Differential Equations Question 153
Question: The solution of the differential equation $ \frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}} $ is
Options:
A) $ ay^{2}={e^{x^{2}/y^{2}}} $
B) $ ay={e^{x/y}} $
C) $ y={e^{x^{2}}}+{e^{y^{2}}}+c $
D) $ y={e^{x^{2}}}+y^{2}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ \frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}} $ . Put $ y=vx $ ; $ \frac{dy}{dx}=v+x.\frac{dv}{dx} $
$ \therefore v+x\frac{dv}{dx}=\frac{(x)(vx)}{x^{2}+v^{2}x^{2}} $
Therefore $ v+x.\frac{dv}{dx}=\frac{v}{1+v^{2}} $
Therefore $ x\frac{dv}{dx}=\frac{-v^{3}}{1+v^{2}} $
Therefore $ \frac{(1+v^{2})}{v^{3}}dv=-\frac{dx}{x} $
Therefore $ ( \frac{1}{v^{3}}+\frac{1}{v} )dv=-\frac{dx}{x} $
Integrating both sides, $ \int _{{}}^{{}}{\frac{dv}{v^{3}}}+\int _{{}}^{{}}{\frac{dv}{v}}=-\int _{{}}^{{}}{\frac{dx}{x}} $
Therefore $ -\frac{1}{2v^{2}}+\log v=-\log x-\log c $
Therefore $ -\frac{x^{2}}{2y^{2}}+\log y=-\log c $
Therefore $ \log cy=\frac{x^{2}}{2y^{2}} $
Therefore $ cy={e^{x^{2}/2y^{2}}} $
Therefore $ c^{2}y^{2}={e^{x^{2}/y^{2}}} $
$ \therefore y^{2}a={e^{x^{2}/y^{2}}} $ , where $ c^{2}=a $ .
BETA
