4y=12sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-3xcos(2x)

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y^{''}+4y=12sin(2x)

y^{^{''}} + 4 y = 12 sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 3 x cos (2 x)

Solution steps

y^{''} (x) + 4 y = 12 sin (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 3 x cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 3 x cos (2 x)

f (x) = \frac{4}{3 x ^{2}} + \frac{1}{x ^{2.1}}

\int \frac{( 8 + x + x ^{2} )}{x} d x

x \to 4 - lim (\frac{6 x}{x - 4})

f (x) = 3 sin^{2} (x)

2 y^{^{''}} + y^{^{'}} - 2 y = 0

What is the solution for y^{''}+4y=12sin(2x) ?
The solution for y^{''}+4y=12sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-3xcos(2x)