JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Differential Equations

If a curve passes through the point (

1, - 2

) and has slope of the tangent at any point ( x ,

y)

on it as

\frac{x ^{2} - 2 y}{x}

, then the curve also passes through the point :

Answer: (B)

Solution

Write the slope condition as

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

, a linear equation with integrating factor

x^{2}

. Then

\frac{d}{d x} (y x^{2}) = x^{3}

, so

y x^{2} = \frac{x ^{4}}{4} + C

. The point

(1, - 2)

gives

- 2 = \frac{1}{4} + C

, so

C = - \frac{9}{4}

and

y x^{2} = \frac{x ^{4}}{4} - \frac{9}{4}

. Testing

(3, 0)

0 = \frac{9}{4} - \frac{9}{4} = 0

, so the curve passes through

(3, 0)

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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