`y^(2)-(dy)/(dx)=x^(2)(dy)/(dx)`A)`y^(-1)+tan^(-1)x=c` B)`x^(-1)+tan^(-1)y=c` C)`y+tan^(-1)x=c` D)`x^(-1)+y^(-1)=tan^(-1)x+c`

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

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

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

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

(dy)/(dx)=1+x tan(y-x)

The solution of the differential equation (1+x^(2))(dy)/(dx)+1+y^(2)=0, is tan^(-1)x-tan^(-1)y=tan^(-1)Ctan^(-1)y-tan^(-1)x=tan^(-1)Ctan^(-1)y=tan^(-1)x=tan Ctan^(-1)y-tan^(-1)x=tan^(-1)C

The solution of the differential equation (1+x^2)(dy)/(dx)+1+y^2=0, is a) tan^(-1)x-tan^(-1)y=tan^(-1)C b) tan^(-1)y-tan^(-1)x=tan^(-1)C c) tan^(-1)y+-tan^(-1)x=tan^(\ )C d) tan^(-1)y+tan^(-1)x=tan^(-1)C

Solve: (1+x^2)(dy)/(dx) = y+tan^(-1)x

Solution of differential equation (x^(2)+1)^(2)(dy)/(dx)+2x(x^(2)+1)y=1 is (A) y=(tan^(-1))/(x^(2)+1)+C( B) y=tan^(-1)x+C(C)y(x^(2)+1)=tan^(-1)x+C(D)y(tan^(-1)x)=x^(2)+C