Another FP2 problemWatch

#1
Use the substitution u=y-x-2 to transform the differential equation dy/dx=(y-x-2)^2 into a differential equation in x and u. By first solving this new equation, find the general solution of the original equation, giving y in terms of x.

I've got the point where I have but continuing this, I end up with something like when the mark scheme gives the answer as . Have I gone wrong and/or how would I get to the right final form?
0
4 years ago
#2
(Original post by Chlorophile)
Use the substitution u=y-x-2 to transform the differential equation dy/dx=(y-x-2)^2 into a differential equation in x and u. By first solving this new equation, find the general solution of the original equation, giving y in terms of x.

I've got the point where I have but continuing this, I end up with something like when the mark scheme gives the answer as . Have I gone wrong and/or how would I get to the right final form?
I do not get square roots
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#3
(Original post by TeeEm)
I do not get square roots
Did you manage to get to the mark scheme's answer from my u^2-1... expression?
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4 years ago
#4
(Original post by Chlorophile)
Did you manage to get to the mark scheme's answer from my u^2-1... expression?
I probably will ... very tedious

But before I do my experience tells me to check one thing in your workings, which I cannot see at the moment.

Partial fractions...

did you get -1/2/(u+1) or 1/2(u+1)
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#5
(Original post by TeeEm)
I probably will ... very tedious

But before I do my experience tells me to check one thing in your workings, which I cannot see at the moment.

Partial fractions...

did you get -1/2/(u+1) or 1/2(u+1)
-1/2(u+1)

I already slipped up on that, I had to redo the entire question twice...
0
4 years ago
#6
(Original post by Chlorophile)
-1/2(u+1)

I already slipped up on that, I had to redo the entire question twice...
well at that stage I get (u-1)/(u+1) =Ae2x

so there will be no square roots
0
4 years ago
#7
(Original post by TeeEm)
well at that stage I get (u-1)/(u+1) =Ae2x
Agreed, this is what I get too

(Original post by Chlorophile)
-1/2(u+1)

I already slipped up on that, I had to redo the entire question twice...
Maybe scan you working and we can tell you where you've gone wrong?
1
#8
(Original post by TeeEm)
well at that stage I get (u-1)/(u+1) =Ae2x

so there will be no square roots
How did you get to that point? I never got (u-1)/(u+1), only 1/(u-1)(u+1) or (u-1)(u+1).
0
4 years ago
#9
(Original post by Chlorophile)
How did you get to that point? I never got (u-1)/(u+1), only 1/(u-1)(u+1) or (u-1)(u+1).
partial fractions 1/(u2-1) = 1/2/(u1) -1/2(u-1)

multipled ODE by 2

integrated to logs

compacted the logs to a division because of the minus sign etc
0
#10
(Original post by usycool1)
Agreed, this is what I get too

Maybe scan you working and we can tell you where you've gone wrong?
This is what I have, sorry that it's potato quality and that part of it's scribbled out!

0
4 years ago
#11
(Original post by Chlorophile)
This is what I have, sorry that it's potato quality and that part of it's scribbled out!

fourth line is wrong

minus!

why?
1
#12
(Original post by TeeEm)
fourth line is wrong

minus!

why?
I am an idiot, that's why! I read -1 as -x...
0
4 years ago
#13
(Original post by Chlorophile)
I am an idiot, that's why!
it happens...

working at this time ...
0
4 years ago
#14
(Original post by Chlorophile)
This is what I have, sorry that it's potato quality and that part of it's scribbled out!

You've integrated 1/(u - 1) wrt 'u' incorrectly - you shouldn't have a minus sign in front following the integration.

EDIT: Too slow
0
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