How do u find the solution of the differential eqtn dy/dx + ycot x = sin2x, 0<x<pi for which y = 1 at x = pi/4

can u please show your working out too, because i am having a real problem with this type of question thanx RB

Answers:
integrating factor method

this method applies to differential equations of the form

dy/dx +g(x)y = h(x)

here, g(x)=cot x, h(x)=sin2x

a) determine the integrating factor p=e^(int(g(x)dx))

= e^( int (cotx)dx))

=e^(ln(sinx)dx) =sinx

b) rewrite the differential equation as d/dx(p(x)y) =p(x)h(x)

>>>>>d/dx( y.sinx)= sinx.sin2x

c)integrate this last equation to obtain

y.sinx =int(sinx.sin2x)dx = int(sinx.2sinx.cosx)dx

=2.int((sinx)^2(cosx))dx =2.int(1-(cosx)^2)cosx)dx

=2.int(cosx-(cos)^3)dx =2.sinx -2(sinx-((sinx)^3)/3)+C

=(2.(sinx)^3)/3 +C

d) y.sinx = (2.(sinx)^3)/3 +C

hence, general solution is y = (2.(sinx)^2)/3 + C/sinx

we are given y=1 at x= pi/4 >>>>> sin (pi/4) = 1/(2)^(1/2)

substitute into general solution

1 = 1/3 + root(2).C >>>>> C = root (2)/3

the particular solution is y = (2(sinx)^2)/3 + root(2)(cosecx)/3

= (2(sinx)^2+root(2)(cosecx))/3

where root is the square root (0<x<pi)
i ask my son he's good at this sort of thing ...means nothing to me
Look up something called The Euler Method (or the Modified Euler Method).

The general principal is that if you re-arrange as:

dy/dx = sin2x - ycot x

then you can find the gradient at any point. So you start by finding the gradient at your known point y=1 at x = pi/4 and you draw a small length of straight line at this point, which can get you to a good estimate for a new x and y, and then you repeat the process. Of course, the further you go, the more error you will get. To reduce the error, reduce the step size.

EDIT: sorry, this method is unnecessarily complicated, please see James L's answer - use the integrating factor
sub them all in and the work it out
Use the integrating factor

exp(antiderivative of cot x) =
exp(ln(sin x)) = sin x

Multiply through by sin x to get

sin x dy/dx + (cos x) y = sin 2x sin x
or
((sin x)y)' = sin 2x sin x

Integrate and divide by sin x:

y(x) = csc x * integral of sin 2x sin x
= csc x * integral of 2 sin^2 x cos x dx
= 2csc x * (sin^3 x / 3 + C)
= (2/3) sin^2 x + C csc x

where C is chosen so that y(pi/4)=1. Plug in x=pi/4 and y=1 and solve for C.

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