JEE Advanced 2022 — Paper 2 — MATH-I solution

JEE Advanced2022Paper 2MATH-I

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 $10 y (2)$ is ______.

Your answer

Solution

Separate the variables: $\frac{d y}{y ( y - 4 )} = \frac{d x}{x}$ .

Partial fractions yield $\frac{1}{4} (\frac{1}{y - 4} - \frac{1}{y}) d y = \frac{d x}{x}$ .

Integrating: $\frac{1}{4} ln \frac{y - 4}{y} = ln ∣ x ∣ + C$ .

Applying $y (1) = 2$ : $\frac{y - 4}{y} = - 1$ at $x = 1$ , giving $C = 0$ . Hence $\frac{y - 4}{y} = x^{4}$ .

Since the slope $\frac{d y}{d x} = \frac{y ( y - 4 )}{x}$ is never zero, $y$ stays in $(0, 4)$ , where $\frac{y - 4}{y} < 0$ . So $\frac{4 - y}{y} = x^{4}$ , giving $y = \frac{4}{1 + x ^{4}}$ .

At $x = 2$ : $y = \frac{4}{1 + 4} = \frac{4}{5}$ . Therefore $10 y (2) = 8$ .

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