If $y=y(x)$ is the solution curve of the differential equation $\left(x^{2}-4\right) d y-\left(y^{2}-3 y\right) d x=0$, $x>2, y(4)=\frac{3}{2}$ and the slope of the curve is never zero, then the value of $\mathrm{y}(10)$ equals :

Differential Equations — JEE Main · 2024 · 27 Jan 2024 · Shift 2 Question with Solution

y = y (x)

is the solution curve of the differential equation

(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0

x > 2, y (4) = \frac{3}{2}

and the slope of the curve is never zero, then the value of

y (10)

equals :

Answer: (A)

(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0

, i.e.

\frac{d y}{y ^{2} - 3 y} = \frac{d x}{x ^{2} - 4}

, for

x > 2

.

Using partial fractions:

\frac{1}{y ( y - 3 )} = \frac{1}{3} (\frac{1}{y - 3} - \frac{1}{y}), \frac{1}{x ^{2} - 4} = \frac{1}{4} (\frac{1}{x - 2} - \frac{1}{x + 2}) .

Integrating both sides:

\frac{1}{3} ln \frac{y - 3}{y} = \frac{1}{4} ln \frac{x - 2}{x + 2} + C .

x = 4, y = \frac{3}{2}

\frac{y - 3}{y} = \frac{- 3/2}{3/2} = - 1

so the left side is

0

;

\frac{x - 2}{x + 2} = \frac{1}{3}

. So

0 = \frac{1}{4} ln \frac{1}{3} + C \Rightarrow C = \frac{1}{4} ln 3.

x = 10

\frac{x - 2}{x + 2} = \frac{8}{12} = \frac{2}{3}

\frac{1}{3} ln \frac{y - 3}{y} = \frac{1}{4} ln \frac{2}{3} + \frac{1}{4} ln 3 = \frac{1}{4} ln 2 \Rightarrow ln \frac{y - 3}{y} = \frac{3}{4} ln 2 = ln 2^{3/4} .

\frac{y - 3}{y} = 8^{1/4}

. Since the slope

\frac{d y}{d x} = \frac{y ( y - 3 )}{x ^{2} - 4}

never vanishes and

y (4) = \frac{3}{2} \in (0, 3)

y

cannot cross

0

3

for

x > 2

, so

y

stays in

(0, 3)

and

\frac{3 - y}{y} = 8^{1/4}

y (1 + 8^{1/4}) = 3 \Rightarrow y = \frac{3}{1 + 8 ^{1/4}} .

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option