The solution of the differential equation $\left(x^{2}+y^{2}\right) d x-5 x y d y=0, y(1)=0$, is :

(A) $\left|x^{2}-4 y^{2}\right|^{5}=x^{2}$

(B) $\left|x^{2}-2 y^{2}\right|^{6}=x$

(C) $\left|x^{2}-4 y^{2}\right|^{6}=x$

(D) $\left|x^{2}-2 y^{2}\right|^{5}=x^{2}$

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Differential Equations

The solution of the differential equation

(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0

, is :

Answer: (A)

Homogeneous: put

y = v x

. Then

(1 - 4 v^{2}) d x = 5 v x d v \Rightarrow \frac{d x}{x} = \frac{5 v}{1 - 4 v ^{2}} d v

. Integrating,

8 ln ∣ x ∣ + 5 ln ∣1 - 4 v^{2} ∣ = const

, i.e.

x^{8} ∣1 - 4 v^{2} ∣^{5} = K

. With

1 - 4 v^{2} = \frac{x ^{2} - 4 y ^{2}}{x ^{2}}

this becomes

∣ x^{2} - 4 y^{2} ∣^{5} = K x^{2}

; and

y (1) = 0 \Rightarrow K = 1

. So

∣ x^{2} - 4 y^{2} ∣^{5} = x^{2}

— option (1).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option

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