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# Simple Differential Equation - Help Me Please

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1. 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
2. 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
3. Sorry, but you are both wrong!

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.
4. (Original post by steve10)
Sorry, but you are both wrong!

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,
5. 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
6. (Original post by member910132)
My mistake was that I took the -1 out of the bracket and forgot to put it back in,
I can confirm this is the correct answer
7. my answer, equated to u/x , then multiplied through by x, is correct.

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

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