JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
If $y\frac{dy}{dx}=x\left[\frac{y^{2}}{x^{2}}+\frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi'\left(\frac{y^{2}}{x^{2}}\right)}\right]$, $x>0$, $\phi>0$, and $y(1)=-1$, then $\phi\left(\frac{y^{2}}{4}\right)$ is equal to :
Differential Equation
Solution
Concept. The equation is built around $v=\frac{y^2}{x^2}$; rewrite it in $v$ and separate. Formulas used. (1) $y^2=vx^2\Rightarrow2y\frac{dy}{dx}=x^2\frac{dv}{dx}+2xv$(2) $\int\dfrac{\phi'(v)}{\phi(v)}\,dv=\ln\phi(v)$ (since $\phi>0$)Step 1 - Substitute. By (1), $y\frac{dy}{dx}=xv+\frac{x^2}{2}\frac{dv}{dx}$. The equation becomes $$xv+\frac{x^2}{2}\frac{dv}{dx}=xv+x\,\frac{\phi(v)}{\phi'(v)}\ \Longrightarrow\ \frac{\phi'(v)}{\phi(v)}\,dv=\frac{2}{x}\,dx.$$ Step 2 - Integrate. By (2), $\ln\phi(v)=2\ln x+c$, so $\phi\left(\frac{y^2}{x^2}\right)=Cx^2$. Step 3 - Use $y(1)=-1$. At $x=1$, $v=1$, so $C=\phi(1)$. Step 4 - Put $x=2$. $\phi\left(\frac{y^2}{4}\right)=4\,\phi(1)$. Answer: (C)