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Year 13 Maths Help Thread

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Original post by kiiten
Now i got 3 + sqrt 17 instead of 3 + sqrt 13??? :s-smilie:


Did you make 2x22x^2 equal the DOUBLE of the path?
Original post by RDKGames
Did you make 2x22x^2 equal the DOUBLE of the path?


No...

so would it be

4x^2 = (2x+2)(x+2) - 2x^2
Original post by kiiten
No...

so would it be

4x^2 = (2x+2)(x+2) - 2x^2


Almost. I see you took a different approach, which is still technically right.

Try: 2x2=2[(2x+2)(x+2)−2x2]2x^2=2[(2x+2)(x+2)-2x^2]

Im not sure where you pulled that 4x24x^2 from.
Original post by RDKGames
Almost. I see you took a different approach, which is still technically right.

Try: 2x2=2[(2x+2)(x+2)−2x2]2x^2=2[(2x+2)(x+2)-2x^2]

Im not sure where you pulled that 4x24x^2 from.


i multiplied the 2x^2 by 2 because it says the area of the lawn (2x^2) is twice that of the path hence 4x^2 ?

EDIT: i misunderstood .... again. Thanks :smile:
Original post by RDKGames
Almost. I see you took a different approach, which is still technically right.

Try: 2x2=2[(2x+2)(x+2)−2x2]2x^2=2[(2x+2)(x+2)-2x^2]

Im not sure where you pulled that 4x24x^2 from.


Sorry i keep going on about this question xD but shouldnt the answer by multiplied by 2 (2x) ??
its asking for the length of the lawn.
Original post by kiiten
Sorry i keep going on about this question xD but shouldnt the answer by multiplied by 2 (2x) ??
its asking for the length of the lawn.


Yep.
Original post by RDKGames
Yep.


So its 6 +2 sqrt 13 (not 3 + sqrt 13)
Original post by kiiten
So its 6 +2 sqrt 13 (not 3 + sqrt 13)


Yep.
Original post by RDKGames
Yep.


yay finally i got something right :smile:
Someone explain what how am I supposed to go about this?

Question asks for a complementary function and a particular integral of dydx−3y=6\frac{dy}{dx}-3y=6 and I've gotten the CF to be y=ae3xy=ae^{3x} but I'm unsure where to go from there because subbing it into the original differential doesn't exactly favour maths as 0≠60\not= 6 so there's no value for aa that will make this work, is there? What does this mean for the PI?
(edited 7 years ago)
Original post by RDKGames
Someone explain what how am I supposed to go about this?

Question asks for a complementary function and a particular integral of dydx−3y=6\frac{dy}{dx}-3y=6 and I've gotten the CF to be y=ae3xy=ae^{3x} but I'm unsure where to go from there because subbing it into the original differential doesn't exactly favour maths as 0≠60\not= 6


Try a PI of the form p p where p p is just a real constant. You don't need to sub your complementary function into the ODE once you have found it. The arbitrary constant a a has to be there for the general solution and you can only find the value of a a if you're given boundary conditions. your general solution will just be y=Ae3x+PI y=\text{A} e^{3x} + PI .
(edited 7 years ago)
Original post by B_9710
Try a PI of the form p p where p p is just a real constant. You don't need to sub your complementary function into the ODE once you have found it. The arbitrary constant a a has to be there for the general solution and you can only find the value of a a if you're given boundary conditions. your general solution will just be y=Ae3x+PI y=\text{A} e^{3x} + PI .


Ah yes I remember that now, y=CF+PIy=CF+PI
So CF=Ae3xCF=Ae^{3x} and since it cancels, it only leaves −3p=6-3p=6
and therefore p=−2=PIp=-2=PI?
(edited 7 years ago)
Original post by RDKGames
Ah yes I remember that now, y=CF+PIy=CF+PI
So CF=Ae3xCF=Ae^{3x} and since it cancels, it only leaves −3p=6-3p=6
and therefore p=−2=PIp=-2=PI?


Yes that's right.
Original post by B_9710
Yes that's right.


Can you explain why the CF of dydx−2y=e2x\frac{dy}{dx}-2y=e^{2x} is yc=Ce2xy_c=Ce^{2x} and not, as I found it, yc=Cxe2xy_c=Cxe^{2x}? Or are they wrong?
Because when I used mine I get PI to be xe2xxe^{2x} which is correct on their answers.
Original post by RDKGames
Can you explain why the CF of dydx−2y=e2x\frac{dy}{dx}-2y=e^{2x} is yc=Ce2xy_c=Ce^{2x} and not, as I found it, yc=Cxe2xy_c=Cxe^{2x}? Or are they wrong?
Because when I used mine I get PI to be xe2xxe^{2x} which is correct on their answers.


The CF is pxe2x pxe^{2x} .
Original post by B_9710
The CF is pxe2x pxe^{2x} .


This book smh... :facepalm:
Am i doing this wrong?

Solve: 5+2sin(2x+1)=6 for 0<=x<=720

rearranged to sin(2x+1)= 1/2

let u = 2x+1
sinu=1/2
u=1/6 pi

change the bounds to 1 <=u<=4pi +1
using a cast circle i got the first answer to be 2.05 but it should be 0.81 ???

*sorry if this doesnt make sense - let me know and ill post the full working
Not quite sure where I'm going wrong.

dudx+2ux=1x2\frac{du}{dx}+\frac{2u}{x}=\frac{1}{x^2}

IF=e∫2x.dx=x2IF=e^{\int \frac{2}{x} .dx}=x^2

but that doesn't seem right as I wouldn't get the left side from ddx(ux2)\frac{d}{dx}(ux^2)
(edited 7 years ago)
Original post by kiiten
Am i doing this wrong?

Solve: 5+2sin(2x+1)=6 for 0<=x<=720

rearranged to sin(2x+1)= 1/2

let u = 2x+1
sinu=1/2
u=1/6 pi

change the bounds to 1 <=u<=4pi +1
using a cast circle i got the first answer to be 2.05 but it should be 0.81 ???

*sorry if this doesnt make sense - let me know and ill post the full working


sin(x)=αsin(x)=\alpha
Then:
x=2nπ+arcsin(α)x=2n\pi + arcsin(\alpha)

and due to symmetry of sine:

x=2nπ+π−arcsin(α)x=2n\pi + \pi - arcsin(\alpha)

For any integer n (try a few around 0 until your solutions go outside the range)

Try that.
Original post by RDKGames
sin(x)=αsin(x)=\alpha
Then:
x=2nπ+arcsin(α)x=2n\pi + arcsin(\alpha)

and due to symmetry of sine:

x=2nπ+π−arcsin(α)x=2n\pi + \pi - arcsin(\alpha)

For any integer n (try a few around 0 until your solutions go outside the range)

Try that.


??? I have no idea what you're talking about.
What is 2n and is arcsin inverse of sin?

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