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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

(d^{2} - 3 d) y = x^{2} + 2 x - 6

y^{^{''}} + y = - 8 x cos (x)

x^{^{''}} - x^{^{'}} - 12 x = 2 sinh^{2} (t)

(D^{3} - 3 D) y = 9 x^{2}

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