1. ## differential equation

dy/dx - y/(x+1) = x y(1)= 0, x>0
integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)]

y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1)

0= 1/2 -(ln2)/2 + C/3

y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1)

2. Originally Posted by twilightstr
dy/dx - y/(x+1) = x y(1)= 0, x>0
integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)]

y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1)

0= 1/2 -(ln2)/2 + C/3

y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1)

You can always check your answer by substituting it into the differential equation and seeing if it works.

3. Originally Posted by twilightstr
dy/dx - y/(x+1) = x y(1)= 0, x>0
integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)]

y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1)

0= 1/2 -(ln2)/2 + C/3

y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1)

No.

$\frac{dy}{dx} - \frac{1}{x + 1}y = x$

The integrating factor is $e^{\int{-\frac{1}{x+1}\,dx}} = e^{-\ln{(x + 1)}} = e^{\ln{(x + 1)^{-1}}} = (x + 1)^{-1}$.

So multiply both sides of the equation by $(x + 1)^{-1}$ and you get

$(x + 1)^{-1}\frac{dy}{dx} - (x + 1)^{-2}y = \frac{x}{x + 1}$

$\frac{d}{dx}[(x + 1)^{-1}y] = 1 - \frac{1}{x + 1}$.

Can you go from here?

4. It is even easier to see that that y does not satisfy y(1)= 0!

5. thanks but do i have the answer now:

y= x^2-xln(x+1)+(-1 + ln2)x +x-ln(x+1)-1+ln2

6. Originally Posted by twilightstr
thanks but do i have the answer now:

y= x^2-xln(x+1)+(-1 + ln2)x +x-ln(x+1)-1+ln2
Let's see...

$\frac{d}{dx}[(x + 1)^{-1}y = 1 - \frac{1}{x + 1}$

$(x + 1)^{-1}y = x - \ln{|x + 1|} + C$

$y = (x + 1)(x - \ln{|x + 1|} + C)$.

We're told $y(1) = 0$ so

$0 = (1 + 1)(1 - \ln{|1 + 1|} + C)$

$0 = 2(1 - \ln{2} + C)$

$0 = 1 - \ln{2} + C$

$C = \ln{2} - 1$.

So $y = (x + 1)(x - \ln{|x + 1|} + \ln{2} - 1)$.