`(1+e^(2x))dy/dx+2(1+y^2)e^x=0 , y(0)=0` ,find `6(f'(0)+(f(logsqrt3))^2)`

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

Solve (1+2e^(x/y))dx+2e^(x/y)(1-x/y)dy=0

If y = f(x) is the solution of the differential equaiton e^(3y) ((dy)/(dx) - 1) = e^(2x) and y(0) = 0 then y(x) = log (Ae^(3x) - Be^(2x))^((1)/(3)) where the value of (A + B) is:

If dy/dx=(y+1)[(y+1)e^(x^2/2)-x] , y(0)=2 then y'(1) is equal to

(dy)/(dx)=2e^(x)y^(3) , when x=0, y=(1)/(2)

x^(2)+y^(2)-4x-6y+3=0 find (dy)/(dx) at (1,6)

Solve: (1+2e^(x/y))dx+2e^(x/y)(1-x/y)dy=0

If y = f (x) is the solution of difierential equation. e ^(y) ((dy)/(dx )-2)=e ^(3x) such that f(0) =0, then f (2) is equal to :

For each of the following initial value problems verify that the accompanying functions is a solution. (i) x(dy)/(dx)=1, y(1)=0 => y=logx (ii) (dy)/(dx)=y , y(0)=1 => y=e^x (iii) (d^2y)/(dx^2)+y=0, y(0)=0, y^(prime)(0)=1 => y=sinx (iv) (d^2y)/(dx^2)-(dy)/(dx)=0, y(0)=2, y^(prime)(0)=1 => y=e^x+1 (v) (dy)/(dx)+y=2, y(0)=3 => y=e^(-x)+2