Question

General solution of $(x^2+y^2)dx-2xydy=0$ is:

• (A) $y^2+x^2=c{x}^2$
• (B) $x^2-y^2=cx$
• (C) $x^2=cx(x^2-y^2)$
• (D) None of these

Answer 1

The Correct Option is C

Explanation:
Given,$(x^2+y^2)dx=2xydy$$\frac{dy}{dx}=\frac{(x^2+y^2)}{2xy}$ is homogeneous.Put, $y=vx$$\frac{dy}{dx}=v+\frac{xdv}{dx}$$\Rightarrow v+x\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2v{x}^2}=\frac{x^2(1+v^2)}{2v{x}^2}$$\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{2v}-v=\frac{1-v^2}{2v}$$\Rightarrow \frac{2v}{1-v^2}dv=\frac{1}{x}dx$By integrating both sides we get,$\int \frac{2v}{1-v^2}dv=\int \frac{1}{x}dx\dots \dots (1)$Now, we integrate $\int \frac{2v}{1-v^2}dv$Let, $1-v^2=a$On differentiating both sides w.r.t. $v$, we get$\frac{d}{dv}(1-v^2)=\frac{d}{dv}a$$-2v=\frac{da}{dv}$$dv=-\frac{da}{2v}$Now,putting these values on equation $(1)$, we get$\int \frac{2v}{a}\times -\frac{da}{2v}=\int \frac{1}{x}dx$$\Rightarrow \int -\frac{da}{a}=\int \frac{1}{x}dx$We know that,$\int \frac{1}{x}dx=\log x$Now,$-\log (1-v^2)=\log x+\log c$$\Rightarrow \log \left[\frac{1}{(1-v^2)}\right]=\log x+\log c$$\Rightarrow \left[\frac{1}{(1-v^2)}\right]=cx$$\Rightarrow \left[\frac{1}{(1-\frac{y^2}{x^2})}\right]=cx$$\Rightarrow \left[\frac{x^2}{(x^2-y^2)}\right]=cx$$\Rightarrow x^2=cx(x^2-y^2)$ is the general solution of the differential equation.Hence, the correct option is (C).