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

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

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y^{''}-3y^'+2y=(1+2x)e^x

y^{^{''}} - 3 y^{^{'}} + 2 y = (1 + 2 x) e^{x}

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

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = (1 + 2 x) e^{x}

Solve linear ODE:

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

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

y^{^{''}} - y = \frac{1}{e ^{x} + 1}

(d^{4} + 4 d^{2}) y = x^{2} + 1

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} (x^{2} - 4 x + 5)

y^{^{''}} + 2 y = e^{t} + 2

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