What is the solution for y^{''}-3y^'+2y=4x ?

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

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y^{''}-3y^'+2y=4x

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x

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

Solution steps

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

Solve linear ODE:

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

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

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

x^{^{''}} - 4 x^{^{'}} + 5 x = cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 e^{2 x}, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{'''}} + 4 y^{^{'}} = t e^{- 2 t} + 3 cos (2 t)

y^{^{''}} - y^{^{'}} = 3 e^{t}

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