What is the solution for y^{'''}-3y^{''}+2y^'=x+e^x ?

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

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

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

y = c_{1} + c_{2} e^{x} + c_{3} e^{2 x} + \frac{x ^{2} + 3 x}{4} - e^{x} x

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{x} + c_{3} e^{2 x} + \frac{x ^{2} + 3 x}{4} - e^{x} x

y = c_{1} + c_{2} e^{x} + c_{3} e^{2 x} + \frac{x ^{2} + 3 x}{4} - e^{x} x

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

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

y^{^{''}} + 9 y = 18 sin (3 t)

y^{^{''}} - y^{^{'}} - 6 y = 3 e^{2 x}

x^{^{''}} + 25 x = cos (t)

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