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C3/C4/FP2/FP3,M1,M2,M4,M5 Resources and links

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Thanks very useful :smile:
Reply 7
In FP2 where are series?
Nevermind they are really easy
Reply 9
Original post by simonli2575
Why is it that in Question 3 of 2nd Order ODE, the particular integral doesn't change?


Posted from TSR Mobile


maybe a typo or a mistake

I will check later

57.jpg
Reply 10
Original post by simonli2575
Why is it that in Question 3 of 2nd Order ODE, the particular integral doesn't change?


Posted from TSR Mobile


I did look at the question and I am afraid I cannot see the problem
Can you be a bit more specific so if it is wrong I can change it
Original post by TeeEm
I did look at the question and I am afraid I cannot see the problem
Can you be a bit more specific so if it is wrong I can change it


In question 5, the particular integral is divided by x.

I also remember that my teacher told me the particular integral doesn't change when doing 2nd order ODEs, but why? And why not in question 5?


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Reply 12
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(edited 7 years ago)
Reply 13
Original post by simonli2575
In question 5, the particular integral is divided by x.

I also remember that my teacher told me the particular integral doesn't change when doing 2nd order ODEs, but why? And why not in question 5?


Posted from TSR Mobile

Sorry but I do not follow
Original post by TeeEm
Sorry but I do not follow


Question 3:
You find that the general solution is
v = y/x = Ae^2x + Be^-2x - e^x
But when you multiply both sides by x, this becomes
y = Axe^2x + Bxe^-2x - e^x
In which e^x is not multiplied by x, and is also a particular integral.

Similarly, in question 5:
You find that the general solution is
u = xy = Ae^-3x + Bxe^-3x + 3x - 2
However, this time when both sides are divided by x, you get
y = A/x*e^-3x + Be^-3x + 3 - 2/x
In which the particular integral, 3x - 2, has been changed.
My question is, do all particular integrals remain unchanged during substitution? If so, why? If not, why not?


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Reply 15
Original post by simonli2575
Question 3:
You find that the general solution is
v = y/x = Ae^2x + Be^-2x - e^x
But when you multiply both sides by x, this becomes
y = Axe^2x + Bxe^-2x - e^x
In which e^x is not multiplied by x, and is also a particular integral.

Similarly, in question 5:
You find that the general solution is
u = xy = Ae^-3x + Bxe^-3x + 3x - 2
However, this time when both sides are divided by x, you get
y = A/x*e^-3x + Be^-3x + 3 - 2/x
In which the particular integral, 3x - 2, has been changed.
My question is, do all particular integrals remain unchanged during substitution? If so, why? If not, why not?


Posted from TSR Mobile


which booklet are you looking at?
is it this one?
http://madasmaths.com/archive/maths_booklets/further_topics/integration/2nd_order_differential_equations_exam_questions.pdf
Reply 19
I found the questions now.
The PI will not change if it is a function say of the independent variable (say x) and the substitution used transforms the dependent variable (say y)
if you get a question like this in your exam you just follow the instructions!

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