Question

27. The integrating factor of the differential equation dy/dx+ycotx=2cosx is
(A) sinx
(B) logsinx
(C) cotx
(D) cosx
a​

Answer 1

Step-by-step explanation:

Comparing the given equation with first order differential equation,

dy/dx+Py = Q(x), we get, P = cotx and Q(x) = 2cosx

So, Integrating factor (I.F) = e^(intcotxdx)

I.F.= e^(ln|sinx|) = sinx

we know, solution of differential equation,

y(I.F.) = intQ(I.F.)dx

:.Our solution will be,

ysinx = int sinx(2cosx)dx

=>ysinx = int sin2xdx

=>ysinx = -cos(2x)/2+c

At y = 0 and x = pi/2, equation becomes

0 = -cospi/2 +c => c = -1/2

So, solution will be,

ysinx = -(cos2x)/2-1/2

=>2ysinx+cos2x+1 = 0