Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

That's a weird one. For what its worth, I get $\lambda = \frac{3}{2}$ with the particular integral .

I'd also be appreciative of anyone who could weigh in on this... since doesn't form part of our complimentary function?

I've seen them do this before as well. I think it's so that when you differentiate to get $dy/dx$, and then when you differentiate again to get $d^2/dx^2$, t is always in your differential value, (if you used $t^2$, then by the time you work out $d^2y/dx^2$, the t term would have disappeared).

I don't see why it's necessary though, but I think that's their logic.

Just had a thought... I think it's to do with the fact that appears in our complimentary function, but also our auxiliary equation has repeated roots, so we multiply it by t to give to account for it appearing in our complimentary function, then t again to give to account for the repeated roots.

Or that could be crazy talk.

@Zacken ? have you got any words of wisdom?

That's a weird one. For what its worth, I get $\lambda = \frac{3}{2}$ with the particular integral .

Just had a thought... I think it's to do with the fact that appears in our complimentary function, but also our auxiliary equation has repeated roots, so we multiply it by t to give to account for it appearing in our complimentary function, then t again to give to account for the repeated roots.Or that could be crazy talk.

Just had a thought... I think it's to do with the fact that appears in our complimentary function, but also our auxiliary equation has repeated roots, so we multiply it by t to give to account for it appearing in our complimentary function, then t again to give to account for the repeated roots.Or that could be crazy talk.

Not so crazy, that's precisely it.

If you find out whether this is right or wrong, can you explain it to me in a little more detail please?

If you find out whether this is right or wrong, can you explain it to me in a little more detail please?

Yes that's right
To find the value of x1, sub only the points where C meets the x-axis into the general equation to get (2-x1)^2 + (0-y1)^2 = r^2 and (-1-x1)^2 + (0-y1)^2 = r^2. If you then equate them, you do not need to find the value for r^2 and the y1^2 terms cancel out.
Do a separate one using the points where C meets the y-axis in order to find the value of y1 and then x1 and y1 will be your centre coordinates.

My logic is this:

We have a complementary function with two constants to find, A and B. However, since our auxiliary equation only gives one solution, we cannot find the two constants as we need two independent solutions to solve for A and B.

So in order to find these two constants, we multiply by t to give us our second independent solution to the general solution.

So if it already appears in the CF, you multiply by t, but then if the CF has two equal roots, then you have to multiply by t again?

Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

In general, if you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t}$ and your CF contains $\alpha t$ and you have repeated roots, it is true that you need to use switch from using $\alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}$.

i.e: the default guess is $\alpha e^{2t}$ and then you multiply by $t$ when you find out that you have repeated roots and then multiply by $t$ again when you find out that your CF contains $\alpha t$.

In this case, you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}$.

Note the extra factor of $t$ on the RHS.

So your default guess is $\alpha t e^{2t}$ since you want to match the form of the RHS.

Then you find out your CF contains that term, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t}$.

Then you find out you have repeated roots, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}$.

Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

In general, if you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t}$ and your CF contains $\alpha t$ and you have repeated roots, it is true that you need to use switch from using $\alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}$.

i.e: the default guess is $\alpha e^{2t}$ and then you multiply by $t$ when you find out that you have repeated roots and then multiply by $t$ again when you find out that your CF contains $\alpha t$.

In this case, you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}$.

Note the extra factor of $t$ on the RHS.

So your default guess is $\alpha t e^{2t}$ since you want to match the form of the RHS.

Then you find out your CF contains that term, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t}$.

Then you find out you have repeated roots, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}$.

Question: https://gyazo.com/3a02dd69a1250ddf22c5f7b51a4946aa
Answer: https://gyazo.com/9b04261b67135f9f0cd792a1b961b03d
2nd ODEs
Basically I've used the transformation y=xv to make the DE go from y wrt. x to v wrt. x and found the GS of v wrt.x. When I sub the v back in since v y/x , which means everything on the RHS is to be times by x. I'm wondering why in the mark scheme only the (Asin3x+Bsin3x) is times by x but the (x^2/9)-(2/81) isn't? Is there a mistake?

Just gonna repost this here cuz I don't think anyone saw it

Just gonna repost this here cuz I don't think anyone saw it

Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

In general, if you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t}$ and your CF contains $\alpha t e^{2t}$ and you have repeated roots, it is true that you need to use switch from using $\alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}$.

i.e: the default guess is $\alpha e^{2t}$ and then you multiply by $t$ when you find out that you have repeated roots and then multiply by $t$ again when you find out that your CF contains $\alpha t$.

In this case, you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}$.

Note the extra factor of $t$ on the RHS.

So your default guess is $\alpha t e^{2t}$ since you want to match the form of the RHS.

Then you find out your CF contains that term, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t}$.

Then you find out you have repeated roots, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}$.

Sorry guys, I misread what Oinkk was saying and thought he'd said what I thought was going on.

In general, if you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 e^{2t}$ and your CF contains $\alpha t e^{2t}$ and you have repeated roots, it is true that you need to use switch from using $\alpha e^{2t} \to \alpha t e^{2t} \to \alpha t^2 e^{2t}$.

i.e: the default guess is $\alpha e^{2t}$ and then you multiply by $t$ when you find out that you have repeated roots and then multiply by $t$ again when you find out that your CF contains $\alpha t$.

In this case, you have $\lambda_1 y'' + \lambda_2 y' + \lambda_3 y = \lambda_4 te^{2t}$.

Note the extra factor of $t$ on the RHS.

So your default guess is $\alpha t e^{2t}$ since you want to match the form of the RHS.

Then you find out your CF contains that term, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t}$.

Then you find out you have repeated roots, so multiply by t. $\alpha t e^{2t} \to \alpha t^2 e^{2t} \to \alpha t^3 e^{2t}$.