Differential Equations

JEE Main 2025 (Online) 22nd January Morning ShiftMCQMedium📈 Topic in 71% papers+4−1

Ques. 1JEE Main 2025, 22 January (Morning Shift)

Let x=x(y) be the solution of the differential equation y2 dx+ (x- 1y )dy=0. If x(1)=1, then x ( 12 ) is :

32+e
12+e
3+e
3-e

Answer

(4) 3-e

Solution

With y as the independent variable the equation is linear: dxdy+ xy2= 1y3, I.F. =e-1/y.

x e-1/y= e-1/yy3dy; put t= 1y: =- te-tdt=(t+1)e-t+C

x= 1y+1+Ce1/y; x(1)=1 C=- 1e

x ( 12 )=2+1- 1e e2=3-e

Answer: (4)

Keep practising — Differential Equations

Ques. 2JEE Main 2025, 22 January (Evening Shift)

If x=f(y) is the solution of the differential equation (1+y2 )+ (x-2e -1y ) dydx=0, y (- 2, 2 ) with f(0)=1, then f ( 1 3 ) is equal to :

e /4
e /12
e /6
e /3

Ques. 3JEE Main 2025, 22 January (Evening Shift)

Let y=f(x) be the solution of the differential equation dydx+ xyx2-1= x6+4x 1-x2, -1<x<1 such that f(0)=0. If 6 -1/21/2f(x) dx=2 - then 2 is equal to .

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