Differential Equations Question 193
Question: If integrating factor of $ x(1-x^{2})dy+(2x^{2}y-y-ax^{3})dx=0 $ is $ _{e}\int pdx $ , then P is equal to
Options:
A) $ \frac{2x^{2}-ax^{3}}{x(1-x^{2})} $
B) $ 2x^{3}-1 $
C) $ \frac{2x^{2}-a}{ax^{3}} $
D) $ \frac{2x^{2}-1}{x(1-x^{2})} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ x(1-x^{2})dy+(2x^{2}y-y-ax^{3})dx=0 $
Or $ x(1-x^{2})\frac{dy}{dx}+2x^{2}y-y-ax^{3}=0 $ Or $ x(1-x^{2})\frac{dy}{dx}+y(2x^{2}-1)=ax^{3} $
Or $ \frac{dy}{dx}+\frac{2x^{2}-1}{x(1-x^{2})}y=\frac{ax^{3}}{x(1-x^{2})} $
Which is of the form $ \frac{dy}{dx}+Py=Q. $ Its integrating factor is $ {e^{\int{Pdx}}} $ .
Here, $ P=\frac{2x^{2}-1}{x(1-x^{2})} $
BETA
