$\displaystyle y'=y+xy^{3} $ need some help!
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$\displaystyle \begin{aligned} y'&=y+xy^{3} \\ y'-y&=xy^{3} \\ \frac{y'}{y^{3}}-\frac{1}{y^{2}}&=x. \end{aligned}$ Make the substitution $\displaystyle z=\frac1{y^2}.$
Originally Posted by skystar $\displaystyle y'=y+xy^{3} $ need some help! Read the section Bernoulli's Equations in my Differential Equations Tutorial. This may help you out a bit... Make the substitution $\displaystyle v=y^{-2} \implies y=v^{-\frac{1}{2}}$ to make it linear.
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