Aspire Faculty ID #17953 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

Let the solution curve of the differential equation

$xdy - ydx = \sqrt{x^2 + y^2},dx,; x>0,; y(1)=0$

be $y = y(x)$. Then $y(3)$ is equal to

Solution

$xdy - ydx = \sqrt{x^2 + y^2},dx$

$\Rightarrow \frac{xdy - ydx}{x^2} = \frac{\sqrt{x^2 + y^2}}{x^2}dx$

$\Rightarrow d\left(\frac{y}{x}\right) = \sqrt{1 + \left(\frac{y}{x}\right)^2}\cdot \frac{dx}{x}$

$\Rightarrow \int \frac{d(y/x)}{\sqrt{1 + (y/x)^2}} = \int \frac{dx}{x}$

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

$\Rightarrow y + \sqrt{x^2 + y^2} = kx^2$

At $x=1,; y=0 \Rightarrow k=1$

$\Rightarrow y + \sqrt{x^2 + y^2} = x^2$

At $x=3$

$y + \sqrt{9 + y^2} = 9$

$\Rightarrow y = 4$
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