JEE PYQ: Differential Equation Question 12
Question 12 - 2021 (25 Feb Shift 2)
If the curve $y = y(x)$ represented by the solution of the differential equation $(2xy^2 - y),dx + x,dx = 0$, passes through the intersection of the lines $2x - 3y = 1$ and $3x + 2y = 8$, then $|y(1)|$ is equal to ______.
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Answer: 1
Solution
Rewrite: $\frac{dx}{dy} + 2y^2 - \frac{y}{x} = 0$. This is $-\frac{1}{y^2}\frac{dy}{dx} + \frac{1}{y}\cdot\frac{1}{x} = 2$. Let $z = \frac{1}{y}$. I.F. $= x$. Solution: $\frac{x}{y} = x^2 + c$. Lines intersect at $(2,1)$: $c = 2-4 = -2$. So $\frac{x}{y} = x^2 - 2$. At $x = 1$: $\frac{1}{y} = -1$, $y = -1$, $|y(1)| = 1$.
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