Differential Equations Question 275
Question: Solution of the differential equation $ \frac{dx}{dy}-\frac{x\log x}{1+\log x}=\frac{e^{y}}{1+\log x’} $ if $ y(1)=0 $ , is
Options:
A) $ x^{x}={e^{ye^{y}}} $
B) $ e^{y}={x^{e^{y}}} $
C) $ x^{x}=y{e^{^{y}}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ (1+logx)\frac{dx}{dy}-x\log x=e^{y} $ putting x log $ x\log x=t\Rightarrow (1+logx)dx=dt $
$ \therefore \frac{dt}{dy}-t=e^{y} $ Now, I.F. $ ={e^{\int{-1dy}}}={e^{-y}} $
$ \Rightarrow t{e^{-y}}=\int{{e^{-y}}e^{y}dy+C} $
$ \Rightarrow t=Ce^{y}+ye^{y}\Rightarrow x\log x=(C+y)e^{y}, $ Since, $ y(1)=0, $ Then $ 0=(C+0)1\Rightarrow C=0 $
$ \therefore ye^{y}=x\log x\Rightarrow x^{x}={e^{ye^{y}}} $
BETA
