`y=e^(x)+1 : y''-y'=0`

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

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

If y=e^(tan^(-1)x) , show that (1+x^(2))y''+(2x-1)y'=0 .

The solution of (e^(x) + 1) y dy + (y+ 1) dx = 0 is

e^x tan y d x+(1-e^x) (sec ^2 y) d y=0

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

Solution of the differential equation (1+e^(x/y))dx + e^(x/y)(1-x/y)dy=0 is

Solution of the differential equation (1+e^(x/y))dx + e^(x/y)(1-x/y)dy=0 is

Given (1+e^(x//y))dx+e^(x//y)(1-x/y)dy=0 Solve the differential equation using x=vy.