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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

$\displaystyle y^{\prime}= \frac{x(x^2+1)}{4y^3}, \quad y(0)=-\frac{1}{\sqrt{2}}$

rewrite

$$\displaystyle\d{y}{x}=\frac{x(x^2+1)}{4y^3}$$

$$\displaystyle y^3\, dy=\frac{x(x^2+1)}{4}\, dx$$

integrate

$$\displaystyle\frac{y^4}{4}= \frac{1}{4}\left(\frac{x^4}{4} +\frac{ x^2}{2}\right)$$

$$\displaystyle y^4=\frac{x^4}{4} +\frac{ x^2}{2}$$

so far hopefully sorta?