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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = x^{2} + 5 cos (x)

y^{^{''}} - y^{^{'}} - 6 y = sin (x)

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

y^{^{''}} - 6 y = 6

y^{^{''}} - 2 y^{^{'}} + 17 y = e x cos (4 x)

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