JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q77

MathematicsSingle correct+4 / -1Question 77 of 90

The solution of the differential equation $y\,dx-(x+2y^2)\,dy=0$ is $x=f(y)$ . If $f(-1)=1$ , then $f(1)$ is equal to :

Answer and worked solution

Official answer: B

$\dfrac{y\,dx-x\,dy}{y^2}=2\,dy$ , i.e. $d\left(\frac xy\right)=2\,dy$ , so $\frac xy=2y+c$ . At $(1,-1)$ : $-1=-2+c$ , so $c=1$ . At $y=1$ : $x=3$ .

JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 77