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

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

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{cos ( 2 ) x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{cos ( 2 ) x}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{cos ( 2 ) x}{4}

(d^{2} - 2 d + 2) y = (x + e^{x}) sin (x)

y^{^{''}} - y = \frac{2 e ^{2 x}}{e ^{2 x} + 1}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 4 sin (x)

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{x}

y^{^{''}} + 4 y^{^{'}} + 5 y = \frac{1}{x} e^{2 x} sin (x)

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