MathQA tutor - Ask your A-level Maths questions here!

[OK Lor]

Hi,


I’m not taking A-level Maths this year, but interested to know the solution for part (b):

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

Thanks.

[iFruit]

OK Lor:

Hi,


I'm not taking A-level Maths this year, but interested to know the solution for part (b):

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

Thanks.

Hi OK Lor,

Is this A-level question? AFAI can see, The equation is a non-homogeneous linear differential equation, so it needs some background to understand the solution..

[OK Lor]

Hi iFruit,

This question is from Malaysia A level paper which I came across while browsing the web :!: .
Thanks.

[iFruit]

OK Lor:

Hi iFruit,

This question is from Malaysia A level paper which I came across while browsing the web :!: .
Thanks.

From what I understand ( the wording is not great), it is asking for a solution of the equation 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t given the initial conditions.

Let me know if you are interested, I can try and and work out a solution (need to recollect my linear diff equations first) but as I said it needs some background theory on linear diff equations first.

[atutor2001]

OK Lor:

Hi,


I'm not taking A-level Maths this year, but interested to know the solution for part (b):

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

Thanks.

diff both statement wrt to t
so the first one is y'' + 2x'' =2x'
second one is y'' - x'' =2y'+1

then simulataneous : statement 1 + 2 x statement 2
so we get 3y''=2x'+4y'+2

then we sub y'-x'=2y+t (2nd given eqn) into it

3y'' = 2(y'-2y-t) +4y' +2
3y'' = 2y'-4y-2t+4y'+2
3y'' = 6y'-4y-2t+2
3y''-6y'+4y =2-2t

Disclaimer : Above working not from me. Was chatting with my kid so took the opportunity to test :lol:

[mathqa]

iFruit:

OK Lor:

Hi,


I'm not taking A-level Maths this year, but interested to know the solution for part (b):

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

Thanks.

Hi OK Lor,

Is this A-level question? AFAI can see, The equation is a non-homogeneous linear differential equation, so it needs some background to understand the solution..

iFruit is right to point out that this question is non-homogeneous linear differential equation. But it is nice to have for A level since it would test students's competence in many areas at once.

The solution is is posted at my blog. Cannot post it here due to oversize images.

http://mathqa.blogspot.com/2010/11/nonhomgeneous-differential-equations.html

MathQA

[OK Lor]

mathqa:

iFruit:

[quote=\"OK Lor\"]Hi,


I'm not taking A-level Maths this year, but interested to know the solution for part (b):

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

Thanks.

Hi OK Lor,

Is this A-level question? AFAI can see, The equation is a non-homogeneous linear differential equation, so it needs some background to understand the solution..

iFruit is right to point out that this question is non-homogeneous linear differential equation. But it is nice to have for A level since it would test students's competence in many areas at once.

The solution is is posted at my blog. Cannot post it here due to oversize images.

http://mathqa.blogspot.com/2010/11/nonhomgeneous-differential-equations.html

MathQA[/quote]Hi,

Thanks to all the Maths Guru here

[iFruit]

mathqa:

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

The solution is is posted at my blog. Cannot post it here due to oversize images.

http://mathqa.blogspot.com/2010/11/nonhomgeneous-differential-equations.html

MathQA

Hi MathQA,

Nice working. But I think there is a small mistake in it. You can't apply initial conditions to the homogeneous equation (equation 6) as they are initial conditions for non-homogeneous equation. That's why your solution doesn't tally back for initial conditions. Also particular solution C = -11/4

The solution after applying initial conditions to the general equation should be (pi = π)

x(t) = ((1-2π)/4) e^t cos(t/√3) - (√3(1+2π)/4) e^t sin(t/√3) - 11/4

Regards.

[atutor2001]

Whoa! So scary. Read so many times also no understand. I didn’t know my poor kids have been subjected to such "torture". Must be nicer to them now.


So funny when comparing this to Pr Math. Parents are making so much noise when kids are in Pr but did not know what their older kids are facing.

[mramk]

iFruit:

mathqa:

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

The solution is is posted at my blog. Cannot post it here due to oversize images.

http://mathqa.blogspot.com/2010/11/nonhomgeneous-differential-equations.html

MathQA

Hi MathQA,

Nice working. But I think there is a small mistake in it. You can't apply initial conditions to the homogeneous equation (equation 6) as they are initial conditions for non-homogeneous equation. That's why your solution doesn't tally back for initial conditions. Also particular solution C = -11/4

The solution after applying initial conditions to the general equation should be (pi = π)

x(t) = ((1-2π)/4) e^t cos(t/√3) - (√3(1+2π)/4) e^t sin(t/√3) - 11/4

Regards.

Thanks MathQA for volunteering your time to help solve this tough question.
Small mistake does not matter. It is just + or - of a constant value.
Overall steps posted on blog are well defined and easily followed.
But may I suggest you try to post the images of your works here. The forum does support [img] tag.

@iFruit: we need steps to understand how to solve the question, not just final answer. How to solve the problem is much more important than final answer. Appreciate if you could be more elaborate next time then :). Thanks!!!

Best regards,
Mr AMK.