The Student Room Group

Differential equation form

Q. By eliminating the arbitrary constant A find the first order differential equation that is equivalent to;

x2+y2=A x^2 + y^2 = A

My answer;
Reply 1
Answer

answer1.1.jpg
d/dx(constant) = 0
Original post by ckfeister
Q. By eliminating the arbitrary constant A find the first order differential equation that is equivalent to;

x2+y2=A x^2 + y^2 = A

My answer;


You forgot to differentiate the constant A in the second line! :smile:
Reply 4
Original post by NotNotBatman
d/dx(constant) = 0


Original post by SlashaRussia
You forgot to differentiate the constant A in the second line! :smile:


dy/dx = y/x (can't use latex as the / (oppose side one) is broken for some annoying reason)

(1/y) dy = (1/x) dx
ln(y) = ln(x) + c
y = x + e^c
:. y = x + A

The answer says it is
y = Ax
???
Original post by ckfeister

ln(y) = ln(x) + c
y = x + e^c


This step is wrong.
Reply 6
Original post by RDKGames
This step is wrong.


Ive bee trying to see where for a few hours with other questions having the same issue, isn't that anti-log? I'm going what I know from textbook only as I'm self-studying.
Original post by ckfeister
dy/dx = y/x (can't use latex as the / (oppose side one) is broken for some annoying reason)

(1/y) dy = (1/x) dx
ln(y) = ln(x) + c
y = x + e^c
:. y = x + A

The answer says it is
y = Ax
???


elnx+c=elnxece^{lnx + c} = e^{lnx} \cdot e^c
Reply 8
Original post by NotNotBatman
elnx+c=elnxece^{lnx + c} = e^{lnx} \cdot e^c


Well thats new...
So
ln[y] = ln[x] + c
y = e^(ln[x] + c)
y = Ax (as e^ln[x] = x)
Original post by ckfeister
Well thats new...
So
ln[y] = ln[x] + c
y = e^(ln[x] + c)
y = Ax (as e^ln[x] = x)


Bit worrying if that's new!

ab+c=abaca^{b+c} = a^b \cdot a^c is something Y10s work with.
Reply 10
Original post by RDKGames
Bit worrying if that's new!

ab+c=abaca^{b+c} = a^b \cdot a^c is something Y10s work with.


In year 10 I got E/Fs which was normal in my school as it is one of the worst as well as the one of the most disadvantaged area in South East England...
(edited 6 years ago)
Original post by ckfeister
Well thats new...
So
ln[y] = ln[x] + c
y = e^(ln[x] + c)
y = Ax (as e^ln[x] = x)


This is a rule you should know, if it's new, go over any other little bits of algebra, to make sure that you can do these type of questions.

Quick Reply

Latest