Differential Equations Question 229
Question: The degree of the differential equation $ \frac{d^{2}y}{dx^{2}}-\sqrt{\frac{dy}{dx}-3}=x $ is
[Orissa JEE 2004]
Options:
A) 2
B) 1
C) ½
D) 3
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{d^{2}y}{dx^{2}}-\sqrt{\frac{dy}{dx}-3}=x\Rightarrow \frac{d^{2}y}{dx^{2}}-x=\sqrt{\frac{dy}{dx}-3} $
Squaring both sides, we get $ {{( \frac{d^{2}y}{dx^{2}}-x )}^{2}}=( \frac{dy}{dx}-3 ) $
Therefore $ {{( \frac{d^{2}y}{dx^{2}} )}^{2}}+x^{2}-2x\frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}-3 $ . Clearly, degree = 2.
BETA
