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

[iFruit]

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.

Hi atutor1,

Actually it is not so difficult as it sounds. It is a well defined procedure to solve the problem. That is why I was mentioning earlier the need to know the background theory.

[atutor2001]

Hi Chief & iFruit


Thank you for your clarifications. Must set resolution for next year - "To study A-level math with my youngest one". If don’t know sure will post here. Piaseh, you all gonna be very busy. Many many thanks in advance first.

Cheers

[iFruit]

atutor2001:

Hi Chief & iFruit


Thank you for your clarifications. Must set resolution for next year - \"To study A-level math with my youngest one\". If don't know sure will post here. Piaseh, you all gonna be very busy. Many many thanks in advance first.

Cheers

Ah atutor2001....Going by the thorough manner in which you answer questions in other forums, I'm sure it will be us asking you for solutions, not the other way... :celebrate:

[atutor2001]

Hi iFruit


Thank you for your kind words. I would like to believe them but I know my limited knowledge. Only very kay kiang and naggy. I hope I would not shock you with how silly my future questions can be.

:celebrate:

[mathqa]

mramk:

iFruit:

[quote=\"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.[/quote]Thanks Mr AMK, and iFruit for your feedbacks and kind words

[OK Lor]

Hi,


Function f is defined by f(x) = { ax² + bx – 2, x ≤ 1; 2 – 1/x², x > 1
where a and b are constants. If f(x) is differentiable at x = 1, find the values of a and b.
Ans: a = -1, b = 4

Thanks.

[mathqa]

OK Lor:

Hi,


Function f is defined by f(x) = { ax² + bx – 2, x ≤1; 2 – 1/x², x > 1
where a and b are constants. If f(x) is differentiable at x = 1, find the values of a and b.
Ans: a = -1, b = 4

Thanks.

1. f(x) is differentiable at x=1 -> f'(1-) = f'(1+) -> 2a+b=2
( 1+ denotes right hand side of 1,
1- denotes left hand side of 1,)

2. since f(x) is differentiable at x=1, it is continous at x=1 (please note that the converse is not true)
f(1-)=f(1+) -> a+b-2=1

3. In summary, 2a+b=2, a+b=3 -> Ans.

[OK Lor]

mathqa:

OK Lor:

Hi,


Function f is defined by f(x) = { ax² + bx – 2, x ≤1; 2 – 1/x², x > 1
where a and b are constants. If f(x) is differentiable at x = 1, find the values of a and b.
Ans: a = -1, b = 4

Thanks.

1. f(x) is differentiable at x=1 -> f'(1-) = f'(1+) -> 2a+b=2
( 1+ denotes right hand side of 1,
1- denotes left hand side of 1,)

2. since f(x) is differentiable at x=1, it is continous at x=1 (please note that the converse is not true)
f(1-)=f(1+) -> a+b-2=1

3. In summary, 2a+b=2, a+b=3 -> Ans.

Thank you mathqa

[OK Lor]

Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

[iFruit]

OK Lor:

Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

y = eˣ / cos x = eˣ sec x

dy/dx = eˣ sec x + eˣ sec x tan x = eˣ sec x (1+ tan x) = y (1+ tan x)

d²y/dx² = dy/dx (1 + tan x) + y ( sec² x) = dy/dx (1 + tan x) + y ( 1 + tan² x) = y(1+ tan x)(1+tan x) + y (1 + tan² x) = 2y (1 + tan x + tan² x) ----(1)

2 tanx dy/dx + 2y = 2 tanx . y (1+ tan x) + 2y = 2y ( 1 + tan x + tan² x)---- (2)

From (1) and (2), d²y/dx² = 2 tanx dy/dx + 2y.

HTH