Differential Equations – Homogeneous Differential Equations – JEE Main 22 Jan 2026 Shift 1

Solution:

Rearranging the given differential equation:
$$ xdy – ydx = \sqrt{x^2+y^2} dx $$
$$ \frac{xdy – ydx}{x^2} = \frac{\sqrt{x^2+y^2}}{x^2} dx $$

Recall that $d\left(\frac{y}{x}\right) = \frac{xdy – ydx}{x^2}$.
Also, inside the square root on the RHS, factor out $x^2$:
$$ \frac{\sqrt{x^2(1+\frac{y^2}{x^2})}}{x^2} dx = \frac{x\sqrt{1+(\frac{y}{x})^2}}{x^2} dx = \frac{1}{x}\sqrt{1+\left(\frac{y}{x}\right)^2} dx $$

So the equation becomes:
$$ d\left(\frac{y}{x}\right) = \frac{1}{x} \sqrt{1+\left(\frac{y}{x}\right)^2} dx $$
$$ \frac{d\left(\frac{y}{x}\right)}{\sqrt{1+\left(\frac{y}{x}\right)^2}} = \frac{dx}{x} $$

Integrating both sides:
$$ \int \frac{dt}{\sqrt{1+t^2}} = \int \frac{dx}{x} $$
$$ \ln|t + \sqrt{1+t^2}| = \ln|x| + \ln C $$
$$ t + \sqrt{1+t^2} = Cx $$

Substitute $t = \frac{y}{x}$:
$$ \frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}} = Cx $$
$$ y + \sqrt{x^2+y^2} = Cx^2 $$

Using the initial condition $y(1) = 0$:
$$ 0 + \sqrt{1^2+0} = C(1)^2 \Rightarrow C = 1 $$

The particular solution is:
$$ y + \sqrt{x^2+y^2} = x^2 $$

We need to find $y(3)$. Let $y(3) = k$.
$$ k + \sqrt{3^2+k^2} = 3^2 $$
$$ \sqrt{9+k^2} = 9 – k $$

Squaring both sides (condition $9-k \ge 0 \Rightarrow k \le 9$):
$$ 9 + k^2 = (9 – k)^2 $$
$$ 9 + k^2 = 81 + k^2 – 18k $$
$$ 9 = 81 – 18k $$
$$ 18k = 72 \Rightarrow k = 4 $$

So, $y(3) = 4$.

Ans. (1)