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

The solution for y^{'''}+4y^'=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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y^{'''}+4y^'=e^x+sin(x)

y^{^{'''}} + 4 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

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

y^{^{''}} + y = 6 sin (2 x)

(D^{2} - 3 D + 2) y = cos (3 x)

2 y^{^{''}} + y = - e^{3} + e^{2 x}

y^{^{'''}} + y^{^{''}} = 6

y^{^{''}} + 3 y^{^{'}} - 4 y = 1, y (0) = 0, y^{^{'}} (0) = 1

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