[email protected]
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#1
Report Thread starter 13 years ago
#1
Using the method of variation of parameters, solve the equation
y'' + y = tanx, 0<x<pi/2

The auxiliary equation is r^2 + 1 = 0 with roots (+/-)i so the solution of y'' + y = 0 is (c1)sinx + (c2)cosx

So we seek a solution of the form
y = (u1)sinx + (u2)cosx, where (u1) and (u2) are arbitrary functions of x.
y' = ((u1)'sinx + (u2)'cosx) + ((u1)cosx - (u2)sinx)

[1] set (u1)'sinx + (u2)'cosx = 0

y'' = (u1)'cosx - (u2)'sinx - (u1)sinx - (u2)cosx

[2] y'' + y = (u1)'cosx - (u2)'sinx = tanx

Now this is the part I don't get:
Solving equations [1] and [2], we get
(u1)'(sin^2(x) + cos^2(x)) = cosxtanx
(u1)' = sinx, (u1) = -cosx

What did they do with equations [1] and [2] to eliminate (u2)' ???
And how did they get (u1)' from (u1)'(sin^2(x) + cos^2(x)) = cosxtanx
How did they even get (u1)'(sin^2(x) + cos^2(x)) = cosxtanx in the first place?!:confused: ....

I bet it's something really silly that I didn't see.......
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dvs
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Report 13 years ago
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(u1)' sin(x) + (u2)' cos(x) = 0
=> (u1)' sin^2(x) + (u2)' cos(x) sin(x) = 0 ... (*)
(multiply by sin(x))

(u1)' cos(x) - (u2)' sin(x) = tan(x)
=> (u1)' cos^2(x) - (u2)' cos(x) sin(x) = cos(x)tan(x) ... (**)
(multiply by cos(x))

Now add (*) and (**) to get:
(u1)' sin^2(x) + (u1)' cos^2(x) = cos(x)tan(x)
=> (u1)' (sin^2(x) + cos^2(x)) = cos(x)tan(x)
=> (u1)' = cos(x) * sin(x)/cos(x) = sin(x)
(since sin^2(x) + cos^2(x) = 1)
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[email protected]
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#3
Report Thread starter 13 years ago
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(Original post by dvs)
(u1)' sin(x) + (u2)' cos(x) = 0
=> (u1)' sin^2(x) + (u2)' cos(x) sin(x) = 0 ... (*)
(multiply by sin(x))

(u1)' cos(x) - (u2)' sin(x) = tan(x)
=> (u1)' cos^2(x) - (u2)' cos(x) sin(x) = cos(x)tan(x) ... (**)
(multiply by cos(x))

Now add (*) and (**) to get:
(u1)' sin^2(x) + (u1)' cos^2(x) = cos(x)tan(x)
=> (u1)' (sin^2(x) + cos^2(x)) = cos(x)tan(x)
=> (u1)' = cos(x) * sin(x)/cos(x) = sin(x)
(since sin^2(x) + cos^2(x) = 1)
You legend! For both helping me out and having a strange sleeping pattern like me , what time do you usually sleep?!
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dvs
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(Original post by [email protected])
You legend! For both helping me out and having a strange sleeping pattern like me , what time do you usually sleep?!
I go to sleep around 4-5am.

Although to be fair, right now it's almost 11pm (different time zones and all that).
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