# Differential equations

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(1-x^2)^(1/2)y'+1+y^2=0

xy(1+x^2)y'-(1+y^2)=0

xyy'=1+x^2+y^2+x^2y^2

sinx(e^y + 1)dx=e^y(1+cosx)dy, Y(0)=0

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##### Solution Summary

This showsh ow to solve several different differential equations.

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1. (1-x^2)^(1/2)y'+1+y^2=0 implies that

y'/(1+y^2)=-1/((1-x^2)^(1/2) ) (1)

We have the formula

The definite integral of

1/((a^2+u^2) is 1/a* [ tan^(-1)(u/a)] +C

The definite integral of

1/((a^2-u^2)^(1/2) is sin^(-1)(u/a)+C.

Integrate both sides of (1) to get ( with a=1)

1/[ tan^(-1)(y)] = - sin^(-1)(x)+C.

Thus

tan(y)= 1/[- ...

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