Hello I was wondering why in this differential equation they did not use double angle identity, I understand how to get the other result also(adsbygoogle = window.adsbygoogle || []).push({});

So

1)Dif eqn:

Dy/dx +ycotx= cosx

Integrating factor IF= e^ (integrand of cotx) = sinx

Solution

yIF= integrand of QIF where Q is cotx from original question)

y sinx= integral of cosxsinx

Now I recognised this as ½ sin2x and the integral would be

ysinx= ½(-cos2x/2+ c)

therefore y= 1/sinx (-cos2x/4+ c/2)

however this could be done by substitution

y sinx= integral of cosxsinx let u=sinx to give

ysinx= integral of cosx./u du/cosx

ysinx= sin^2x +c

y=1/sinx (sin^2x +c)

why are they not the same ?? or are they really the same

I mean if we differentiate either sin^2x or -cos2x/2 we would get back to cosxsinx

2)

(x-2)dy/dx –y = (x-2)^3

dy/dx – y/(x-2)= (x-2)^2

IF= 1/(x-2)

yIF= intergral of QIF

y/(x-2)= integral of (x-2)^2 .1/(x-2)

y/(x-2)= integral of (x-2)

If I were to integrate (x-2) why is it (x-2)^2/2 +c is it the same as x^2/2-2x+c

(I am confused about this one as from expansion I can see its similar but if I put any value for c into either say 1 I get different answers)

3)Finally integral of 1/2x I can do this by dividing by a half

i.e ½ integral of 1/x =1/2 lnx

but is itthe same as 1/2ln2x as this wway I say what must I do to denominator (2x) to get to numerator and then simply multiply this coefficient to the natural log of the bottom

i.e 1/4x+3 = ½ ln 4x+3

Can I only do that if I cannot factorise out a coeff?

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# Integration of 1/2x & sinxcosx & x+1

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