Differential Equations Question 56
Question: The solution of $ {y}’-y=1,\ y(0)=-1 $ is given by $ y(x)= $
[MP PET 2000]
Options:
A) $ -\exp (x) $
B) $ -\exp (-x) $
- 1
D) $ \exp (x)-2 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{dy}{dx}-y=1 $
Therefore $ \frac{dy}{1+y} = dx $
Integrating both sides $ \ln (1+y)=x+c $
Therefore $ 1+y=e^{x}.e^{c} $
$ \because x=0,y=-1 $ .Then, $ 0=e.e^{c} $
Therefore $ e^{c} \neq 0 $
Therefore solution $ 1+y=e^{x}\times 0\Rightarrow y(x)=-1 $ .
BETA
