Differential Equations Question 270
Question: The general solution the differential equation $ \frac{dy}{dx}-\frac{\tan y}{1+x}={{(1+xe)}^{x}}\sec y $ is
Options:
A) $ \sin (1+x)=y(e^{x}+c) $
B) $ y\sin (1+x)=ce^{x} $
C) $ (1+x)\sin y=e^{x}+c $
D) $ \sin y=(1+x)(e^{x}+c) $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Divide the equation by sec y $ \cos y\frac{dy}{dx}-\frac{\sin y}{1+x}=(1+x)e^{x} $
Put $ \sin y=z\Rightarrow \cos y\frac{dy}{dx}=\frac{dz}{dx} $
then $ \frac{dz}{dx}-( \frac{1}{1+x} )z=(1+x)e^{x} $ I.E. $ ={e^{-\int{\frac{1}{1+x}dx}}}={e^{-\log (1+x)}}=\frac{1}{1+x} $
The solution is $ z( \frac{1}{1+x} )=e^{x}+c\Rightarrow \sin y=(1+x)(e^{x}+c) $
BETA
