Mathematics · Differential Equations

JEE Main 2024 — 8 April, Shift 2 — Question 27

Lei α∣x∣=∣y∣exy−β,α,β∈N\alpha|x|=|y| e^{x y-\beta}, \alpha, \beta \in N be the solution of the differential equation

xdy−ydx+xy(xdy+ydx)=0x d y-y d x+x y(x d y+y d x)=0, y(1)=2y(1)=2. Then α+β\alpha+\beta is equal to

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

a∣x∣=∣y∣eyx−β,a,b∈Na|x|=|y| e^{y x-\beta}, a, b \in N

xdy−ydx+xy(xdy+ydx)=0x d y-y d x+x y(x d y+y d x)=0

dyy−dxx+(xdy+ydx)=0\frac{d y}{y}-\frac{d x}{x}+(x d y+y d x)=0

ℓnn⁡∣y∣−ln⁡∣x∣+xy=c\ell \operatorname{nn}|\mathrm{y}|-\ln |\mathrm{x}|+\mathrm{xy}=\mathrm{c}

y(1)=2\mathrm{y}(1)=2

ℓn∣2∣−0+2=c\ell \mathrm{n}|2|-0+2=\mathrm{c}

c=2+ℓn⁡2\mathrm{c}=2+\operatorname{\ell n} 2

ℓn∣y∣−ℓn∣x∣+xy=2+ℓn2\ell \mathrm{n}|\mathrm{y}|-\ell \mathrm{n}|\mathrm{x}|+\mathrm{xy}=2+\ell \mathrm{n} 2

ln⁡∣x∣=ln⁡∣y2∣−2+xy\ln |x|=\ln \left|\frac{y}{2}\right|-2+x y

∣x∣=∣y2∣exy−2|x|=\left|\frac{y}{2}\right| e^{x y-2}

2∣x∣=∣y∣exy−22|\mathrm{x}|=|\mathrm{y}| \mathrm{e}^{\mathrm{xy}-2}

α=2,β=2,α+β=4\alpha=2, \quad \beta=2, \quad \alpha+\beta=4

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Differential Equations
Topic
Methods of solving a First Order,First Degree Differential
Lei α x = y e x y-β , α, β in N be the solution of the differential… | JEE Main 2024 PYQ with Solution · DhiX AI