What is the solution for (x^2+2)y^'+2xy=x^3 ?

The solution for (x^2+2)y^'+2xy=x^3 is y=(x^4)/(4(x^2+2))+(c_{1})/(x^2+2)

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(x^2+2)y^'+2xy=x^3

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

y = \frac{x ^{4}}{4 ( x ^{2} + 2 )} + \frac{c _{1}}{x ^{2} + 2}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{4}}{4 ( x ^{2} + 2 )} + \frac{c _{1}}{x ^{2} + 2}

y = \frac{x ^{4}}{4 ( x ^{2} + 2 )} + \frac{c _{1}}{x ^{2} + 2}

\int_{0}^{5} 4 x^{2} d x

\int 2 z^{2} d z

\int \frac{1}{3 - 6 x - x ^{2}} d x

x \to 0 lim (5 x + 4)

\frac{\partial}{\partial x} (ln (5 x^{2} + 4 y^{2} + 7))

What is the solution for (x^2+2)y^'+2xy=x^3 ?
The solution for (x^2+2)y^'+2xy=x^3 is y=(x^4)/(4(x^2+2))+(c_{1})/(x^2+2)