`(e^x+1)y dy+(y+1)dx=0`

answer the foll. Questions 3. solve : (e^x+1)y dy-(y^2+1) e^x dx=0 , given x=0 , y=0.

if ( dy )/(dx) =1 + x +y +xy and y ( -1) =0 , then the value of ( y (0) +3 - e^(1//2))

if ( dy )/(dx) =1 + x +y +xy and y ( -1) =0 , then the value of ( y (0) +3 - e^(1//2))

The solution of the differential equation (1+e^(x))y(dy)/(dx) = e^(x) when y=1 and x = 0 is

Solve the following differential equation: (1+ e ^(y/x)) dy + e^(y/x) (1- y/x) dx = 0 (x ne 0)

Solve the following differential equation: (1+ e ^(y/x)) dy + e^(y/x) (1- y/x) dx = 0 (x ne 0)

Solve the following differential equation: (1+ e ^(y/x)) dy + e^(y/x) (1- y/x) dx = 0 (x ne 0)

If e^x+e^y=e^(x+y) , prove that (dy)/(dx)=-(e^x(e^y-1))/(e^y(e^x-1)) or, (dy)/(dx)+e^(y-x)=0

If e^(x) + e^(y) = e^(x + y) , then prove that (dy)/(dx) = (e^(x)(e^(y) - 1))/(e^(y)(e^(x) - 1)) or (dy)/(dx) + e^(y - x) = 0 .

(dy)/(dx) -y =e^(x ) " when" x=0 and y=1