Q20 / 54 · JEE Advanced 2022 (Paper 2) · Mathematics

If $y (x)$ is the solution of the differential equation $x d y - (y^{2} - 4 y) d x = 0$ for $x > 0, y (1) = 2$ , and the slope of the curve $y = y (x)$ is never zero, then the value of $10 y (\sqrt{2})$ is _____ .

Correct answer: 8

Solution

Given differential equation is $\frac{d y}{y^{2} - 4 y} = \frac{d x}{x}$
$\Rightarrow \frac{1}{4} (\frac{1}{y - 4} - \frac{1}{y}) d y = \frac{d x}{x}$
Integrating both sides, we get,
$\frac{1}{4} \ln |\frac{y - 4}{y}| = \ln |x| + c$
Given, $y (1) = 2$
i.e. $c = 0$
So, $\frac{y - 4}{y} = \pm x^{4}$ or, $y = \frac{4}{1 \pm x^{4}}$
Considering $y = \frac{4}{1 + x^{4}}$
So, $10 y (\sqrt{2}) = 10 (\frac{4}{1 + 4}) = 8$

Hence, the correct answer is 8.

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Solution

Relevant Questions

If $y (x)$ is the solution of the differential equation $x d y - (y^{2} - 4 y) d x = 0$ for $x > 0, y (1) = 2$ , and the slope of the curve $y = y (x)$ is never zero, then the value of $10 y (\sqrt{2})$ is _____ .