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Question

Find the general solution of the differential equation (1+x2)dy+2xydx=cot x dx(x0)


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Solution

Given differential equation :

(1+x2)dy+2xydx=cot x dx

(1+x2)dydx+2xydx=cot x

dydx+2x(1+x2) y=cot x(1+x2)

Given differential equation is of the form
dydx+Py=Q

By comparing both the equations, we get

P=2x(1+x2) y and Q=cot x(1+x2)

The general solution of the given differential equation is

y(I.F)=(Q×I.F.)dx+c ....(i)

Firstly, we need to find I.F.

I.F.=epdx

I.F.=e2x(1+x2) y dx

Let 1+x2=t
dt=2xdx

I.F.=edtt

I.F.=elog t (elog x=x)

I.F.=t

Substituting t=(1+x2), we get

I.F.=(1+x2)

Substituting the value of I.F in (i), we get

y(1+x2)=cot x(1+x2).(1+x2)dx+c

y(1+x2)=cot x dx+c

y(1+x2)=log |sin x|+c

y=(1+x2)1log |sin x|+c(1+x2)1


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