Select Page

# Function

Tagged:

Viewing 2 posts - 1 through 2 (of 2 total)
• Author
Posts
• #29479
Aniruddha Bardhan
Participant

###### Attachments:
You must be logged in to view attached files.
#29491
Alpha Beta
Participant

Are you sure it’s not g'(x)-g(x)-2e^x=0 ?

Partially differentiating the equation with respect to x and y separately, we get

g'(x+y)=g'(x)e^y+g(y)e^x

and, g'(x+y)=g(x)e^y+g'(y)e^x

so,  g'(x)e^y+g(y)e^x=g(x)e^y+g'(y)e^x

or, [g'(x)-g(x)]e^y=[g'(y)-g(y)]e^x

suppose, for some x, g'(x)=g(x).

Then,  for all y,  0=[g'(y)-g(y)]e^x.

Since, e^x is always positive, g'(y)-g(y)=0,

implying g(y)=e^(y+c) for all y, and some

constant c.

From g'(0)=2, we get g(y)=2e^y for all y.

From the equation, 2e^(x+y)=4e^(x+y) for all

x,y. Or, 2e^(x+y)=0, which is impossible.

Therefore, g'(x)-g(x) is non-zero for all x.

or, [g'(y)-g(y)]/[g'(x)-g(x)]=e^y/e^x, for all x,y.

So, g'(x)-g(x)=ke^x for all x, where k is a constant.

Solving, and imposing the 2 conditions ( g'(0)=2, and g(x+y)=g(x)e^y+g(y)e^x), we get

g(x)=2xe^x, implying, g'(x)-g(x)-2e^x=0.

Viewing 2 posts - 1 through 2 (of 2 total)
• You must be logged in to reply to this topic.