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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 12 x^{^{'}} + 11 x = 8 cos (2 t)

y^{^{''}} - 6 y^{^{'}} + 8 y^{^{'}} = x^{2} - 1

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

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0, y^{^{''}} (0) = 1

y^{^{''}} - 4 y^{^{'}} - 32 y = - 80 t + 64 t^{2}

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