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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Polar Coordinates is probably the most difficult subject in FP2, anyone have any guides or strategies to learn it?
Reply 161
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edothero
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15
Gonna go ahead and bump this because people seem to have forgotten about it :frown:
Revising this at the moment, doing differential equations, first order subs.
Reply 163
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edothero
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15
Had an FP2 mock, messed up on one of the complex numbers transformation questions (question was worth 6 marks - should get at least 2). I hate transformations:colonhash:
Apart from that I'm fairly confident I got the rest correct. Including the last question, which was a beautiful second order question :biggrin:

Here's the question if anyone wants to have a go.

Show that the substitution x=ez\displaystyle x=e^{z} transforms the differential equation


x2d2ydx2+2xdydx−2y=3lnx      −x^{2}\dfrac{d^{2}y}{dx^{2}} + 2x\dfrac{dy}{dx} - 2y = 3lnx\ \ \ \ \ \ - (I)

into the equation


d2ydz2+dydz−2y=3z      −\dfrac{d^{2}y}{dz^{2}}+\dfrac{dy}{dz}-2y = 3z\ \ \ \ \ \ - (II)

Hence find the general solution to the differential equation (II) and obtain the general solution of the differential equation (I). Giving your answer in the form y=f(x).
(14 marks)
(edited 8 years ago)
Original post by edothero
Had an FP2 mock, messed up on one of the complex numbers transformation questions (question was worth 6 marks - should get at least 2). I hate transformations:colonhash:
Apart from that I'm fairly confident I got the rest correct. Including the last question, which was a beautiful second order question :biggrin:

Here's the question if anyone wants to have a go.

Show that the substitution x=ez\displaystyle x=e^{z} transforms the differential equation


x2d2ydx2+2xdydx−2y=3lnx      −x^{2}\dfrac{d^{2}y}{dx^{2}} + 2x\dfrac{dy}{dx} - 2y = 3lnx\ \ \ \ \ \ - (I)
into the equation


d2ydz2+dydz−2y=3z      −\dfrac{d^{2}y}{dz^{2}}+\dfrac{dy}{dz}-2y = 3z\ \ \ \ \ \ - (II)

Hence find the general solution to the differential equation (II) and obtain the general solution of the differential equation (I). Giving your answer in the form y=f(x).
(14 marks)


This is a nice question, started a minute ago and I'm struggle :colondollar:
Reply 165
Avatar for edothero
edothero
OP
15
Original post by kkboyk
This is a nice question, started a minute ago and I'm struggle :colondollar:


Hint: dydx= ?×?\dfrac{dy}{dx} =\ ? \times ?

Using zz :biggrin:
Original post by edothero
Hint: dydx= ?×?\dfrac{dy}{dx} =\ ? \times ?

Using zz :biggrin:


I had to revisit 2nd ODE to know which form of particular integral my answer must have. But it took the piss.
Original post by edothero
Had an FP2 mock, messed up on one of the complex numbers transformation questions (question was worth 6 marks - should get at least 2). I hate transformations:colonhash:
Apart from that I'm fairly confident I got the rest correct. Including the last question, which was a beautiful second order question :biggrin:

Here's the question if anyone wants to have a go.

Show that the substitution x=ez\displaystyle x=e^{z} transforms the differential equation


x2d2ydx2+2xdydx−2y=3lnx      −x^{2}\dfrac{d^{2}y}{dx^{2}} + 2x\dfrac{dy}{dx} - 2y = 3lnx\ \ \ \ \ \ - (I)
into the equation


d2ydz2+dydz−2y=3z      −\dfrac{d^{2}y}{dz^{2}}+\dfrac{dy}{dz}-2y = 3z\ \ \ \ \ \ - (II)

Hence find the general solution to the differential equation (II) and obtain the general solution of the differential equation (I). Giving your answer in the form y=f(x).
(14 marks)


Might as well. We have: dxdz=ez\frac{\mathrm{d}x}{\mathrm{d}z} = e^z so since ezdydx=dydze^z \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}z} , we then have, by taking second derivatives:

and d2ydx2=(e−zy′′−y′e−z)dzdx=e−2z(y′′−y′)\frac{d^2y}{\mathrm{d}x^2} = (e^{-z} y'' - y'e^{-z})\frac{\mathrm{d}z}{\mathrm{d}x} = e^{-2z}(y'' - y' )

So our DE becomes:

y′′−y′+2y′−2y=3z  ⟺  y′′+y′−2y=3z\displaystyle y'' - y' + 2y' - 2y = 3z \iff y'' + y' - 2y = 3z as required.

Now this is easily solvable, the auxiliary quadratic is α2+α−2=0⇒α=−2,1\alpha^2 + \alpha - 2 = 0 \Rightarrow \alpha = -2, 1

So complementary solution is Ae−2z+BezAe^{-2z} + Be^{z} and the particular solution −32-\frac{3}{2}, so:

y=Ae−2z+Bez−3z2−34\displaystyle y = Ae^{-2z} + Be^{z} - \frac{3z}{2} - \frac{3}{4}, then

y=Ax2+Bx−3ln⁡x2−34\displaystyle y = \frac{A}{x^2} + Bx - \frac{3\ln x}{2} - \frac{3}{4}
(edited 8 years ago)
Reply 168
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edothero
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15
Original post by Zacken
...


Yep very nice :biggrin:
Reply 169
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edothero
OP
15
Original post by kkboyk
I had to revisit 2nd ODE to know which form of particular integral my answer must have. But it took the piss.


What have you got up to? You might want to look at Zacken's solution. If you don't understand anything I can elaborate
Reply 170
Avatar for edothero
edothero
OP
15
Original post by Zacken
So complementary solution is Ae−2z+BezAe^{-2z} + Be^{z} and the particular solution −32-\frac{3}{2}, so:

y=Ae−2z+Bez−32\displaystyle y = Ae^{-2z} + Be^{z} - \frac{3}{2}, then

y=Ax2+Bx−32\displaystyle y = \frac{A}{x^2} + Bx - \frac{3}{2}


Hang on, just realised, your particular solution is incorrect

Spoiler

(edited 8 years ago)
Original post by edothero
Hang on, just realised, your particular solution is incorrect

Spoiler



Lol yeah, that's what I get for doing it in my head thanks for pointing it out.
Original post by edothero
Hang on, just realised, your particular solution is incorrect

Spoiler



Yeah I realised my mistake here thanks :tongue:
Hi, please could someone explain how to do question 8c on this paper https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/Mock%20QP%20-%20FP2%20Edexcel.pdf in particular how you get arc sin 3/13 as I'm not sure how this is found! Thanks :smile:
Original post by economicss
Hi, please could someone explain how to do question 8c on this paper https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/Mock%20QP%20-%20FP2%20Edexcel.pdf in particular how you get arc sin 3/13 as I'm not sure how this is found! Thanks :smile:


this should help, find the angle from the origin to the point (5,12) then use the info on the diagram to find the min and max tangents

Original post by DylanJ42
this should help, find the angle from the origin to the point (5,12) then use the info on the diagram to find the min and max tangents



Thank you so much!
Show that the locus of z=x+yi,arg⁡(z−3iz−5)=π4\displaystyle z=x+yi, \arg\left(\frac{z-3i}{z-5}\right)=\frac{\pi}{4}, has the cartesian equation (x−1)2+(y+1)2=17\displaystyle (x-1)^2+(y+1)^2=17, subject to 5y+3x<15\displaystyle 5y+3x<15.

Spoiler

(edited 8 years ago)
Original post by EricPiphany
Show that the locus of z=x+yi,arg⁡(z−3iz−5)=π4\displaystyle z=x+yi, \arg\left(\frac{z-3i}{z-5}\right)=\frac{\pi}{4}, has the cartesian equation (x−1)2+(y+1)2=17\displaystyle (x-1)^2+(y+1)^2=17, subject to 5y+3x<15\displaystyle 5y+3x<15.

Spoiler


Nice questions :tongue:
Original post by Joshthemathmo
Nice questions :tongue:


Thanks :colondollar:
The question is actually in the Edexcel textbook, 3F 7 d, as a 'sketch' problem, I just decided to try it algebriacally.
(edited 8 years ago)
Original post by EricPiphany
Thanks :colondollar:
The question is actually in the Edexcel textbook, 3F 7 d, as a 'sketch' problem, I just decided to try it algebriacally.


Fair enough, i haven't really thoroughly gone through the textbook questions. It's nice to see problems tackled from different positions :h:

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