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Question

Find the particular solution of the differential equation (1+x2)dydx=(emtan1 xy), given that y=1, when x=0.

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Solution

(1+x2)dydx=(em tan1 xy)

dydx+11+x2y=em tan1x1+x2 ...(i)

Here, P=11+x2 and Q=em tan1x1+x2

I.F. = e11+x2dx=etan1x

y.etan1x= etan1x.em tan1x1+x2dx+c

= I+c

Put tan1x=t

11+x2dx=dt

I= et.emtdt

= e(m+1)tdt

= e(m+1)tm+1=e(m+1)tan1xm+1

y.etan1x=e(m+1)tan1xm+1+c ...(ii)

When x=0, y=1, we have

1.e0=e0m+1+c

c=mm+1

Put the value of c in eq. (ii)
y.etan1x=e(m+1)tan1xm+1+mm+1

y.(m+1)etan1x=e(m+1)tan1x+m

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