FP3 Differential Equations Help

A curve passing through the point (1,1) has the property that, at each point P of the curve, the gradient of the curve is 1 less than the gradient of OP. Find the equation of the curve.

Would really appreciate some help on this guys, thanks xx

Reply 1

joostan

Original post by tbhlouise

You'd have got a faster response in the Maths Forum.
As for your question, it will help to let:

P

be the variable point

(x,y)

Then, define the gradient of

OP

in terms of

x

and

y

.
More advice:

Spoiler

If you have any trouble just quote me :smile:

(edited 9 years ago)

Reply 2

MathsNerd1

Original post by tbhlouise

Moved to Maths forum :smile:

Reply 3

tbhlouise

Original post by joostan

You'd have got a faster response in the Maths Forum.
As for your question, it will help to let:

P

be the variable point

(x,y)

Then, define the gradient of

OP

in terms of

x

and

y

.
More advice:

Spoiler

If you have any trouble just quote me :smile:

Thanks for the guidance! :smile:

Got the right answer of y=x(1-lnx)

Reply 4

joostan

Original post by tbhlouise

Thanks for the guidance! :smile:

Got the right answer of y=x(1-lnx)

Excellent, no worries :smile:

Reply 5

tbhlouise

Bit stuck on this one as well :s-smilie:

Got up to x.qu^q-1 du/dx + 2u^q= x^2.u^3q but don't know what to do next :/

(edited 9 years ago)

image.jpg361.4KB

Reply 6

davros

Original post by tbhlouise

Bit stuck on this one as well :s-smilie:

Got up to x.qu^q-1 du/dx + 2u^q= x^2.u^3q but don't know what to do next :/

I haven't checked your algebra, but I'd suggest dividing both sides of the equation by

u^{q-1}

and then work out what value of q makes the RHS a lot simpler :smile:

Reply 7

tbhlouise

Original post by davros

I haven't checked your algebra, but I'd suggest dividing both sides of the equation by

u^{q-1}

and then work out what value of q makes the RHS a lot simpler :smile:

Thanks davros! I kind of had to use trial and error to find out q :biggrin:

. After a few guesses, I managed to get q=-1/2 which is correct :smile:

. I'm pretty worried about FP3 cos this is only the first chapter! :eek:

Reply 8

davros

Original post by tbhlouise

Thanks davros! I kind of had to use trial and error to find out q :biggrin:

. After a few guesses, I managed to get q=-1/2 which is correct :smile:

. I'm pretty worried about FP3 cos this is only the first chapter! :eek:

Well done! Remember that you're trying to make the DE as simple as possible so you choose q to get rid of the u term on the RHS then you can divide by qx and get a standard-looking DE that you should have seen before :smile:

Reply 9

TeeEm

Original post by davros

forgive me for buzzing in but I think q=-2

(I could be wrong ....)

Reply 10

davros

Original post by TeeEm

forgive me for buzzing in but I think q=-2

(I could be wrong ....)

I haven't worked through the question myself but if the OP's earlier post was correct, you basically need to divide

u^{3q}

u^{q-1}

which would mean you want make 2q + 1 = 0 giving q = -1/2.

Do you disagree with the OP's earlier algebra where she posted the transformed equation in u and x?

Reply 11

TeeEm

Original post by davros

I haven't worked through the question myself but if the OP's earlier post was correct, you basically need to divide

u^{3q}

u^{q-1}

which would mean you want make 2q + 1 = 0 giving q = -1/2.

Do you disagree with the OP's earlier algebra where she posted the transformed equation in u and x?

As I said I could be wrong (very tired at the moment) but this type of ODE is known in the trade as a Bernoulli equation.

This is an ODE of the form

y'+yf(x) = yⁿg(x)

If I am not mistaken the standard transformation is u=1/y^n-1.

I could be wrong

EDIT: IGNORE MY COMMENT/ SEE POST 16

(edited 9 years ago)

Reply 12

joostan

Original post by TeeEm

The standard transformation is actually:

u=y^{1-n}

y=u^n

like the textbook suggested.
so you were pretty close.

Reply 13

TeeEm

Original post by joostan

The standard transformation is actually:

u=y^{1-n}

y=u^n

like the textbook suggested.
so you were pretty close.

that is what I meant (long day)

in this case y³ so y=1/u²=u^-2

so I cannot see how -1/2 could be right.

Reply 14

joostan

Original post by TeeEm

that is what I meant (long day)

in this case y³ so y=1/u²=u^-2

so I cannot see how -1/2 could be right.

Spoiler

EDIT: The initial comment is slightly unnecessary, I was gonna type out more, but didn't fancy all that Tex.

(edited 9 years ago)

Reply 15

TeeEm

Original post by joostan

Spoiler

EDIT: The initial comment is slightly unnecessary, I was gonna type out more, but didn't fancy all that Tex.

I think you are correct (as far as the question asked it)and I am not.

the substitution is u = y^-2 and the question was asking it as y=f(u) which the same substitution as y=u^-1/2 on rearrangement.

I just did not read the question carefully as it was sideways, I saw Bernoulli and I was quoting what you saw.

Apologies. :smile:

(edited 9 years ago)

Reply 16

joostan