Differential Equations Question 316
Question: A solution of the differential equation $ {{( \frac{dy}{dx} )}^{2}}-x\frac{dy}{dx}+y=0 $ is
[IIT 1999; Karnataka CET 2002]
Options:
A) $ y=2 $
B) $ y=2x $
C) $ y=2x-4 $
D) $ y=2x^{2}-4 $
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation can be written as $ y=x\frac{dy}{dx}-{{( \frac{dy}{dx} )}^{2}} $
If $ \frac{dy}{dx}=p, $ then $ y=px-p^{2} $
Differentiating w.r.t. x, we get
$ p=p+x\frac{dp}{dx}-2p\frac{dp}{dx} $
Therefore $ \frac{dp}{dx}(x-2p)=0 $
Therefore $ \frac{dp}{dx}=0 $
Integrating w.r.t. x, we get $ p=c $
$ \therefore \frac{dy}{dx}=c $ ;
$ \therefore y=cx-c^{2} $
If $ c=2 $ , then $ y=2x-4 $ .
BETA
