Differential Equations Question 198
Question: The solution of differential equation $ \frac{x+y\frac{dy}{dx}}{y-x\frac{dy}{dx}}=\frac{x{{\cos }^{2}}(x^{2}+y^{2})}{y^{3}} $ is
Options:
A) $ \tan (x^{2}+y^{2})=\frac{x^{2}}{y^{2}}+c $
B) $ cot(x^{2}+y^{2})=\frac{x^{2}}{y^{2}}+c $
C) $ tan(x^{2}+y^{2})=\frac{y^{2}}{x^{2}}+c $
D) $ cot(x^{2}+y^{2})=\frac{y^{2}}{x^{2}}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The given equation can be written as $ \frac{xdx+ydy}{(ydx-xdy)/y^{2}}=y^{2}\frac{x}{y^{3}}{{\cos }^{2}}(x^{2}+y^{2}) $ Or $ \frac{xdx+ydx}{{{\cos }^{2}}(x^{2}+y^{2})}=\frac{x}{y}( \frac{ydx-xdy}{y^{2}} ) $ Or $ \frac{1}{2}{{\sec }^{2}}(x^{2}+y^{2})d(x^{2}+y^{2})=\frac{x}{y}d( \frac{x}{y} ) $
On integrating, we get $ \frac{1}{2}\tan (x^{2}+y^{2})=\frac{1}{2}{{( \frac{x}{y} )}^{2}}+\frac{c}{2} $ Or $ \tan (x^{2}+y^{2})=\frac{x^{2}}{y^{2}}+c $
BETA
