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Question

# Solve differential equation (1+y2)dx=(tan−1y−x)dy.

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Solution

## (1+y2)dx=(tan−1y−x)dx∴dxdy=tan−1y−x1+y2∴dxdy=tan−1y1+y2−x1+y2∴dxdy+x1+y2=tan−1y1+y2 (1)dxdy+P(y)x=Q(y) which is linear differential equation of the type∴P=1+y2 Q(A)=tan−1y1+y2∴ Integrating factor =e∫P(y)dy=e∫11+y2dy∴ Multiplying equation (1) by etan−1y∴etan−1ydxdy+etan−1y1+y−−=etan−1y(tan−1y1+y2)∴ddx(etan−1y⋅x)=etan−1y⋅(tan−1y1+y2)∴ Integrating both sides pairwise∴xetan−1y=∫etan−1y⋅(tan−1y1+y2)dy .(2)∴ Now, finding ∫etan−1y⋅(tan−1y1+y2)dyTaking tan−1y=t∴11+y2=dy=dt∴xetan−1y=∫etdt∴∫et[(t−1)+1]dt∴ Here taking f(t)=1∴xetan−1y=∫et(f(t)+f′(t))dtetf(t)+c∴xetan−1y=et(t−1)+c∴ Result (2) becames as follows∴xetan−1y=etan−1y(tan−1y−1)+c∴x=(tan−1y−1)+c⋅e−tan−1y∴ Which is required solution of given differential equation.

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