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

Question

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

sub them all in and the work it out

Answer 5

i ask my son he's good at this sort of thing .....means nothing to me