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[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

[OK Lor]

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

Hi iFruit,

Thanks for your help. Was stuck and coundn't continue from (1) to get the answer without (2), which it was just one more step to go, the reverse of (2) :lol:

[iFruit]

OK Lor:

iFruit:

[quote=\"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

Hi iFruit,

Thanks for your help. Was stuck and coundn't continue from (1) to get the answer without (2), which it was just one more step to go, the reverse of (2) :lol:[/quote]Hi OK Lor,

Sure. Yeah, sometimes it is easier to simplify both sides to a common third.

[Mathducator]

@atutor2001


Primary school mathematics is harder in a sense: you are like fighting with bare hands (e.g. you normally don’t use algebra, but bar model diagrams). Secondary and JC mathematics is like fighting with knives and guns – it’s easier to ‘kill’, but you have to learn how to use the weapons first.

[OK Lor]

Hi,


Please evaluate
http://www.postimage.org/

Thanks.