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First order differential equation

I need to solve:

Ln(y^x). dy/dx = 3x^2 . y
Rearrange to get

X.lny (dy) = 3x^2 (dx)
Int (X.lny (dy)) = Int (3x^2 (dx))

To get

(lny)^2/2 = 3x^2/2 + C

According to my mark scheme I should get C/2
But I don't know why?
Reply 1
Avatar for starkrush
starkrush
1
∫ln⁡y≠(ln⁡y)22\int \ln y \not= \frac{(\ln y)^2}{2} - try integrating by parts (where u = ln y, and dv/dy = 1)

Also, I don't see where your y from the first line has gone. And if you're finding c, then you'll have been given x and y values which you haven't stated in your post.

Edit: And ignore the first two parts as they have been corrected below - next time try and keep the y in your working throughout, but equally (if not more so) it's my bad for not trying with it first :smile: ).
(edited 12 years ago)
Reply 2
Avatar for f1mad
f1mad
13
Original post by claret_n_blue
....


To integrate ln(y)/y use substitution:

let u= lny

du/dy = 1/y

du= dy/y

-> I u du = u^2/2= (lny)^2/2.
Reply 3
Avatar for f1mad
f1mad
13
http://www.wolframalpha.com/input/?i=xln%28y%29*+dy%2Fdx+%3D+3x%5E2y

Wolfram says no constant/2.

Not quite sure where that constant/2 has come from, I certainly don't get that either.
Reply 4
Avatar for f1mad
f1mad
13
Original post by starkrush
...


Wrong.

OP is correct, if I'm interpreting correctly.
(edited 12 years ago)
Reply 5
Avatar for starkrush
starkrush
1
Original post by f1mad
Wrong.

OP is correct, if I'm interpreting correctly.


Ninja edited 1 minute before you replied :tongue:

And edited to reflect this clearer
(edited 12 years ago)
Reply 6
Avatar for f1mad
f1mad
13
Original post by starkrush
Ninja edited 1 minute before you replied :tongue:

And edited to reflect this clearer


Last edited by f1mad; 13 Minutes Ago at 16:46

Last edited by starkrush; 1 Minute Ago at 16:57.

:tongue:
Reply 7
Avatar for starkrush
starkrush
1
That was to strikethrough the text!

(Yes, this is very constructive maths help)

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