JEE Main 2026 Jan 22 Shift 1, Mathematics Q19: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Jan 22, Shift 1Mathematics

Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

Solution

Divide both sides by $x^{2}$ : $\frac{x d y - y d x}{x ^{2}} = \frac{1 + ( y / x ) ^{2}}{x} d x$ .

The left side is $d (y / x)$ . Substitute $u = y / x$ : $\frac{d u}{1 + u ^{2}} = \frac{d x}{x}$ .

Integrate: $ln (u + 1 + u^{2}) = ln (k x)$ , i.e. $u + 1 + u^{2} = k x$ . Multiplying by $x$ : $y + x^{2} + y^{2} = k x^{2}$ .

Use $y (1) = 0$ : $0 + 1 = k$ , so $k = 1$ . Solution: $y + x^{2} + y^{2} = x^{2}$ .

At $x = 3$ : $9 + y^{2} = 9 - y$ . Square: $9 + y^{2} = 81 - 18 y + y^{2} \Rightarrow 18 y = 72 \Rightarrow y = 4$ .

Concept behind this question

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