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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d t} = k y (1 - y)

\int x x^{2} + 5 d x

\int (- x + 3)^{2} d x

(ln (ln (x)))^{^{'}}

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

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