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# URGENT!!) C4 differential equations!!! watch

1. Hi I am stuck with this question and I can't seem to get to the answer its so frustrating.

x(x-1)dy/dx =y (y=1, at x=2)

do i expand the bracket into x^2-x and move it across
or do i do x/x-1 ?

Thank you!
2. (Original post by liemluji)
Hi I am stuck with this question and I can't seem to get to the answer its so frustrating.

x(x-1)dy/dx =y (y=1, at x=2)

do i expand the bracket into x^2-x and move it across
or do i do x/x-1 ?

Thank you!
Regardless of whether you expand it or not (you'll find that if you expand it and do what you have to do you should simplify it so no point really..) what is the aim, how do you integrate it?
3. (Original post by SeanFM)
Regardless of whether you expand it or not (you'll find that if you expand it and do what you have to do you should simplify it so no point really..) what is the aim, how do you integrate it?
It says 'find the particular solution to each of these differential equations.'
so I have to change the form into y=mx+c

but before I do that i have to arrange them into integral of f(y) dy= Integral of f(x)dx

and this is where I'm stuck.
4. (Original post by liemluji)
It says 'find the particular solution to each of these differential equations.'
so I have to change the form into y=mx+c

but before I do that i have to arrange them into integral of f(y) dy= Integral of f(x)dx

and this is where I'm stuck.
What have you tried? Have you got any workings?
5. (Original post by zetamcfc)
What have you tried? Have you got any workings?
I tried expanding the bracket into x^2-x

and got integral 1/y dy= integral 1/x^2-x dx

but i'm not sure if it's right and I don't know how to integrate 1/x^2-x

i've tried substituion but it didn't work
6. (Original post by liemluji)
I tried expanding the bracket into x^2-x

and got integral 1/y dy= integral 1/x^2-x dx

but i'm not sure if it's right and I don't know how to integrate 1/x^2-x

i've tried substituion but it didn't work
Partial fractions my friend
7. (Original post by zetamcfc)
Partial fractions my friend
but if I made 1/x^2-x into partial fractions I get A=0 which is impossible

so 1/x(x-1) =a/x+b/x-1

so 1= A(x-1) + B(x)

so A=0 and B=1?
8. (Original post by liemluji)
but if I made 1/x^2-x into partial fractions I get A=0 which is impossible

so 1/x(x-1) =a/x+b/x-1

so 1= A(x-1) + B(x)

so A=0 and B=1?
A=-1 B=1
9. (Original post by zetamcfc)
A=-1 B=1
oh thank you so much

I got ln y= -lnx + ln(x-1) +c and when i subbed y=1 and x=2 i got c=2

which is the same as the answer in the back of the book

Sorry but how would i change that in terms of y=mx+c?

is it y= (x-1)/x +2?

the asnwer in the back of the book is y=2-2/x

thank you
10. (Original post by liemluji)
oh thank you so much

I got ln y= -lnx + ln(x-1) +c and when i subbed y=1 and x=2 i got c=2

which is the same as the answer in the back of the book

Sorry but how would i change that in terms of y=mx+c?

is it y= (x-1)/x +2?

the asnwer in the back of the book is y=2-2/x

thank you
Ok, so when we have integrated we have ln(y)=ln((x-1)/x) +C

So to get y we put e to the power of both sides. To get y=A((x-1)/x), then from here you know what you are doing
11. (Original post by zetamcfc)
Ok, so when we have integrated we have ln(y)=ln((x-1)/x) +C

So to get y we put e to the power of both sides. To get y=A((x-1)/x), then from here you know what you are doing
I don't know how to thank you enough

Thank you so much for helping me

I finally got the answer thank you!!!!

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