JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
The solution curve of the differential equation $$y\frac{dx}{dy} = x(\log_e x - \log_e y + 1), x > 0, y > 0$$ passing through the point $$(e, 1)$$ is
$$y\frac{dx}{dy}=x(\ln x-\ln y+1)=x\ln\frac{ex}{y}$$.
Let $$v=x/y$$: $$x=vy$$, $$\frac{dx}{dy}=v+y\frac{dv}{dy}$$.
$$y(v+y\frac{dv}{dy})=vy\ln(ev)$$. $$v+y\frac{dv}{dy}=v\ln(ev)=v(1+\ln v)$$.
$$y\frac{dv}{dy}=v\ln v$$. $$\frac{dv}{v\ln v}=\frac{dy}{y}$$. $$\ln(\ln v)=\ln y+C$$.
$$\ln v=ky$$. At $$(e,1)$$: $$v=e/1=e$$, $$\ln e=k(1)$$, $$k=1$$.
$$\ln(x/y)=y$$.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
