Differential Equations Question 20
Question: The solution of $ \frac{dy}{dx}=\frac{e^{x}({{\sin }^{2}}x+\sin 2x)}{y(2\log y+1)} $ is
[AISSE 1990]
Options:
A) $ y^{2}(\log y)-e^{x}{{\sin }^{2}}x+c=0 $
B) $ y^{2}(\log y)-e^{x}{{\cos }^{2}}x+c=0 $
C) $ y^{2}(\log y)+e^{x}{{\cos }^{2}}x+c=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=\frac{e^{x}({{\sin }^{2}}x+\sin 2x)}{y(2\log y+1)} $
Therefore $ \int _{{}}^{{}}{(2y\log y+y)dy=\int _{{}}^{{}}{e^{x}({{\sin }^{2}}x+\sin 2x})dx} $
On integrating by parts, we get $ y^{2}(\log y)=e^{x}{{\sin }^{2}}x+c $ .
BETA
