What is the solution for y^'+2y=x^2+2x ?

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

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y^'+2y=x^2+2x

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

y = \frac{x ^{2}}{2} + \frac{x}{2} - \frac{1}{4} + \frac{c _{1}}{e ^{2 x}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2}}{2} + \frac{x}{2} - \frac{1}{4} + \frac{c _{1}}{e ^{2 x}}

y = \frac{x ^{2}}{2} + \frac{x}{2} - \frac{1}{4} + \frac{c _{1}}{e ^{2 x}}

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