initial value DE problem

**harveyo** · Dec 15th 2009, 02:01 PM

dy/dx= (x+1)y/2x^2 where y(1)=1, for all x>=1

i ended up with

y^2/2= 1/2 ln (2x^2)+1/4 tan-1 (x/2) + K

anythoughts?

**pickslides** · Dec 15th 2009, 02:19 PM

I think this DE is separable

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

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

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

$\int\frac{1}{y}~dy= \int \frac{x}{2x^2}+\frac{1}{2x^2}~dx$

$\int\frac{1}{y}~dy= \int \frac{1}{2x}+\frac{1}{2x^2}~dx$

Did you use this method?

**harveyo** · Dec 15th 2009, 02:38 PM

i did it that way before and came up with:

ln [y] = ln [2x] + 1/4 tan-1 (x/2) + K

still doesnt seem right

**pickslides** · Dec 15th 2009, 02:43 PM

I think that solution is better, from here raise each side of the equation to the power of $e$ . This will eliminate the $\ln(y)$ on the left hand side.

Before this, may I ask how you arrived at the $\tan^{-1}(x)$ term? I can't see this being correct.

**harveyo** · Dec 15th 2009, 02:49 PM

table of integrals du / a^2 + u^2 = 1/a tan-1 u/a + C

**pickslides** · Dec 15th 2009, 02:59 PM

I see, I'm not sure if you have applied that the correct way here.

consider

$\int\frac{1}{y}~dy= \int \frac{1}{2x}+\frac{1}{2x^2}~dx$

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

$\ln(y)= \frac{\ln(x)}{2}- \frac{x^{-1}}{2}+C$

**harveyo** · Dec 15th 2009, 03:10 PM

i definitely overcomplicated that one obviously

cheers

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