Question

integration factor following linear differential equation
DY + y cot x- sin x
dx

​

Answer 1

Answer:

Comparing the given equation with first order differential equation,

dy

dx

+Py=Q(x), we get, P=cotxandQ(x)=2cosx

So, Integrating factor (I.F)=e∫cotxdx

I.F.=eln|sinx|=sinx

we know, solution of differential equation,

y(I.F.)=∫Q(I.F.)dx

∴Our solution will be,

ysinx=∫sinx(2cosx)dx

⇒ysinx=∫sin2xdx

⇒ysinx=-

cos(2x)

2

+c

At y=0andx=

π

2

, equation becomes

0=-

cosπ

2

+c⇒c=-

1

2

So, solution will be,

ysinx=-

cos2x

2

-

1

2

⇒2ysinx+cos2x+1=0

Step-by-step explanation:

hope it will help you