1. ## [SOLVED] separable equation

hey there.. i'm stuck in the calculation below..

dy/dx = ye^x y(0) = 2e
∫1/y dy = ∫e^x dx
ln y + C = (e^x) + C
how to proceed so that my answer is y = 2e^(e^x)

2. Originally Posted by nameck
hey there.. i'm stuck in the calculation below..

dy/dx = yex
Is that y*e*x or y*e^x?

3. Since you have ln y = (e^x)+c, you need to free the y from the ln. How you do that raise everything to the power of e.

From that, you obtain e^(ln y) = e^[(e^x)+c].

y = e^[(e^x)+c]

The last step before substitution is understanding that e^(a+b) can be written as (e^a)*(e^b) and that when you have e^c is just another constant and can be written just as C

4. Ok… so I proceed…
y = e^ [(e^x) + C]
when x = 0; y =2e

hence, 2e = e^ [(e^0) + C]
2e = (e^1) + C

Apply ln to both sides
2 = 1 + ln C
C = e^1
C = 0

Hence, y = e^(e^x)

Still didn’t get the answer provided..

5. Originally Posted by nameck
Ok… so I proceed…
y = e^ [(e^x) + C]
I made sure to mention how exponents work in my post but here is it again.

For example if we have 2^(2+x), then we could write (2^2)*(2^x). This works because 2*2=(2^1)*(2^1)=2^(1+1)=2^2=2*2.

Therefore, e^[e^x + c]= e^(e^x)*e^c and e^c is just another constant so e^c=C. Now we have (e^(e^x))*c=C*(e^(e^x).

y= C*(e^(e^x))

y(0)=C*(e^(e^0))=2*e

e^0=1

I think you should be able to obtain the correct solution from there.

6. got it noW!!
thanks dwsmith!!