You are Here: Home >< Maths

# another differential Watch

1. I'm stuck on part iii

for ii) I got the partial fractions as 1/(x-1) - 1/(x+1)

for part iii)
the last line of the question is:
"given that y=1 when x=3. Express y as a function of x" (photo cut it off)

these are my workings so far:

I integrated
1/y dy = -[(x^2+1)/(x^2-1)] dx

and got:
ln y= integrate -1 + 2/x^-1 dx

*2/x^-1 is from the part ii*

ln y = integrate -1 + 1/(x-1) - 1/(x+1)

ln y = -x + ln (1-x) - ln (x+1)
I e'd both sides
y= e^(-x) + [(1-x)/(x+1)]

but the answer is supposed to be y= (x+1)/(2(x-1)) * e^(3-x)

I have my fractions the wrong way up and I'm not sure how the 2 got there?
and where the 3 came from?
2. (Original post by 0utdoorz)

I'm stuck on part iii

for ii) I got the partial fractions as 1/(x-1) - 1/(x+1)

for part iii)

these are my workings so far:

I integrated
1/y dy = -[(x^2+1)/(x^2-1)] dx

and got:
ln y= integrate -1 + 2/x^-1 dx

*2/x^-1 is from the part ii*

ln y = integrate -1 + 1/(x-1) - 1/(x+1)

ln y = -x + ln (1-x) - ln (x+1)
I e'd both sides
y= e^(-x) + [(1-x)/(x+1)]

but the answer is supposed to be y= (x+1)/(2(x-1)) * e^(3-x)

I have my fractions the wrong way up and I'm not sure how the 2 got there?
and where the 3 came from?
Where's your value for +C? Did they give any values of x or y to sub in to find this C?
3. you have forgotten that, when you expanded the RHS to partial fractions, it was multipled by -1 (you haven`t done that)

(I,e the signs of your partial fractions for part 3 are not correct)
4. (Original post by 0utdoorz)

I'm stuck on part iii

for ii) I got the partial fractions as 1/(x-1) - 1/(x+1)

for part iii)

these are my workings so far:

I integrated
1/y dy = -[(x^2+1)/(x^2-1)] dx

and got:
ln y= integrate -1 + 2/x^-1 dx

*2/x^-1 is from the part ii*

ln y = integrate -1 + 1/(x-1) - 1/(x+1)

ln y = -x + ln (1-x) - ln (x+1)
I e'd both sides
y= e^(-x) + [(1-x)/(x+1)]

but the answer is supposed to be y= (x+1)/(2(x-1)) * e^(3-x)

I have my fractions the wrong way up and I'm not sure how the 2 got there?
and where the 3 came from?
How confident are you that the answer is correct, because I'm not getting it!
5. (Original post by metaltron)
How confident are you that the answer is correct, because I'm not getting it!
pretty sure, the back of the book says it is?

6. (Original post by 0utdoorz)
pretty sure, the back of the book says it is?

How far have you got now?
7. still stuck

I tried substituting in the values y=1 and x=3 into ln y=-x-ln(1-x)+ln(x+1) +c

I got
0 = -3-ln (2)+ln(4)+c
ln (2/4) = -3 +c
1/2 = -3+c
7/2 = c

but I don't think that's right?
8. (Original post by 0utdoorz)
still stuck

I tried substituting in the values y=1 and x=3 into ln y=-x-ln(1-x)+ln(x+1) +c

I got
0 = -3-ln (2)+ln(4)+c
ln (2/4) = -3 +c
1/2 = -3+c
7/2 = c

but I don't think that's right?
Did you get told it goes through the points (1,3) somewhere. Maybe I'm being blind!
9. (Original post by metaltron)
Did you get told it goes through the points (1,3) somewhere. Maybe I'm being blind!
yeah! sorry!! I'm such a fail, I just realized my photo cut off the last line!

"given that y=1 when x=3. Express y as a function of x"
10. (Original post by 0utdoorz)
still stuck

I tried substituting in the values y=1 and x=3 into ln y=-x-ln(1-x)+ln(x+1) +c

I got
0 = -3-ln (2)+ln(4)+c
ln (2/4) = -3 +c
1/2 = -3+c
7/2 = c

but I don't think that's right?
How about you try getting it in terms of y before subbing in for the c value. So start with:

and you should get to (don't look until you've done it yourself please):

Spoiler:
Show

Then find a and you're home.
11. (Original post by 0utdoorz)
yeah! sorry!! I'm such a fail, I just realized my photo cut off the last line!

"given that y=1 when x=3. Express y as a function of x"
OK don't worry, see my post above. Finally, my answer agrees with your book. Phew!
12. (Original post by metaltron)
How about you try getting it in terms of y before subbing in for the c value. So start with:

and you should get to (don't look until you've done it yourself please):

Spoiler:
Show

Then find a and you're home.
13. (Original post by 0utdoorz)
I got

ln y= (ln x+1/ln x-1) -x +c

e both sides: y=(x+1/x-1) *e^-x *A

e^-x = 1/e^x

so
y= (x+1)A/e^x(x-1)

I substituted x=3 and y=1

1= 4A/2e^3
2e^3 = 4A

A= e^3/2 ???

This is right Just sub back into the y equation.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 11, 2013
Today on TSR

### Is this person a genius?

...with these A Level results?

### I think I'm transgender AMA

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.