Differential Equations Question 65
Question: The solution of the equation $ (x+2y^{3})\frac{dy}{dx}-y=0 $ is
[MP PET 1998; 2002]
Options:
A) $ y(1-xy)=Ax $
B) $ y^{3}-x=Ay $
C) $ x(1-xy)=Ay $
D) $ x(1+xy)=Ay $ Where A is any arbitrary constant
Show Answer
Answer:
Correct Answer: B
Solution:
$ (x+2y^{3})\frac{dy}{dx}=y $
Therefore $ \frac{dy}{dx}=\frac{y}{x+2y^{3}} $
Therefore $ \frac{dx}{dy}=\frac{x+2y^{3}}{y} $ or $ \frac{dx}{dy}-\frac{x}{y}=2y^{2} $ , which is a linear equation of the form $ \frac{dx}{dy}+Px=Q $
So, integrating factor (I.F.) $ ={e^{-\int _{{}}^{{}}{\frac{1}{y}dy}}} $ and solution is $ x\frac{1}{y}=\int _{{}}^{{}}{\frac{1}{y}2y^{2}dy+A=y^{2}+A} $
Therefore $ x=y^{3}+Ay $
Therefore $ y^{3}-x=Ay; $ where A can be $ -ve $ or $ +ve $ .
BETA
