Question

$Solutionoftheequation:(1-x^2)dy+xydx=x{y}^{2}dx$

• A. $\left({y-1}^{)2}(1-x^2\right)=0$
• B. $\left({y-1}^{)2}(1-x^2\right)=c^2{y}^2$
• C. $\left({y-1}^{)2}(1+x^2\right)=c^2{y}^2$
• D. None of these

Answer 1

The correct option is B $\left({y-1}^{)2}(1-x^2\right)=c^2{y}^2$

$(1-x^2)dy+xydx=x{y}^{2}dx$

Dividing by $y^{2}dx$, we get(and also divisible by $(1-x^2)$

$\frac{1}{y^2}\frac{dy}{dx}+\frac{x}{1-x^2}\frac{1}{y}=\frac{x}{1-x^2}$

Now by putting $\frac{-1}{y}=z$

$\frac{1}{y^2}dy=dz$

Therefore, equation becomes

$\frac{dz}{dx}-\frac{x}{1-x^2}dz=\frac{x}{(1-x^2)}$

Now ,$IF=e^{\int \frac{-x}{1-x^2}}=e^{+\frac{1}{2}\log (1-x^2)}=\sqrt{1-x^2}$

Multiplying by IF , we get

$\frac{d}{dx}\left(\right(\sqrt{1-x^2}\left)z\right)$=$\frac{x}{\sqrt{1-x^2}}$

Now integrating on both the sides and taking

$\sqrt{1-x^2}=P$ on RHS, we get

$\frac{1}{2\sqrt{1-x^2}}\times (-2x)dx=dP$

$\frac{-x}{\sqrt{1-x^2}}dx=dP$

$\frac{x}{\sqrt{1-x^2}}dx=-dP$

$\int d\left(\right(1-\sqrt{1-x^2}\left)dz\right)=-\int -dP$

$(1-\sqrt{1-x^2})z=-P+C=-\sqrt{1-x^2}+C$

$\left(\sqrt{1-x^2)}\right(z+1)=C$

$\int \left(\sqrt{1-x^2)}\right(\frac{-1}{y}+1)=C$

Now squaring on both the sides, we get

$(y-1{)}^2(1-x^2)=C^2{y}^2$