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

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

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y^'+2y=x^2

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{2 x} + sin (x) d x

\frac{d y}{d x} + \frac{2}{x} y = \frac{1}{x ^{4}}

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