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

Solve the differential equation: x(dy)/(dx)-y=(x-1)e^x,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 .

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

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

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

(1+3e^((y)/(x)))dy+3e^((y)/(x))(1-(y)/(x))dx=0

(1+3e^((y)/(x)))dy+3e^((y)/(x))(1-(y)/(x))dx=0 , given that y=0 and x=1

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