JEE PYQ: Differential Equation Question 40
Question 40 - 2019 (12 Apr Shift 2)
The general solution of the differential equation $(y^2-x^3),dx - xy,dy = 0$ $(x \neq 0)$ is (where $c$ is a constant of integration):
(1) $y^2 - 2x^2 + cx^3 = 0$
(2) $y^2 + 2x^3 + cx^2 = 0$
(3) $y^2 + 2x^2 + cx^3 = 0$
(4) $y^2 - 2x^3 + cx^2 = 0$
Show Answer
Answer: (2)
Solution
$\frac{dy}{dx} = \frac{y^2-x^3}{xy} = \frac{y}{x} - \frac{x^2}{y}$. Homogeneous-like. Actually: $y,dy - \frac{y^2}{x},dx = -x^2,dx$. $\frac{d(y^2/x)}{dx}\cdot x = …$. From the solution: $\frac{y^2}{x} = -x^2 + c_1$. Actually: $d\left(\frac{y^2}{x}\right) = -dx$… Let me just verify answer (2): $y^2 + 2x^3 + cx^2 = 0$, i.e. $\frac{y^2}{x^2} = -2x - c$. Differentiating $y^2 = -2x^3 - cx^2$: $2yy’ = -6x^2 - 2cx$. And from original: $xyy’ = y^2 - x^3$. Consistent with $c = 2c_1$.
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