Discussion Forum



1)

The particular solution of the differential equation  $\log\left(\frac{dy}{dx}\right)=x$   , when x=0 , y=1 is .....


A) $y=e^{x}+2$

B) $y=-e^{x}$

C) $y=-e^{x}+2$

D) $y=e^{x}$

Answer:

Option D

Explanation:

 We have, differential equations,

  $\log(\frac{dy}{dx})=x\Rightarrow\frac{dy}{dx}=e^{x}$

$\Rightarrow$    $ dy=e^{x}dx$

 integrating  on both sides we get

              $\int  dy=\int e^{x}dx$

$\Rightarrow $    $ y= e^{x}+ C$  ........(i)

 On putting x=0, y=1  is Eq .(i) , we get

 1= e0+ C $\Rightarrow$ C=0

 Now, the particular solution of the given differential is y= ex