What is the solution for (d^3+4d)y=e^x+sin(x) ?

The solution for (d^3+4d)y=e^x+sin(x) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)+1/5 e^x-1/3 cos(x)

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(d^3+4d)y=e^x+sin(x)

(d^{3} + 4 d) y = e^{x} + sin (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{5} e^{x} - \frac{1}{3} cos (x)

Solution steps

(d^{3} + 4 d) y = e^{x} + sin (x)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{5} e^{x} - \frac{1}{3} cos (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{5} e^{x} - \frac{1}{3} cos (x)

2 y^{^{''}} + 18 y^{^{'}} = 6 tan (3 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = e^{3} x + e^{4} x

y^{^{''}} + 4 y^{^{'}} = 10 t + 1

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

y^{^{''}} + y^{^{'}} = 2 + 2 x + x^{2}

What is the solution for (d^3+4d)y=e^x+sin(x) ?
The solution for (d^3+4d)y=e^x+sin(x) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)+1/5 e^x-1/3 cos(x)