Differential Equations Question 192
Question: The solution of $ \frac{dy}{dx}=\frac{x^{2}+y^{2}+1}{2xy} $ satisfying y(1)=1 is given by
Options:
A) a system of parabolas
B) a system of circles
C) $ y^{2}=x(1+x)-1 $
D) $ {{(x-2)}^{2}}+{{(y-3)}^{2}}=5 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Rewriting the given equation as $ 2xy\frac{dy}{dx}-y^{2}=1+x^{2} $
Or $ 2y\frac{dy}{dx}-\frac{1}{x}y^{2}=\frac{1}{x}+x $ Putting $ y^{2}=u, $
we have $ \frac{du}{dx}-\frac{1}{x}u=\frac{1}{x}+x $ I.F. $ ={e^{-\int{\frac{1}{x}dx}}}=\frac{1}{x} $
Thus, solution is u $ \frac{1}{x}=\int{( \frac{1}{x^{2}}+1 )}dx=-\frac{1}{x}+x+C $ Or $ y^{2}=(x^{2}-1)+Cx $
Since y(1) =1, we get C=1. Hence, $ y^{2}=x(1+x)-1 $ which represents a systems of hyperbola.
BETA
