# differential equations

• Jan 5th 2009, 11:19 AM
holly123
differential equations
find a general solution to the differential equation

y*ln(x)-xy' =0

this is my work so far:
ylnx=x dy/dx
ylnx dx= xdy
(lnxdx)/x= dy/y
integral of (lnx)/x dx = integral of dy/y

???? = lny

i need help finding the integral of (lnx)/x dx
and then i need to solve for y

thank you!
• Jan 5th 2009, 11:21 AM
Jester
Quote:

Originally Posted by holly123
find a general solution to the differential equation

y*ln(x)-xy' =0

this is my work so far:
ylnx=x dy/dx
ylnx dx= xdy
(lnxdx)/x= dy/y
integral of (lnx)/x dx = integral of dy/y

???? = lny

i need help finding the integral of (lnx)/x dx
and then i need to solve for y

thank you!

Try $\displaystyle u = \ln x$ in $\displaystyle \int \frac{\ln x}{x}\, dx$
• Jan 5th 2009, 11:22 AM
holly123
oh thank you! i was on winter break for 2 weeks and i guess i'm fuzzy on integrating (Giggle)
• Jan 5th 2009, 11:24 AM
holly123
wait so what do i do after i get u=lnx and du= 1/x dx
is it 1/2 u^2 +c
1/2 (lnx)^2 +c ??
• Jan 5th 2009, 11:26 AM
Jester
Quote:

Originally Posted by holly123
wait so what do i do after i get u=lnx and du= 1/x dx
is it 1/2 u^2 +c
1/2 (lnx)^2 +c ??

Yep - that's right!
• Jan 5th 2009, 11:28 AM
holly123
thanks!! so i have to inject e to get rid of ln and solve for y
would that be y= 1/2 x^2 + e^c
• Jan 5th 2009, 11:40 AM
Jester
Quote:

Originally Posted by holly123
thanks!! so i have to inject e to get rid of ln and solve for y
would that be y= 1/2 x^2 + e^c

You can't simplify like that. In general

$\displaystyle \left( \ln x \right)^n \ne \ln \left( x^n \right)$

and

$\displaystyle a \ln (x) \ne \ln (a x)$
• Jan 5th 2009, 11:41 AM
holly123
for some reason this is really confusing me...i tried again and got y=x+ e^c
(Doh)
• Jan 5th 2009, 11:58 AM
Jester
Quote:

Originally Posted by holly123
for some reason this is really confusing me...i tried again and got y=x+ e^c
(Doh)

Since $\displaystyle \ln y = \frac{1}{2} \left( \ln x \right)^2 + c$

then the best you're going to get is

$\displaystyle y = e^{\frac{1}{2} \left( \ln x \right)^2} \cdot e^c = C e^{\frac{1}{2} \left( \ln x \right)^2}$ where $\displaystyle C = e^c$