Differential Equations Solved

Solve

(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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y= arctan{1/[- sin^(-1)(x)+C.]}

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

yy'/((1+y^2)=1/ [x(1+x^2)] (2)

Now 1/ [x(1+x^2)]= x/ [x^2(1+x^2)]

Let v=y^2, then dv=2ydy, thus

yy'/((1+y^2)dy=1/2 1/(1+v) dv

Let u=x^2, then du=2x dx, thus

x/ [x^2(1+x^2)]dx= ½ 1/[u(1+u)] du

We have the formula

The definite integral ...