We have been given a differential equation for which we have to find the equation of curve . I tried it a lot but not able to proceed .
Can someone give me some hints .
$yy'\sin x = \cos x(\sin x-y^2)$
Let $z= y^2$, then $z' = 2yy'$. The equation becomes $z'\sin x/2 +(\cos x)z = \cos x \sin x,$ which is linear.
I think this question has a nice solution (I think, because I am not expert at differantial equations). Anyway, let $y^2=u$ and $sinx=v$ then rewrite the equation as $$\frac{v.du}{2}=v.dv-u.dv$$. This doesn't look familiar, maybe $$v(\frac {v.du}{2}+u.dv)=v^2.dv$$ makes sense. $$ \int \frac {v^2.du}{2}+uv.dv=\int v^2.dv \Rightarrow \frac{v^3}{3}+C=\frac {v^2.u}{2}$$ (from partial differantial) so we rewrite the equation with $x$ and $y$ variables, we get $\displaystyle x=\arcsin\left ( \frac {2y^6+6C}{3y^4}\right)$
$$ ... $$
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. And it is fine.
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Oct 26, 2016 at 8:53