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Simple Differential Equation - Help Me Please

Question:

Given that y=u−1 \displaystyle y = u^{-1}

And dudx−ux−1=−x−1ln⁡x \displaystyle \dfrac{du}{dx} - ux^{-1} = -x^{-1}\ln x

And y(1)=12 \displaystyle y(1)=\frac12

Find y=f(x) \displaystyle y=f(x)

My pdf textbook is notorious for getting things wrong so I would appreciate it if someone could verify my answer, I won't post my full working but:

Integrating factor: I=x−1 \displaystyle I = x^{-1}

Hence ux−1=∫−x−2ln⁡x dx \displaystyle ux^{-1} = \int -x^{-2}\ln x \ dx

After using integration by parts on the RHS: ux−1=−x−1ln⁡x−x−1+C \displaystyle ux^{-1} = -x^{-1}\ln x - x^{-1} +C

Now I get u=−ln⁡x−1+C \displaystyle u = -\ln x -1 + C

And the book gets u=ln⁡x+1+C \displaystyle u = \ln x +1 +C

Without solving for y, which is just the reciprocal, can anyone confimr who is correct ?

Thnx
(edited 11 years ago)
Reply 1
Avatar for Hasufel
Hasufel
16
the book is correct.

the integration by parts is:

Integral[ {-1/x^2}{Log[x]} = - (-1/x)(Log[x]) + Integral[(-1/x)(1/x)]dx

= 1/x+(log[x])/x +C =u/x
(edited 11 years ago)
Reply 2
Avatar for steve10
steve10
7
Sorry, but you are both wrong!

The correct answer is,

u = ln(x) - 1 + Cx

In your IBP, you should have ux−1=+x−1ln(x)−x−1+Cux^{-1}=+x^{-1}ln(x)-x^{-1}+C

If you work out dudx−ux−1=−x−1ln(x)\frac{du}{dx}-ux^{-1}=-x^{-1}ln(x)

using my answer, you will see that it satisfies the DE.
Reply 3
Avatar for member910132
member910132
OP
Original post by steve10
Sorry, but you are both wrong!

The correct answer is,

u = ln(x) - 1 + Cx

In your IBP, you should have ux−1=+x−1ln(x)−x−1+Cux^{-1}=+x^{-1}ln(x)-x^{-1}+C

If you work out dudx−ux−1=−x−1ln(x)\frac{du}{dx}-ux^{-1}=-x^{-1}ln(x)

using my answer, you will see that it satisfies the DE.


My mistake was that I took the -1 out of the bracket and forgot to put it back in,
The correct answer is u=ln⁡x+1+cx u = \ln x +1 +cx .
Reply 4
Avatar for Rennit
Rennit
Some really wrong answers on here.. I agree with most of what you've said and Steve10 is the closest.

The correct answer would be u=ln(x)+x+cx
Reply 5
Avatar for Rennit
Rennit
Original post by member910132
My mistake was that I took the -1 out of the bracket and forgot to put it back in,
The correct answer is u=ln⁡x+1+cx u = \ln x +1 +cx .


I can confirm this is the correct answer
Reply 6
Avatar for Hasufel
Hasufel
16
my answer, equated to u/x , then multiplied through by x, is correct.

(k = 1) and y=1/(x+Log(x)+1)
(edited 11 years ago)

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