# Thread: Tricky seperable DE. At least it's tricky to me

1. ## Tricky seperable DE. At least it's tricky to me

Solve the sepereable D.E.

xdy = y*ln(y)dx , x = 2 , y = e

I started by dividing both sides by y*ln(y) and x to wind up with

dx/x = dy/(y*ln(y))

Then I go to integrate and end up with

ln(x) + C = ln(ln(y)) + C

And then I'm stuck. I'm entirely unsure of where to go with this. Any ideas? Thank you

- Charles

2. Originally Posted by Charles Dexter Ward

Then I go to integrate and end up with

ln(x) + C = ln(ln(y)) + C
Just consider one constant, make it $\displaystyle k,$ hence $\displaystyle \ln|x|=\ln|\ln y|+k.$

We can also set $\displaystyle \ln|x|=\ln|\ln y|+\ln|k_1|,$ and from there you can easily isolate $\displaystyle y.$

3. Originally Posted by Krizalid
Just consider one constant, make it $\displaystyle k,$ hence $\displaystyle \ln|x|=\ln|\ln y|+k.$

We can also set $\displaystyle \ln|x|=\ln|\ln y|+\ln|k_1|,$ and from there you can easily isolate $\displaystyle y.$
Ok. So I can ln the constant because it doesn't change anything, it's still a constant, right? So from

ln|x| = ln|lny| + ln|k|

I e everything right? e^ln|x| = e^ln|lny| + e^ln|k|

which leaves

x = lny + k

lny = x - k

y = e^x - e^k => e^x - k

k = e^x - y , x = 2, y = e

k = e^2 - e^1 = 4.6708

Is that correct?

Am I allowed to e everything like that?

4. Anyone able to confirm if my solution is correct or not? Thank you.

5. Originally Posted by Charles Dexter Ward
Solve the sepereable D.E.

xdy = y*ln(y)dx , x = 2 , y = e
$\displaystyle \frac{dy}{y\ln(y)}=\frac{dx}{x}\implies \ln|\ln(y)|+C_1=\ln|x|+C_2\implies \ln|\ln(y)|=\ln|x|+C; C=C_2-C_1$
$\displaystyle \implies \ln(y)=Ae^{\ln|x|}; A=\pm e^C$
$\displaystyle \therefore y=e^{A|x|}$

When x=2, y=e:

$\displaystyle e=e^{2A}\implies 1=2A\implies A=\frac{1}{2}$

Therefore, $\displaystyle \color{red}\boxed{y=e^{\frac{1}{2}|x|}}$

6. I see. My method also gets the same answer to both ways are correct! Thank you!

Originally Posted by Chris L T521
$\displaystyle \frac{dy}{y\ln(y)}=\frac{dx}{x}\implies \ln|\ln(y)|+C_1=\ln|x|+C_2\implies \ln|\ln(y)|=\ln|x|+C; C=C_2-C_1$
$\displaystyle \implies \ln(y)=Ae^{\ln|x|}; A=\pm e^C$
$\displaystyle \therefore y=e^{A|x|}$

When x=2, y=e:

$\displaystyle e=e^{2A}\implies 1=2A\implies A=\frac{1}{2}$

Therefore, $\displaystyle \color{red}\boxed{y=e^{\frac{1}{2}|x|}}$