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solve the differential equation y ' = (1/2) sin(2x)
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Hi Akpolome
Have you been taught how the differentiate and integrate 'trig' functions yet?
eg. y' = sinx
If you can tell me this, we've got the right basis for your question next.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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solution of Y' =sinX
Bob
integrate both side of the equation the result will be
Y=-cosX + K
where K is constant of integration
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Hi Akpolome,
OK, now I know where we're starting from.
The 1/2 at the start can simply be ignored until you write your answer as it's just a multiplier.
So what about the sin(2x) ?
If you wanted to differentiate cos(2x) you'd have to use the 'chain rule' as it's a 'function of a function' .
ie cos u where u = 2x
So you have to differentiate cos u (=-sin u) and multiply by u differentiated with respect to x (= 2)
So d (cos2x) /dx = -sin u . 2 = -2sin(2x).
Now back to the integration, ask what must I have differentiated to get (1/2). sin (2x)
It must have a cos and must start minus so that the result ends up positive and it shouldn't have a 2 in it, so better half at the start and finally another 1/2 for the multiplier.
............................... - (1/2) . (1/2) . cos (2x) + C = -(1/4)cos(2x) + C
and that's the answer.
When I have to do these I usually start with a guess at the function ( cos something) and then adjust my answer so that when it is differentiated it gives the desired result.
Hope that helps,
Bob
Last edited by Bob (2010-08-24 04:51:47)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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