dy/dx+(e^((y^2)+x))/y=0 How to put C in solving differential equation

dy/dx+(e^((y^2)+x))/y=0 How to put C in solving differential equation

dy/dx=-[e^(y^2)*e^x]/y
-ye^(-y^2)dy=e^xdx
∫-ye^(-y^2)dy=∫e^xdx
1/2*∫e^(-y^2)d(-y^2)=∫e^xdx
e^(-y^2)=2e^x+C
-y^2=ln(2e^x+C)
y^2=-ln(2e^x+C)
y=±√[-ln(2e^x+C)]
Where C is an arbitrary constant

Y + xcosy + x = 0, then dy / DX =?

y+xcosy+x=0
dy+cosydx-xsinydy+dx=0
dy/dx=(1+cosy)/(xsiny-1)

The solution of the differential equation "dy / DX = YX satisfies the condition y (0) = 1" is

dy/dx=yx
dy/y=xdx
ln|y|=x^2/2+C
y=ce^(x^2/2)
From Y (0) = 1, C = 1
So the solution is y = e ^ (x ^ 2 / 2)

Let the equation E ^ y-xy ^ 2 = e ^ 2 determine the function y = y (x), and find the value of dy / DX (x = 0)
Detailed steps to help master!!!

The derivation of X on both sides is: (e ^ y) * y '- (y ^ 2 + X * 2Y * y') = 0
The solution is y '= y ^ 2 / (e ^ y-2xy)
When x = 0, we get y = 2
So y '(x = 0) = 4 / (e ^ 2-0)