Solving 2xy′(x−y2)+y3=0

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Brennan Thompson · Accepted Answer

Step 1Well, we are trying to solve:1)&amp;nbsp;nxy′(x)(x−y(x)2)+y(x)3=0Solving for&amp;nbsp;y′(x), gives:2)&amp;nbsp;y′(x)=−y(x)3nx(x−y(x)2)Now, we can write the DE in terms of x, since&amp;nbsp;dydxdxdy=1dy(x)dx=1dx(y)dy. So we get:3)&amp;nbsp;1dx(y)dy=−y3nx(y)(x(y)−y2)Raise both sides to the power -1, expand, subtract&amp;nbsp;nx(y)y&amp;nbsp;from both sides and divide both sides by&amp;nbsp;−x(y)24)&amp;nbsp;nyx(y)−dx(y)dyx(y)2=ny3Let&amp;nbsp;μ(y)&amp;nbsp;=1x(y). This gives&amp;nbsp;dμ(y)dy=−dx(y)dyx(y)25)&amp;nbsp;dμ(y)dy+nμ(y)y=ny3Let:6)&amp;nbsp;ξ(y)&amp;nbsp;=exp{∫ny&amp;nbsp;dy}=ynMultiply both sides by&amp;nbsp;ξ(y), subsitute&amp;nbsp;nyn−1=ddy(yn)&amp;nbsp;and apply the reserve product rule. This leads to:7)&amp;nbsp;ddy(ynμ(y))=nyn−3Integrate both sides with respect to y and divide both sides by&amp;nbsp;ξ(y)8)&amp;nbsp;μ(y)=ny2(n−2)+CynSovling for x(y), gives:9)&amp;nbsp;1x(y)=ny2(n−2)+CynSo:10)&amp;nbsp;1x=ny(x)2(n−2)+Cy(x)n

Leonardo Mcpherson · Accepted Answer

Step 1As I noted in my above hint in the comments, it would be helpful to instead try solving this as if it was a differential equation in x=x(y) not y=y(x). To that end, suppose the function is smooth and we can writey3x′+2x(x−y2)=0y3u′+2y2u=2,  via substitution  x=u−1u′+2y−1u=2y−3(y2u)′=2y−1y2u=2ln⁡y+Cy2=2xln⁡y+CxYou can check that this is indeed a solution of the original differential equation.

Solving \(\displaystyle{2}{x}{y}'{\left({x}-{y}^{{2}}\right)}+{y}^{{3}}={0}\)

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