Question

Solve the following initial value problem:
 (x² + y²) dx = 2xy dy, y (1) = 0

Answer 1

(x² + y²)dx = 2xy dy, y(1) = 0
We have,
(x² + y²) dx = 2xy        .....(i)
This is a homogenous equation, so let us take y = vx
\[\text{ Then, }\frac{dy}{dx} = v + x\frac{dv}{dx}\]
Putting y = vx in equation (i)
\[\left( x^2 + v^2 x^2 \right) = 2v x^2 \left( v + x\frac{dv}{dx} \right)\]
\[ x^2 \left( 1 + v^2 \right) = 2v x^2 \left( v + x\frac{dv}{dx} \right)\]
\[\left( 1 + v^2 \right) = 2 v^2 + 2vx\frac{dv}{dx}\]
\[1 - v^2 = 2vx\frac{dv}{dx}\]
\[\frac{dx}{x} = \frac{2v dv}{1 - v^2}\]
On integrating both sides, we get
\[\int\frac{1}{x}dx = \int\frac{2v}{1 - v^2}dv\]
\[\text{ Let, }\left( 1 - v^2 \right) = t\]
\[ \Rightarrow - 2v dv = dt\]
\[ \log_e x = - \int\frac{dt}{t} \]
\[ \log_e x = - \log_e t + c\]
\[ \log_e x = - \log_e \left( 1 - \frac{y^2}{x^2} \right) + c\]
\[ \log_e \left[ x\left( \frac{x^2 - y^2}{x^2} \right) \right] = c\]
\[ \log_e \left( \frac{x^2 - y^2}{x} \right) = c\]
\[\text{ As }y\left( 1 \right) = 0\]
\[ \Rightarrow c = 0\]
\[ \therefore \log_e \left( \frac{x^2 - y^2}{x} \right) = 0\]
\[ \Rightarrow \frac{x^2 - y^2}{x} = 1\]
\[ \Rightarrow x^2 - y^2 = x\]