Differential Equations Question 311
Question: The solution of $ \frac{dy}{dx}=\frac{e^{x}({{\sin }^{2}}x+\sin 2x)}{y(2\log y+1)} $ is
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:
[a] $ \frac{dy}{dx}=\frac{e^{x}(sin^{2}x+sin2x)}{y(2logy+1)} $
$ \Rightarrow \int{(2ylogy+y)dy=\int{e^{x}(sin^{2}x+sin2x)dx}} $
On integrating by parts, we get $ y^{2}(logy)=e^{x}{{\sin }^{2}}x+c. $
BETA
