`xdx+(xdy-ydx)/(x^2+y^2)=0`
`(A) (x^(2))/(2)+tan^(-1)(x/y)=k`
`(B) (x^(2))/(2)+tan^(-1)(y/x)=k`
`(C) (x^(2))/(2)-tan^(-1)(x/y)=k`
`(D) (x^(2))/(2)-tan^(-1)(y/x)=k`

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

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

The solution of the differential equation (1+y^2)+(x-e^(tan^-1y))dy/dx=0 is (A) x e^(2 tan^-1y)=e^(tan^-1y)+k (B) (x-2)=k e^(-tan^-1y) (C) x e^(tan^-1y)=e^(2 tan^-1y)+k (D) x e^(tan^-1y)=tan^-1y+k

tan ^ (- 1) ((1) / (x + y)) + tan ^ (- 1) ((y) / (x ^ (2) + xy + 1)) = cot ^ (- 1) x

If y=e^(tan^(-1)x), prove that (1+x^(2))y_(2)+(2x-1)y_(1)=0

The solution of the differential equation (1+y^2)+(xe^(tan^-1)y) (dy)/(dx)=0, is (1) (x-2)=ke^(tan^) -1y) (2) 2xe^(2tan^-1y) =e^(2tan^-1y)+k (3) xe^(tan^-1y)=tan^-1y+k (4) xe^(2tan ^-1y)=e^(tan^-1y)+k

If x and y are of same sign,then the value of (x^(3))/(2)cos ec^(2)((1)/(2)(tan^(-1)x)/(y))+(y^(3))/(2)sec^(2)((1)/(2)(tan^(-1)y)/(x)) is equal to (x-y)(x^(2)+y^(2)) (b) (x+y)(x^(2)-y^(2))(x+y)(x^(2)+y^(2))*(d)(x-y)(x^(2)-y^(2))

If y=e^(tan^^)((-1)x), prove that (1+x^(2))y_(2)+(2x-1)y_(1)=0

int(2x^(2)+a^(2))/(x^(2)(x^(2)+a^(2)))dx=(k)/(x)+(1)/(a)Tan^(-1)((x)/(a))+c then k

If logy=tan^(-1)x,"then "(1+x^(2))y_(2)+(2x-1)y_(1)+4=