[2*f(x+y+1)(dy)/(dx)=1uu I]

Solve the differential equation: (i) (1+y^(2))+(x-e^( tan ^(-1)y))(dy)/(dx)=0 (ii) x(dy)/(dx)+cos^(2)y=tan y(dy)/(dx)

Let y=f (x) and x/y (dy)/(dx) =(3x ^(2)-y)/(2y-x^(2)),f(1)=1 then the possible value of 1/3 f(3) equals :

Let y=f (x) and x/y (dy)/(dx) =(3x ^(2)-y)/(2y-x^(2)),f(1)=1 then the possible value of 1/3 f(3) equals :

A: If y = x ^(y) then (dy)/(dx) = (y ^(2))/(x(1- log y )) If y = f (x) ^(y), then (dy)/(dx) = (y ^(2) f '(x))/(f (x) [1- ylog f (x)])= (y ^(2) f'(x))/(f (x) [1- log y])

If x+y=tan^(-1)y" and "(d^(2)y)/(dx^(2))=f(y)(dy)/(dx) , then f(y)=

If x+y=tan^(-1)y" and "(d^(2)y)/(dx^(2))=f(y)(dy)/(dx) , then f(y)=

If x+y= tan^(-1)y and (d^(2)y)/(dx^(2))=f(y)(dy)/(dx) then f(y)=

Solve the following differential equations (i) (1+y^(2))dx = (tan^(-1)y - x)dy (ii) (x+2y^(3))(dy)/(dx) = y (x-(1)/(y))(dy)/(dx) + y^(2) = 0 (iv) (dy)/(dx)(x^(2)y^(3)+xy) = 1

The solution of differential equation x^2=1 +(x/y)^(-1)(dy)/(dx)+((x/y)^-2((dy)/(dx))^2)/(2!)+((x/y)^(-3)((dy)/(dx))^3)/(3!)+... i s