Let $y=y(x)$ be the solution curve of the differential equation, $\left(y^{2}-x\right) \frac{d y}{d x}=1$, satisfying $y(0)=1$. This curve intersects the $x$-axis at a point whose abscissa is:

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution curve of the differential equation,

(y^{2} - x) \frac{d y}{d x} = 1

, satisfying

y (0) = 1

. This curve intersects the

x

-axis at a point whose abscissa is:

Answer: (D)

The equation

(y^{2} - x) \frac{d y}{d x} = 1

is awkward in

y

but linear in

x

: rewrite it as

\frac{d x}{d y} + x = y^{2}

. The integrating factor is

e^{y}

, so

\frac{d}{d y} (x e^{y}) = y^{2} e^{y}

, and integrating the right side by parts twice gives

x e^{y} = e^{y} (y^{2} - 2 y + 2) + C

. The condition

y = 1

x = 0

gives

0 = e (1) + C

, so

C = - e

. The curve meets the

x

-axis where

y = 0

, at which

x = (0 - 0 + 2) - e = 2 - e

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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