Differential Equations Question 208
Question: The solution to the differential equation $ \frac{dy}{dx}=\frac{x+y}{x} $ satisfying the condition y(1)=1 is
Options:
A) $ y=\ln x+x $
B) $ y=x\ln x+x^{2} $
C) $ y=x{e^{(x-1)}} $
D) $ y=x\ln x+x $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given equation is $ \frac{dy}{dx}=1+\frac{y}{x}. $ Let $ y=vx $
$ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} $
$ \Rightarrow v+x\frac{dv}{dx}=1+v $
$ \Rightarrow dv=\frac{dx}{x} $
$ \therefore v=\ln x+c $
$ \Rightarrow \frac{y}{x}=\ln x+c $ Since, y (1) =1, c=1, so we have $ y=x\ln x+x $
BETA
