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Reply 760
I guess I'm in for some maths on a Friday night.
dy/dx + y/x = x


Anybody?
Reply 762
Original post by Mr Dangermouse
dy/dx + y/x = x


Anybody?


Need to solve it using the integrating factor method, where if you have dy/dx + P(x)y = Q(x) (which is purely a function of x).

Then:
Unparseable latex formula:

e^\int{P(x)dx}

is your integrating factor. Multiply both sides of the equation by this integrating factor and see if you can think where to go from there.

The LaTex isn't clear but that's e raised to the integral of P(x).

If you get stuck.

Spoiler

(edited 12 years ago)
So it's a first order linear differential equation of form:

dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)

What's P(x)? What's Q(x)?

You can then use P(x) to find the integrating factor that's of the form eP(x)dxe^{\int P(x)dx}, which you multiply the entire original equation by, surely you can remember where to go from here?

My solution:

Spoiler

Erm... wut?
I normally solve them by splitting the variables.
Original post by Mr Dangermouse
I normally solve them by splitting the variables.


You can't do that with this. Variables separable only applies to some first order differential equations. Integrating factor is the other method used to solve this type of first order differential equations.

Steps for the Integrating factor method:

1. Ensure the equation is in standard form - dy/dx + p(x)y = Q(x)

2. Identify P(x) and P(x)dx\int P(x)dx

3. Write out the integrating factor eP(x)dxe^{\int P(x)dx}

4.Multiply both sides by the integrating factor

5. The LHS will be ddx(I.Fy)\frac{d}{dx}(I.F\cdot y)

6. Integrate both sides.

7. Substitute in initial conditions/rearrange to make y the subject.
(edited 12 years ago)
Reply 767
Original post by Hype en Ecosse
So it's a first order linear differential equation of form:

dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)

What's P(x)? What's Q(x)?

You can then use P(x) to find the integrating factor that's of the form eP(x)dxe^{\int P(x)dx}, which you multiply the entire original equation by, surely you can remember where to go from here?

My solution:

Spoiler


Spoiler


Original post by Mr Dangermouse
I normally solve them by splitting the variables.

Aren't able to split the variables here, have to use the integrating factor method. The separation of variables can only be done if you've got the derivative of a function equal to a function of x multiplied by a function of y.
(edited 12 years ago)
Original post by Mr Dangermouse
I normally solve them by splitting the variables.


We don't have a differential equation where we can do that, I can't for the life of me remember what the first type that you're talking about are called, but I know what you mean. Where you'd have:

dy/dx = xy

And you'd make it dy/y = xdx and integrate. But you don't have that here, have you done first-order linear differential equations?


Original post by JordanR

Spoiler




c is a completely arbitrary constant, so I thought one didn't need to divide the c? :tongue:
(edited 12 years ago)
Reply 769
I don't know if they have a specific name actually. I don't think I've been told one if they do.

But if they're like: ddxP(x)=Q(x)R(y)\frac{d}{dx}P(x) = Q(x)R(y) then we can split the variables up.


Original post by Hype en Ecosse


c is a completely arbitrary constant, so I thought one didn't need to divide the c? :tongue:

Hence why it's semantics! Although it would become important if one was given an initial condition and one was also asked to find the value of one's constant, so it's definitely good practice for one to do so.
(edited 12 years ago)
Original post by JordanR
I don't know if they have a specific name actually. I don't think I've been told one if they do.

But if they're like: ddxP(x)=Q(x)R(y)\frac{d}{dx}P(x) = Q(x)R(y) then we can split the variables up.



Hence why it's semantics! Although it would become important if one was given an initial condition and one was also asked to find the value of one's constant, so it's definitely good practice for one to do so.


Original post by Hype en Ecosse
c is a completely arbitrary constant, so I thought one didn't need to divide the c? :tongue:


Yes, but they're not the same arbitrary constant, so you need to either show the division or rename it. Same reason x=k↛12x=kx = k \not\rightarrow \tfrac{1}{2}x = k. It's not just semantics: your last step implies c = 0.
Original post by TheUnbeliever
Yes, but they're not the same arbitrary constant, so you need to either show the division or rename it. Same reason x=k↛12x=kx = k \not\rightarrow \tfrac{1}{2}x = k. It's not just semantics: your last step implies c = 0.


When I first started doing differential equations, I always divided my c, too. The answer schemes never divided the c, and the teacher told me I didn't need to divide it through; I made the same argument as you, but the teacher was having none of it...

This was shortly after I got a row about "remember that she's done a degree and you've not." So... :tongue:
Reply 773
I hate it when teachers try and argue from authority without justifying it at all.
Reply 774
Hmm, this is quite a nice question. Quite a few different techniques required for it:

Solve by use of standard integrals.

2x+3x2+4x+8 dx\int \frac{2x+3}{x^2+4x+8}\ dx

Also, I have a few questions:

What does "find the term independent of x" mean and are we supposed to be able to do this? Is it just the term in the binomial expansion that has an x term that is to the power of 0?

Are related rates examinable? Because I don't think we covered them in class, but I'm sure I seen them in an AH book (but not 100% sure).
Reply 775
Original post by Quintro
Hmm, this is quite a nice question. Quite a few different techniques required for it:

Solve by use of standard integrals.

2x+3x2+4x+8 dx\int \frac{2x+3}{x^2+4x+8}\ dx

Also, I have a few questions:

What does "find the term independent of x" mean and are we supposed to be able to do this? Is it just the term in the binomial expansion that has an x term that is to the power of 0?

Are related rates examinable? Because I don't think we covered them in class, but I'm sure I seen them in an AH book (but not 100% sure).


Very nice indeed. On my phone but it seems like long division and then a couple of u-subs? I'll try it later.

Yes, I'd think so. So if you had (x+y)^5 you'd find the term that only had y in it.

Probably, but I've never seen them in any past papers. Not looked at many though.
Original post by Quintro
Is it just the term in the binomial expansion that has an x term that is to the power of 0?


Yes.

Are related rates examinable? Because I don't think we covered them in class, but I'm sure I seen them in an AH book (but not 100% sure).


Yes.

(Not repeating Jordan needlessly - just adding certainty! :smile:)
Original post by JordanR
Very nice indeed. On my phone but it seems like long division and then a couple of u-subs? I'll try it later.


It's not no - the degree at the top is less than the degree on the bottom so it's reduced already. It's a ln integral and then an arctan integral.
Reply 778
Original post by ukdragon37
It's not no - the degree at the top is less than the degree on the bottom so it's reduced already. It's a ln integral and then an arctan integral.


Oops, silly me. Think it has to be rewritten in another form somehow. Split the numerator up into two parts (2x+4) and (-1)?
Reply 779
Original post by JordanR
Oops, silly me. Think it has to be rewritten in another form somehow. Split the numerator up into two parts (2x+4) and (-1)?


Yep!