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Solve differential equation $$y^2y'^2+2axyy'+(1-a)y^2+ax^2+(a-1)b=0,$$ where $y=y(x)$ and $a,b \in \mathbb{R}.$

My work: Let $y^2=z, \,\,\,z=z(x).$ Then $2yy'=z'$ and our differential equation become $$\frac{z'^2}{4}+axz'+(1-a)z+ax^2+(a-1)b=0.$$

But, what now?


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1 Answer 1

up vote 2 down vote accepted

HINT: $$(yy'+ax)^2-a^2x^2+(1-a)y^2+ax^2+(a-1)b=0$$

$$(yy'+ax)^2+(1-a)(y^2+ax^2-b)=0$$ $P=y^2+ax^2$

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That was nice. Thank you! – Cortizol Jan 20 '13 at 20:14

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