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):
dxdy​=yx2​+x2y​
Let y=ux2
dxdy​=2xu+x2dxdu​
Plugging this in gives:
2xu+x2dxdu​=u1​+2xu
x2dxdu​=u1​
∫u du=∫x21​ dx
21​u2=−x1​+c
Plugging in boundary conditions of y = 2, x= 1 (so u = 2) we get c = 3, and:
21​u2=3−x1​
u=6−x2​​ (Note: We take the positive root as x>0, y>0 so u>0)
y=ux2=x26−x2​​
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).
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):
dxdy​=yx2​+x2y​
Let y=ux2
dxdy​=2xu+x2dxdu​
Plugging this in gives:
2xu+x2dxdu​=u1​+2xu
x2dxdu​=u1​
∫u du=∫x21​ dx
21​u2=−x1​+c
Plugging in boundary conditions of y = 2, x= 1 (so u = 2) we get c = 3, and:
21​u2=3−x1​
u=6−x2​​ (Note: We take the positive root as x>0, y>0 so u>0)
y=ux2=x26−x2​​
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 Now if only I could get an average for my other ones
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):
dxdy​=yx2​+x2y​
Let y=ux2
dxdy​=2xu+x2dxdu​
Plugging this in gives:
2xu+x2dxdu​=u1​+2xu
x2dxdu​=u1​
∫u du=∫x21​ dx
21​u2=−x1​+c
Plugging in boundary conditions of y = 2, x= 1 (so u = 2) we get c = 3, and:
21​u2=3−x1​
u=6−x2​​ (Note: We take the positive root as x>0, y>0 so u>0)
y=ux2=x26−x2​​
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
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
I don't think the actual substitution is wrong, but when you get to the separable differential equation you have integrated it incorrectly.
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
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.
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
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
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).
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
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 I did the first two integrals just know using ibp and got the answers for both
IMO, so long as you showed how you got the final answer, then it should be fine I did the first two integrals just know using ibp and got the answers for both
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
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.