हल कीजिए `(x+1)(dy)/(dx) = 2e^(-y) -1, y(0)=0`

Solve the initial value problem: (x+1)(dy)/(dx)=2e^(-y)-1,\ y(0)=0

Solve the following initial value problem: (x+1)(dy)/(dx)=2e^(-y)-1,y=(0)=0

Find the particular solution of the following : (x+1)(dy)/(dx)=2e^(-y)-1,y(0)=0 .

Solve (dy)/(dx)-y=e^(x) ,y(0)=1

(x+1)(dy)/(dx) -1 = 2e^(-y) , y=0, " when " x=1

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 .

The solution of (dy)/(dx) -y=e^(x) , y(0) = 1 , is

The solution of (dy)/(dx) -y=e^(x) , y(0) = 1 , is

(1+e^(x))/(y)(dy)/(dx)=e^(x), when y=1, x=0