What is the solution for y^{''}-3y^'+2y=4x+6 ?

The solution for y^{''}-3y^'+2y=4x+6 is y=c_{1}e^{2x}+c_{2}e^x+2x+6

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y^{''}-3y^'+2y=4x+6

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x + 6

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x + 6

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 4 x + 6

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x + 6

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x + 6

y^{^{''}} + y^{^{'}} - 6 y = x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (x)

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 24

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

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

What is the solution for y^{''}-3y^'+2y=4x+6 ?
The solution for y^{''}-3y^'+2y=4x+6 is y=c_{1}e^{2x}+c_{2}e^x+2x+6