Host of Differential Equations Watch

Zacken
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I've been biting into STEP for about a week now, but I'm unfortunately terrified of Differential Equations, so I've always skipped them on the STEP paper.

So, to remedy this situation, I've learnt up on solving the equations and the different types; the only thing I need now is a host of standard equations, nothing special, just a simple "Solve this (a)-(z)" kinda thing to get me familiar with my material and confident with it.

I've tried googling, but it comes up with past papers from Uni's and lecture notes of which have a few problem, but I feel very unconfident whilst doing it in case it's a uni-level question that I have no experience solving.

So, basically. Can any of you point me towards a library of questions with their corresponding solutions?

Thank you!
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TeeEm
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(Original post by Zacken)
I've been biting into STEP for about a week now, but I'm unfortunately terrified of Differential Equations, so I've always skipped them on the STEP paper.

So, to remedy this situation, I've learnt up on solving the equations and the different types; the only thing I need now is a host of standard equations, nothing special, just a simple "Solve this (a)-(z)" kinda thing to get me familiar with my material and confident with it.

I've tried googling, but it comes up with past papers from Uni's and lecture notes of which have a few problem, but I feel very unconfident whilst doing it in case it's a uni-level question that I have no experience solving.

So, basically. Can any of you point me towards a library of questions with their corresponding solutions?

Thank you!
what level exactly?
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Zacken
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(Original post by TeeEm)
what level exactly?
Well, I suppose A-level?

I'm actually facepalming really hard right now, why did I not think of using your website? :eek:
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morgan8002
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When I learned the AQA FP3 book was good practice.
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TeeEm
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(Original post by Zacken)
Well, I suppose A-level?

I'm actually facepalming really hard right now, why did I not think of using your website? :eek:
I am not allowed to advertise ...
:mad:
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Zacken
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(Original post by morgan8002)
When I learned the AQA FP3 book was good practice.
I was considering textbooks, but unfortunately I'm self studying and live elsewhere. Mauritius, you might've heard of it. Tiny island. Anyways, not much in the way of textbooks sold here, and shipping is horrendous when buying online. We don't do A-levels in our education system, but I decided I'd self-study and do it myself.

Thanks anyway!
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morgan8002
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(Original post by Zacken)
I was considering textbooks, but unfortunately I'm self studying and live elsewhere. Mauritius, you might've heard of it. Tiny island. Anyways, not much in the way of textbooks sold here, and shipping is horrendous when buying online. We don't do A-levels in our education system, but I decided I'd self-study and do it myself.

Thanks anyway!
It's a free download
http://filestore.aqa.org.uk/subjects...3-TEXTBOOK.PDF
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Zacken
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(Original post by TeeEm)
I am not allowed to advertise ...
:mad:
I've downloaded your 1st_order_differential_equations _practice.pdf and odes_separable_no_context.pdf files, anything else I should get? Only first order, I'll deal with second order another day.
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Krollo
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STEP I 2010, Q6 is a fairly nice one - it leads you through most of the question. Let me know if you want any hints.


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Zacken
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(Original post by Krollo)
STEP I 2010, Q6 is a fairly nice one - it leads you through most of the question. Let me know if you want any hints.


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Ooh, looks tasty! I'll have a go at it, expect a reply in a bit.
Thanks.

+Repped.
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TeeEm
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(Original post by Zacken)
I've downloaded your 1st_order_differential_equations _practice.pdf and odes_separable_no_context.pdf files, anything else I should get? Only first order, I'll deal with second order another day.
  • elementary separable ODES are in STANDARD TOPICS/INTEGRATION (Usually in C4 of standard A levels)
  • more advanced ODES are in FURTHER TOPICS/INTEGRATION (Usually in Further Maths A levels) This includes various ODE transformations.Look at exam question files.
  • More advanced techniques in ADVANCED TOPICS (beyond A level)
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Zacken
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Thanks a lot!

+Rep.
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Zacken
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(Original post by Krollo)
STEP I 2010, Q6 is a fairly nice one - it leads you through most of the question. Let me know if you want any hints.
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If y=e^x, then \displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} = e^x and \displaystyle \frac{\mathrm{d^2}y }{\mathrm{d} x^2} = e^x

Substituting these into *, yields:

L.H.S = (x-1)e^x - xe^x + e^x = xe^x - xe^x + xe^x - xe^x = 0 = R.H.S and therefore, y=e^x is a solution to *.


Spoiler:
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Now if we let \displaystyle y=ue^x, then \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = e^x\frac{\mathrm{d} u}{\mathrm{d} x} + ue^x and style \displaystyle \frac{\mathrm{d^2} y}{\mathrm{d} x^2} = e^x \bigg(\frac{\mathrm{d^2} u}{\mathrm{d} x^2} + 2\frac{\mathrm{d} u}{\mathrm{d} x} + u\bigg).

Substituting these into * yields:

\displaystyle (x-1)e^x \bigg(\frac{\mathrm{d^2} u}{\mathrm{d} x^2} + 2\frac{\mathrm{d} u}{\mathrm{d} x} + u\bigg) - xe^x\frac{\mathrm{d} u}{\mathrm{d} x} - xue^x + ue^x = 0

\displaystyle \iff xe^x\frac{\mathrm{d^2} u}{\mathrm{d} x^2} - e^x\frac{\mathrm{d^2} u}{\mathrm{d} x^2} + xe^x\frac{\mathrm{d} u}{\mathrm{d} x} - 2e^x\frac{\mathrm{d} u}{\mathrm{d} x} - ue^x + ue^x - xe^x\frac{\mathrm{d} u}{\mathrm{d} x} - xue^x + xue^x

\displaystyle \iff e^x(x-1)\frac{\mathrm{d^2} u}{\mathrm{d} x^x} + e^x(2x-2-x)\frac{\mathrm{d} u}{\mathrm{d} x} = 0

We can divide through by e^x since \forall x \in \mathbb{R}, e^x > 0, hence:

\displaystyle (x-1)\frac{\mathrm{d^2} u}{\mathrm{d} x^x} + (x-2)\frac{\mathrm{d} u}{\mathrm{d} x} = 0


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Now we let \displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = v \iff \frac{\mathrm{d^2} u}{\mathrm{d} x^2} = 1

Substituting into ** yields:

x-1 + (x-2)v = 0 \iff v=\frac{1-x}{x-2}

So we arrive at the separable differential equation:

\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x}=\frac{1-x}{x-2} \iff \int \mathrm{d}u = \int \frac{1-x}{x-2} \mathrm{d}x

Therefore:

\displaystyle u= - \int \frac{(x-2) + 1}{x-2} \mathrm{d}x \iff u = -\int \mathrm{d}x - \int\frac{1}{x+2} \mathrm{d}x

Hence:

u = -(x + \ln \mid x-2 \mid) + c.


That's all I've done so far, is it correct?

Also, I have no clue how to begin the last part!
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Krollo
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(Original post by Zacken)
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If y=e^x, then \displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} = e^x and \displaystyle \frac{\mathrm{d^2}y }{\mathrm{d} x^2} = e^x

Substituting these into *, yields:

L.H.S = (x-1)e^x - xe^x + e^x = xe^x - xe^x + xe^x - xe^x = 0 = R.H.S and therefore, y=e^x is a solution to *.


Spoiler:
Show

Now if we let \displaystyle y=ue^x, then \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = e^x\frac{\mathrm{d} u}{\mathrm{d} x} + ue^x and style \displaystyle \frac{\mathrm{d^2} y}{\mathrm{d} x^2} = e^x \bigg(\frac{\mathrm{d^2} u}{\mathrm{d} x^2} + 2\frac{\mathrm{d} u}{\mathrm{d} x} + u\bigg).

Substituting these into * yields:

\displaystyle (x-1)e^x \bigg(\frac{\mathrm{d^2} u}{\mathrm{d} x^2} + 2\frac{\mathrm{d} u}{\mathrm{d} x} + u\bigg) - xe^x\frac{\mathrm{d} u}{\mathrm{d} x} - xue^x + ue^x = 0

\displaystyle \iff xe^x\frac{\mathrm{d^2} u}{\mathrm{d} x^2} - e^x\frac{\mathrm{d^2} u}{\mathrm{d} x^2} + xe^x\frac{\mathrm{d} u}{\mathrm{d} x} - 2e^x\frac{\mathrm{d} u}{\mathrm{d} x} - ue^x + ue^x - xe^x\frac{\mathrm{d} u}{\mathrm{d} x} - xue^x + xue^x

\displaystyle \iff e^x(x-1)\frac{\mathrm{d^2} u}{\mathrm{d} x^x} + e^x(2x-2-x)\frac{\mathrm{d} u}{\mathrm{d} x} = 0

We can divide through by e^x since \forall x \in \mathbb{R}, e^x > 0, hence:

\displaystyle (x-1)\frac{\mathrm{d^2} u}{\mathrm{d} x^x} + (x-2)\frac{\mathrm{d} u}{\mathrm{d} x} = 0


Spoiler:
Show

Now we let \displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = v \iff \frac{\mathrm{d^2} u}{\mathrm{d} x^2} = 1

Substituting into ** yields:

x-1 + (x-2)v = 0 \iff v=\frac{1-x}{x-2}

So we arrive at the separable differential equation:

\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x}=\frac{1-x}{x-2} \iff \int \mathrm{d}u = \int \frac{1-x}{x-2} \mathrm{d}x

Therefore:

\displaystyle u= - \int \frac{(x-2) + 1}{x-2} \mathrm{d}x \iff u = -\int \mathrm{d}x - \int\frac{1}{x+2} \mathrm{d}x

Hence:

u = -(x + \ln \mid x-2 \mid) + c.


That's all I've done so far, is it correct?

Also, I have no clue how to begin the last part!
First two parts are excellent, for the third part I think you have assumed v to be a single variable, as opposed to a function of x.


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Zacken
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(Original post by Krollo)
First two parts are excellent, for the third part I think you have assumed v to be a single variable, as opposed to a function of x.


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Urgh, true that!

So instead:

\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = v \iff \frac{\mathrm{d^2} u}{\mathrm{d} x^2} = \frac{\mathrm{d} v}{\mathrm{d} x}

Substituting this in gives us a separable differential equation:

\displaystyle (x-1)\frac{\mathrm{d} v}{\mathrm{d} x} + (x-2)v = 0 \iff \frac{1}{v}\frac{\mathrm{d} v}{\mathrm{d} x} = - \frac{x-2}{x-1}

Then, we solve this to get:

\displaystyle \ln |v| = -\int\frac{x-2}{x-1} \mathrm{d}x \iff \ln |v| = -x + \ln |x-1| + c

Leading to:

\displaystyle |v| = e^{(-x + ln |x-1| + c)} \iff |v| = e^c|x-1|e^{-x}.

Now we have that \displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = e^c|x-1|e^{-x} \iff u = \int e^c(x-1)e^{-x} \mathrm{d}x.

Using IBP on the integral yields:

\displaystyle u = -e^cxe^{-x} + d

And hence \displaystyle y = -e^cx + de^x is our general solution.
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Zacken
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(Original post by Hasufel)
might these help to get you going?

(attached - the plots are my own - done with either Maple or Mathematica)

Sorry about the big-ass post, but i don`t know how to stop it doing that!
+Repped.

Goly, thanks for that!
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Hasufel
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Oops! seem to have deteled them - here i`ll post again...
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Zacken
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(Original post by Hasufel)
Oops! seem to have deteled them - here i`ll post again...
Got 'em!
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