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#1 2010-08-24 01:51:39

Akpolome
Member
Registered: 2010-08-23
Posts: 4

Hi Maths Guru I need assistance to the differential below

solve the differential equation y ' = (1/2) sin(2x)

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#2 2010-08-24 01:54:46

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,092

Re: Hi Maths Guru I need assistance to the differential below

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


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#3 2010-08-24 03:17:06

Akpolome
Member
Registered: 2010-08-23
Posts: 4

Re: Hi Maths Guru I need assistance to the differential below

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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#4 2010-08-24 04:50:13

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,092

Re: Hi Maths Guru I need assistance to the differential below

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 bundy (2010-08-24 04:51:47)


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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