The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=-\left(\frac{\mathrm{x}^{2}+3 \mathrm{y}^{2}}{3 \mathrm{x}^{2}+\mathrm{y}^{2}}\right), \mathrm{y}(1)=0$ is

(C) $\log _{\mathrm{e}}|\mathrm{x}+\mathrm{y}|+\frac{2 \mathrm{xy}}{(\mathrm{x}+\mathrm{y})^{2}}=0$

(A) $\log _{\mathrm{e}}|\mathrm{x}+\mathrm{y}|-\frac{\mathrm{xy}}{(\mathrm{x}+\mathrm{y})^{2}}=0$

(B) $\log _{\mathrm{e}}|\mathrm{x}+\mathrm{y}|+\frac{\mathrm{xy}}{(\mathrm{x}+\mathrm{y})^{2}}=0$

(D) $\log _{\mathrm{e}}|\mathrm{x}+\mathrm{y}|-\frac{2 \mathrm{xy}}{(\mathrm{x}+\mathrm{y})^{2}}=0$

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Differential Equations

The solution of the differential equation

\frac{dy}{dx} = - (\frac{x ^{2} + 3 y ^{2}}{3 x ^{2} + y ^{2}}), y (1) = 0

Answer: (C)

The equation is homogeneous, so set

y = v x

. Then

v + x \frac{d v}{d x} = - \frac{1 + 3 v ^{2}}{3 + v ^{2}}

, giving

x \frac{d v}{d x} = - \frac{1 + 3 v ^{2}}{3 + v ^{2}} - v = - \frac{( v + 1 ) ^{3}}{3 + v ^{2}}

. Separating variables and writing

u = v + 1

\int \frac{u ^{2} - 2 u + 4}{u ^{3}} d u + \int \frac{d x}{x} = c

, which integrates to

ln ∣ v + 1∣ + \frac{2}{v + 1} - \frac{1}{( v + 1 ) ^{2}} + ln ∣ x ∣ = c

. Replacing

v = \frac{y}{x}

, the two rational terms combine as

\frac{2 x y}{( x + y ) ^{2}}

and the logarithms as

ln ∣ x + y ∣

, so

ln ∣ x + y ∣ + \frac{2 x y}{( x + y ) ^{2}} = c

. The condition

y (1) = 0

gives

c = 0

, hence

lo g_{e} ∣ x + y ∣ + \frac{2 x y}{( x + y ) ^{2}} = 0

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option