# First Order Derivative II

• Feb 18th 2009, 11:40 AM
Simplicity
First Order Derivative II
$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = y - 2, \ \ \ y(0)=1$

Is this solveable to get $\displaystyle y(x)=...$. When I used seperation of variable I got $\displaystyle c=\ln (-1)$ which isn't possible. Can someone solve it? Thanks in advance.
• Feb 18th 2009, 11:45 AM
running-gag
Hi

Don't forget the absolute value !

Integrating $\displaystyle \frac{u'(x)}{u(x)}$ gives $\displaystyle \ln(|u(x)|)$
• Feb 18th 2009, 11:49 AM
Simplicity
Quote:

Originally Posted by running-gag
Hi

Don't forget the absolute value !

Integrating $\displaystyle \frac{u'(x)}{u(x)}$ gives $\displaystyle \ln(|u(x)|)$

Is this even case when:

$\displaystyle -1 = e^c$

$\displaystyle \ln (-1) = c$

It's not from the integration so it is still $\displaystyle \ln (-1) = \ln (1)$?
• Feb 18th 2009, 11:53 AM
running-gag
No, look :

$\displaystyle \frac{dy}{dx} = y - 2, \ \ \ y(0)=-1$

$\displaystyle \frac{dy}{y-2} = dx, \ \ \ y(0)=-1$

$\displaystyle \ln(|y(x)-2|) = x + c, \ \ \ y(0)=-1$

$\displaystyle \ln(|y(0)-2|) = c, \ \ \ y(0)=-1$

$\displaystyle \ln(3) = c$
• Feb 18th 2009, 12:00 PM
Simplicity
Quote:

Originally Posted by running-gag
No, look :

$\displaystyle \frac{dy}{dx} = y - 2, \ \ \ y(0)=-1$

$\displaystyle \frac{dy}{y-2} = dx, \ \ \ y(0)=-1$

$\displaystyle \ln(|y(x)-2|) = x + c, \ \ \ y(0)=-1$

$\displaystyle \ln(|y(0)-2|) = c, \ \ \ y(0)=-1$

$\displaystyle \ln(3) = c$

No, Sorry, I wrote the question wrong:

$\displaystyle \frac{dy}{dx} = y - 2, \ \ \ y(0)=1$

It's 1 and not -1.
• Feb 18th 2009, 12:32 PM
Simplicity
And $\displaystyle \ln 1 = 0$.

Which would give equation $\displaystyle y=x+2$ which when you differentiate does not give the required solution.

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = y - 2, \ \ \ y(0)=1$

Can someone please solve my confusion. (Worried)
• Feb 19th 2009, 07:48 AM
running-gag
$\displaystyle \frac{dy}{dx} = y - 2, \ \ \ y(0)=1$

$\displaystyle \frac{dy}{y-2} = dx, \ \ \ y(0)=1$

$\displaystyle \ln(|y(x)-2|) = x + c, \ \ \ y(0)=1$

$\displaystyle \ln(|y(0)-2|) = c, \ \ \ y(0)=1$

$\displaystyle \ln(1) = c$

$\displaystyle c = 0$

$\displaystyle \ln(|y(x)-2|) = x, \ \ \ y(0)=1$

$\displaystyle |y(x)-2| = e^x, \ \ \ y(0)=1$

$\displaystyle y(x)-2 = e^x$ or $\displaystyle -y(x)+2 = e^x$

$\displaystyle y(x) = e^x + 2$ or $\displaystyle y(x) = -e^x + 2$

$\displaystyle y(0) = 1$ therefore $\displaystyle y(x) = -e^x + 2$