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}}

\int_{0}^{1} \frac{2 x}{( x ^{2} + 2 ) ^{3}} d x

\frac{\partial}{\partial y} (tan (4 + 2 x^{2} y^{3} z^{2}))

\int \frac{( x ^{4} - 3 x ^{2} + 5 )}{x ^{4}} d x

\int (ln (x))^{3} \frac{1}{x} d x

10 x^{1.5} - 6 x^{0.5}

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)}