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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} + y = \frac{sin ( t )}{3} - 2 e^{2 t}

x^{^{''}} + 4 x^{^{'}} + 29 = 0

x^{^{''}} + 3 x^{^{'}} + 2 x = cos (2 t)

y^{^{''}} + 4 y = 10 e^{5 x}

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

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