Question

(1+x^2)dy/dx+y^2=1

Answer 1

(1+x²)dy/dx=1-y²,

dy/(1-y²)=dx/(1+x²) (separation of variables).

1/(1-y²)=A/(1-y)+B/(1+y) (partial fractions).

A+Ay+B-By=1, so A+B=1, A-B=0, A=B=½.

½∫dy/(1-y)+½∫dy/(1+y)=∫dx/(1+x²)=tan^-1(x)+C, where C is a constant.

ln((1+y)/(1-y))=2tan^-1(x)+K,

(1+y)/(1-y)=e^{2tan⁻¹(x)+K},

1+y=e^{2tan⁻¹(x)+K}-ye^{2tan⁻¹(x)+K},

y(1+e^{2tan⁻¹(x)+K})=e^{2tan⁻¹(x)+K}-1,

y=(e^{2tan⁻¹(x)+K}-1)/(e^{2tan⁻¹(x)+K}+1), where K is a constant.