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

The solution for y^{''}-3y^'+2y=4x+3 is y=c_{1}e^{2x}+c_{2}e^x+2x+9/2

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x + \frac{9}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x + \frac{9}{2}

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 x + \frac{9}{2}

y^{^{''}} + y = 9 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 5 y = - 9 e^{2 x}

y^{^{''}} + 18 y^{^{'}} + 81 y = x, y (0) = 0, y^{^{'}} (0) = 0

20 y^{^{''}} + 9 y^{^{'}} + x = 0, y (0) = 0, y^{^{'}} (0) = 3

d, d (x) = \frac{2 x ^{3} + 1}{x ^{6}}

What is the solution for y^{''}-3y^'+2y=4x+3 ?
The solution for y^{''}-3y^'+2y=4x+3 is y=c_{1}e^{2x}+c_{2}e^x+2x+9/2