What is the solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 ?

The solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 is y=c_{1}e^{-3x}+c_{2}cos(x)+c_{3}sin(x)+1/8 e^x+4/3

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

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

y = c_{1} e^{- 3 x} + c_{2} cos (x) + c_{3} sin (x) + \frac{1}{8} e^{x} + \frac{4}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} cos (x) + c_{3} sin (x) + \frac{1}{8} e^{x} + \frac{4}{3}

y = c_{1} e^{- 3 x} + c_{2} cos (x) + c_{3} sin (x) + \frac{1}{8} e^{x} + \frac{4}{3}

y^{^{''}} + 4 y = 2 t

y^{^{''}} + 9 y = cos (2 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = x

d (x) = - 3 x^{2} - 15 x + 146

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

What is the solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 ?
The solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 is y=c_{1}e^{-3x}+c_{2}cos(x)+c_{3}sin(x)+1/8 e^x+4/3