What is the solution for 2(d^2y)/(dx^2)+8y=cos(2x) ?

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

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2(d^2y)/(dx^2)+8y=cos(2x)

2 \frac{d ^{2} y}{d x ^{2}} + 8 y = cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{8}

Solution steps

2 \frac{d ^{2} y}{d x ^{2}} + 8 y = cos (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{8}

\frac{\partial}{\partial x} (2 (x - 1))

\int \frac{1}{( x + 3 ) ^{3}} d x

\int_{- 1}^{2} (y + 2) - (y^{2}) d y

\int tan (x) x d x

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 7 )} d x

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