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Topic: Can't find my trig/DE mistake.
Replies: 1   Last Post: Nov 10, 2012 10:37 PM

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dy/dx

Posts: 6
Registered: 11/10/12
Can't find my trig/DE mistake.
Posted: Nov 10, 2012 6:52 PM
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Trying to solve this differential equation:

dt/dr = k(cos^2 t - sin^2 t)/r^2

I seem to have run into a snag. Here's what I've got so far:

dt/dr = k(cos^2 t - sin^2 t)/r^2
= k cos 2t/r^2

which is separable:

dt/cos 2t = k dr/r^2

One change of variables later we have

int_a^t(R) dt/cos 2t = int_0^R k dr/r^2

and, integrating,

ln(tan(t + pi/4))/2 |_a^t(R) = -k/R

ln(tan(t(R) + pi/4)) - ln(tan(a + pi/4)) = -2k/R

Now it's all over except for the algebra:

ln(tan(t(R) + pi/4)) = ln(tan(a + pi/4)) - 2k/R

tan(t(R) + pi/4) = e^[ln(tan(a + pi/4)) - 2k/R]
= tan(a + pi/4)/e^(2k/R)

tan(t(R) + pi/4)cot(a + pi/4) = e^(-2k/R)

tan(t(R) + pi/4)tan(pi/4 - a) = e^(-2k/R)

t(R) = tan^-1(e^(-2k/R)/tan(pi/4 - a)) - pi/4

Checking this as R goes to zero:

t(0) -> -pi/4

Shouldn't t(0) go to a?



I also did the following in an attempt to get rid of the
tan^-1(...tan(...)):


tan(t(R) + pi/4)tan(pi/4 - a) = e^(-2k/R)

[sin(t(R) + pi/4)sin(pi/4 - a)]/[cos(t(R) + pi/4)cos(pi/4 - a)] = e^(-2k/R)

[sin(t(R) + pi/4)sin(pi/4 - a) + cos(t(R) + pi/4)cos(pi/4 - a) - cos(t(R) +
pi/4)cos(pi/4 - a)]/[cos(t(R) + pi/4)cos(pi/4 - a)] = e^(-2k/R)

[cos(t(R) + a) - cos(t(R) + pi/4)cos(pi/4 - a)]/[cos(t(R) + pi/4)cos(pi/4 -
a)] = e^(-2k/R)

(cos(t(R) + a))/[cos(t(R) + pi/4)cos(pi/4 - a)] = e^(-2k/R) + 1

2(cos(t(R) + a))/[cos(t(R) + a) + cos(t(R) + pi/2 - a)] = e^(-2k/R) + 1

[cos(t(R) + a) + cos(t(R) + pi/2 - a)]/cos(t(R) + a) = 2/(e^(-2k/R) + 1)

sin(t(R) - a)/cos(t(R) + a) = 1 - 2/(e^(-2k/R) + 1)

sin(t(R) + a - 2a)/cos(t(R) + a) = 1 - 2/(e^(-2k/R) + 1)

[sin(t(R) + a)cos(2a) - cos(t(R) + a)sin(2a)]/cos(t(R) + a) = 1 -
2/(e^(-2k/R) + 1)

tan(t(R) + a)cos(2a) - sin(2a) = 1 - 2/(e^(-2k/R) + 1)
= (e^(-2k/R) - 1)/(e^(-2k/R) + 1)

t(R) = tan^-1((e^(-2k/R) - 1)/(e^(-2k/R) + 1)cos(2a) + tan(2a)) - a

A little messy, and now there are three a's, but at least there's no tan
nested in tan^-1.

t(0) -> tan^-1((sin(2a) - 1)/cos(2a)) - a
= tan^-1(-(cos a - sin a))/(cos a + sin a)) - a
= tan^-1(-(cos a + cos(a + pi/2))/(sin a + sin(a + pi/2))) - a
= tan^-1(-1/(tan(a + pi/4))) - a
= tan^-1(-cot(a + pi/4)) - a
= tan^-1(-cot(pi/2 - (pi/4 - a))) - a
= tan^-1(-tan(pi/4 - a)) - a
= tan^-1(tan(a - pi/4)) - a
= -pi/4

At least that agrees with the above.



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