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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{4} + d^{3} + d^{2}) y = 5 x^{2} + cos (x)

(d^{4} + d^{3} + d^{2}) y = 5 x^{2}

(d^{3} - 3 d^{2} + 3 d - 1) y = sin (x) + x^{3}

(d^{3} + d) y = cos (x)

(d^{3} + 2 d^{2} + d) y = x^{2} + x

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