Differential Equations Question 290
Question: Which one of the following differential equations represents the family of straight lines which are at unit distance from the origin-
Options:
A) $ {{( y-x\frac{dy}{dx} )}^{2}}=1-{{( \frac{dy}{dx} )}^{2}} $
B) $ {{( y+x\frac{dy}{dx} )}^{2}}=1+{{( \frac{dy}{dx} )}^{2}} $
C) $ {{( y-x\frac{dy}{dx} )}^{2}}=1+{{( \frac{dy}{dx} )}^{2}} $
D) $ {{( y+x\frac{dy}{dx} )}^{2}}=1-{{( \frac{dy}{dx} )}^{2}} $
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Answer:
Correct Answer: C
Solution:
[c] $ y=mx+c $ (Equation of straight lie) $ \frac{dy}{dx}=m $ and $ mx-y+c=0 $ is at unit distance from origin.
$ \therefore \frac{| m(0)-(0)+c |}{\sqrt{m^{2}+{{(-1)}^{2}}}}=1\Rightarrow c=\sqrt{1+m^{2}} $ Now: $ {{[ y-x\frac{dy}{dx} ]}^{2}}={{[mx+c-xm]}^{2}}=c^{2}=1+m^{2} $ also, $ {{[ y+x\frac{dy}{dx} ]}^{2}}={{[mx+c+mx]}^{2}}={{[2mx+\sqrt{1+m^{2}}]}^{2}} $ also, $ 1-{{( \frac{dy}{dx} )}^{2}}=1-m^{2} $ and $ 1+{{( \frac{dy}{dx} )}^{2}}=1+m^{2} $
$ \Rightarrow {{[ y-x\frac{dy}{dx} ]}^{2}}=1+{{( \frac{dy}{dx} )}^{2}} $
BETA
