What is the solution for (d^3+d)y=cos(x) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

(d^{2} - 4) y = 2 cos^{2} (x)

(d - 1)^{2} y = 2 lo g_{10} (x)

y^{^{''}} + 10 y^{^{'}} + 9 y = 2 e - 2 x

y^{^{'''}} - y^{^{''}} = 12 x^{2} + 6 x

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