Differential Equations Question 71
Question: Solution of $ \frac{dy}{dx}=\frac{x\log x^{2}+x}{\sin y+y\cos y} $ is
[EAMCET 2003]
Options:
A) $ y\sin y=x^{2}\log x+c $
B) $ y\sin y=x^{2}+c $
C) $ y\sin y=x^{2}+\log x+c $
D) $ y\sin y=x\log x+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=\frac{x\log x^{2}+x}{\sin y+y\cos y} $ . Separating the variables and integrating $ \int{(\sin y+y\cos y)dy=\int{(x\log x^{2}+x)dx}} $
Therefore $ -\cos y+y\sin y+\cos y $
$ =\frac{x^{2}}{2}\log x^{2}-\int{\frac{x^{2}}{2}.\frac{1}{x^{2}}.2xdx+\int{xdx+c}} $
Therefore $ y\sin y=\frac{x^{2}}{2}2\log x-\int{xdx+\int{xdx+c}} $
Therefore $ y\sin y=x^{2}\log x+c $ .
BETA
