IIT-JEE 2005 ScreeningQuestion 1
For the primitive integral equation $y\,dx+y^{2}\,dy=x\,dy$; $x\in R$, $y>0$, $y=y(x)$, $y(1)=1$, then $y(-3)$ is
Differential Equation
Solution
Concept. $y\,dx-x\,dy=y^2\,d\left(\frac xy\right)$. Formulas used. (1) $d\left(\dfrac xy\right)=\dfrac{y\,dx-x\,dy}{y^2}$Step 1 - Rewrite. $y\,dx-x\,dy=-y^2\,dy$. By (1), $d\left(\dfrac xy\right)=-dy$. Step 2 - Integrate. $\dfrac xy=-y+C$. $y(1)=1$ gives $C=2$. Step 3 - At $x=-3$. $-\dfrac3y=-y+2$, so $y^2-2y-3=0$, i.e. $(y-3)(y+1)=0$. With $y>0$: $y=3$. Answer: (A)