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STEP 2013 Solutions

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Original post by Mastermind2
STEP I 2013 Question 7:

Spoiler


Please let me know if I've made any mistakes.


I agree with i) and ii), but I disagree with your solution to iii) (As does Wolfram, it gets a different general solution). Here is my working to iii):

dydx=x2y+2yx \frac{dy}{dx} = \frac{x^2}{y} + \frac{2y}{x}

Let y=ux2 y = ux^2

dydx=2xu+x2dudx \frac{dy}{dx} = 2xu + x^2 \frac{du}{dx}

Plugging this in gives:

2xu+x2dudx=1u+2xu 2xu + x^2 \frac{du}{dx} = \frac{1}{u} + 2xu

x2dudx=1u x^2 \frac{du}{dx} = \frac{1}{u}

∫u du=∫1x2 dx \int u \ du = \int \frac{1}{x^2} \ dx

12u2=−1x+c \frac{1}{2} u^2 = -\frac{1}{x} + c

Plugging in boundary conditions of y = 2, x= 1 (so u = 2) we get c = 3, and:

12u2=3−1x \frac{1}{2} u^2 = 3 - \frac{1}{x}

u=6−2x u = \sqrt{6- \frac{2}{x}} (Note: We take the positive root as x>0, y>0 so u>0)

y=ux2=x26−2x y = ux^2 = x^2 \sqrt{6- \frac{2}{x}}

On another note; the solution to part i) also has x>e^-2. I think that this means we may be expected (Only for a couple of marks, mind) to state the set of valid values of x for both ii and iii; that is, x^2 > 1/5 for ii) and x > 1/3 for iii).
Original post by DJMayes
I agree with i) and ii), but I disagree with your solution to iii) (As does Wolfram, it gets a different general solution). Here is my working to iii):

dydx=x2y+2yx \frac{dy}{dx} = \frac{x^2}{y} + \frac{2y}{x}

Let y=ux2 y = ux^2

dydx=2xu+x2dudx \frac{dy}{dx} = 2xu + x^2 \frac{du}{dx}

Plugging this in gives:

2xu+x2dudx=1u+2xu 2xu + x^2 \frac{du}{dx} = \frac{1}{u} + 2xu

x2dudx=1u x^2 \frac{du}{dx} = \frac{1}{u}

∫u du=∫1x2 dx \int u \ du = \int \frac{1}{x^2} \ dx

12u2=−1x+c \frac{1}{2} u^2 = -\frac{1}{x} + c

Plugging in boundary conditions of y = 2, x= 1 (so u = 2) we get c = 3, and:

12u2=3−1x \frac{1}{2} u^2 = 3 - \frac{1}{x}

u=6−2x u = \sqrt{6- \frac{2}{x}} (Note: We take the positive root as x>0, y>0 so u>0)

y=ux2=x26−2x y = ux^2 = x^2 \sqrt{6- \frac{2}{x}}

On another note; the solution to part i) also has x>e^-2. I think that this means we may be expected (Only for a couple of marks, mind) to state the set of valid values of x for both ii and iii; that is, x^2 > 1/5 for ii) and x > 1/3 for iii).


This is the form of my final answer so I'm glad you also got it like this as it means I'm confident of getting that question along with question 1 completely correct :biggrin: Now if only I could get an average for my other ones :redface:
Original post by DJMayes
I agree with i) and ii), but I disagree with your solution to iii) (As does Wolfram, it gets a different general solution). Here is my working to iii):

dydx=x2y+2yx \frac{dy}{dx} = \frac{x^2}{y} + \frac{2y}{x}

Let y=ux2 y = ux^2

dydx=2xu+x2dudx \frac{dy}{dx} = 2xu + x^2 \frac{du}{dx}

Plugging this in gives:

2xu+x2dudx=1u+2xu 2xu + x^2 \frac{du}{dx} = \frac{1}{u} + 2xu

x2dudx=1u x^2 \frac{du}{dx} = \frac{1}{u}

∫u du=∫1x2 dx \int u \ du = \int \frac{1}{x^2} \ dx

12u2=−1x+c \frac{1}{2} u^2 = -\frac{1}{x} + c

Plugging in boundary conditions of y = 2, x= 1 (so u = 2) we get c = 3, and:

12u2=3−1x \frac{1}{2} u^2 = 3 - \frac{1}{x}

u=6−2x u = \sqrt{6- \frac{2}{x}} (Note: We take the positive root as x>0, y>0 so u>0)

y=ux2=x26−2x y = ux^2 = x^2 \sqrt{6- \frac{2}{x}}

On another note; the solution to part i) also has x>e^-2. I think that this means we may be expected (Only for a couple of marks, mind) to state the set of valid values of x for both ii and iii; that is, x^2 > 1/5 for ii) and x > 1/3 for iii).

Thinking back, I'm pretty sure I got the same solution as you for iii in the actual exam, but I re-did it differently as I was typing it up here. What is wrong with my solution? I can't see why it doesn't work :confused:
Original post by Mastermind2
Thinking back, I'm pretty sure I got the same solution as you for iii in the actual exam, but I re-did it differently as I was typing it up here. What is wrong with my solution? I can't see why it doesn't work :confused:


I don't think the actual substitution is wrong, but when you get to the separable differential equation you have integrated it incorrectly.
Original post by DJMayes
I don't think the actual substitution is wrong, but when you get to the separable differential equation you have integrated it incorrectly.

Edited, I've now got the same answer as you. I'm sure I did it the correct way in the exam, so I'm not kicking myself :tongue:
I think this is OK for Q2
Reply 126
Original post by tinto99
Seeing that solution for iii scared the hell out of me! Surely there is no e or ln term in the solution??? My answer to part iii was sort of the same format as my other ones


I forgot what I wrote during the exam...nonetheless I think I got the answer because i was very focused after I failed the 1st attempt by y=ux and used y=ux^2 for second trial

good luck if you are taking III
I think this is OK for Q5
Original post by cpdavis
Q4

1 Integral



2 Integral



3 Integral



4 Integral



Hi, for some reason I tried using IPB for the first two. No idea why I just dived in without thinking about it first, but I did end up with the right answer after a good page of working out. They wouldn't deduct marks for that, would they? For the third and fourth integrals I didn't use the first two results, but I definitely got (iv) right and possibly (iii); do you know whether the question said 'hence', or could you do it any way which worked? Thanks.
Reply 129
Original post by mikelbird
I think this is OK for Q2

For the last part you must separately consider what happens when n=0, when n=1 as well as when c=0 and 1. Remember that they are asking for necessary and sufficient conditions which means they need to be exhaustive :smile:
Original post by Jkn
For the last part you must separately consider what happens when n=0, when n=1 as well as when c=0 and 1. Remember that they are asking for necessary and sufficient conditions which means they need to be exhaustive :smile:


Firstly the conditions are that n is greater or equal 1, secondly c=1 the are an infinite number of solutions...im not sure that you need to consider this.

I suppose it just needs c is not equal to 0 or one (to eliminate the infinite solutions).
(edited 10 years ago)
Reply 131
Original post by mikelbird
Firstly the conditions are that n is greater or equal 1, secondly c=1 the are an infinite number of solutions...im not sure that you need to consider this.

I suppose it just needs c is not equal to 0 or one (to eliminate the infinite solutions).

Oh my bad, didn't see that is specified that! And, as it uses n, we can assume it is not asking for infinities. So yeah, looks spot on in that case :smile:
Original post by Jkn
Oh my bad, didn't see that is specified that! And, as it uses n, we can assume it is not asking for infinities. So yeah, looks spot on in that case :smile:


perhaps this would be better.....
Reply 133
Original post by Math-Magician
Hi, for some reason I tried using IPB for the first two. No idea why I just dived in without thinking about it first, but I did end up with the right answer after a good page of working out. They wouldn't deduct marks for that, would they? For the third and fourth integrals I didn't use the first two results, but I definitely got (iv) right and possibly (iii); do you know whether the question said 'hence', or could you do it any way which worked? Thanks.


IMO, so long as you showed how you got the final answer, then it should be fine :yep: I did the first two integrals just know using ibp and got the answers for both :yep:

How did you solve the last two?
Original post by cpdavis
IMO, so long as you showed how you got the final answer, then it should be fine :yep: I did the first two integrals just know using ibp and got the answers for both :yep:

How did you solve the last two?


For the last two I think I first used by parts to get a nicer looking integral, although I can't remember how I split them up unfortunately. I probably worked out the resulting integrals again using IBP, using exact values rather than leaving them in terms of n, whereas I could have saved time and used the first two results :rolleyes:

Your answers look very familiar so hopefully I still got them right, but I can't remember if the question just said 'evaluate' or 'hence...' so I don't know if I'd still get the marks for them. Basically I raced through this question without really thinking about it so I didn't have a very efficient method.
Step I, Q13.

Spoiler

When are STEP III solutions going to be posted?
Reply 137
The Q8 paper I can someone explain to my why the range of the very last part is greater than or equals to 0 , i thought it was just > thanks!
Reply 138
Original post by jack.hadamard
When are STEP III solutions going to be posted?


The exam was yesterday, so... imminently!
I will do the odd ones. :tongue:

If you find the solutions inelegant (or incorrect), then say so. :smile:

Question 3 (either I missed something, or this one is a slog)

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Question 4 (I typed it in anyway)

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Question 5

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Question 6

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Question 8

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Please, don't quote the entire post, since I plan to edit it.
(edited 10 years ago)

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