[member910132]

Question:

Given that

And

And

Find

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:

Hence

After using integration by parts on the RHS:

Now I get

And the book gets

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

Thnx

[Hasufel]

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

[steve10]

Sorry, but you are both wrong!

The correct answer is,

u = ln(x) - 1 + Cx

In your IBP, you should have

If you work out

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

[member910132]

(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

If you work out

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 .

[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

[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 .

I can confirm this is the correct answer

[Hasufel]

my answer, equated to u/x , then multiplied through by x, is correct.

(k = 1) and y=1/(x+Log(x)+1)