JEE Main · 2005 · 08 May 2005 · Offline · Differential Equations

x \frac{d y}{d x} = y (lo g y - lo g x + 1)

, then the solution of the equation is

Answer: (C)

Solution

Write the equation as

\frac{d y}{d x} = \frac{y}{x} (ln \frac{y}{x} + 1)

and substitute

y = v x

, so that

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

. This gives

v + x \frac{d v}{d x} = v (ln v + 1)

, which reduces to

x \frac{d v}{d x} = v ln v

. Separating variables,

\frac{d v}{v ln v} = \frac{d x}{x}

; integrating the left side with

ln v

as the new variable gives

ln (ln v) = ln x + ln c

, so

ln v = c x

. Returning to

v = \frac{y}{x}

, the solution is

ln \frac{y}{x} = c x

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0$ , $y (0) = 1$ . Then the value of $y (1)$ is :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation $sec x \frac{dy}{dx} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2})$ , $y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to:
JEE Main 2026View solution →
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
JEE Main 2026View solution →
If $y = y (x)$ satisfies the differential equation
$16 (x + 9 x) (4 + 9 + x) cos y d y = (1 + 2 sin y) d x$ , $x > 0$ and $y (256) = \frac{π}{2}$ , $y (49) = α$ , then $2 sin α$ is equal to :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation
$x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0$ , $x > 0$ , $y (\frac{π}{2}) = 0$ .

Then $π^{4} y (\frac{π}{3})$ is equal to :
JEE Main 2026View solution →
If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0, x > 2$ , passes through the point (3,15), then the local maximum value of f is
JEE Main 2026View solution →