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)

(Bow)

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- Mar 26th 2010, 06:10 PMnameck[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)

(Bow) - Mar 26th 2010, 06:12 PMdwsmith
- Mar 26th 2010, 06:18 PMdwsmith
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 - Mar 26th 2010, 06:28 PMnameck
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.. - Mar 26th 2010, 06:35 PMdwsmith
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. - Mar 26th 2010, 06:42 PMnameck
got it noW!! (Rofl)

thanks dwsmith!!(Bow)