## anonymous 5 years ago solve the DE: 2xy^3dx=(1-x^2)dy

Notice that you can manipulate this DE into a separable DE by dividing both sides by (1-x^2) and y^3 which gives you the following: $(2x/(1-x^{2}))dx=(1/y^{3})dy$this DE now has the separable form $M(x)dx=N(y)dy$ Now integrate the left side of the DE with respect to x and the right side with respect to y.$\int\limits_{}^{}(2x/(1-x^{2}))dx=\int\limits_{}^{}y^{-3}dy$Using u=1-x^2 and performing u-substitution, the integral on the left hand side is $-\ln(1-x^{2})+k$and the right hand side integral is $-1/2y^{-2}+c$ So,$-\ln(1-x^{2}) + k = -1/2y^{-2} + c$The c absorbs the -k when it is subtracted over to the right hand side of the equation giving you: $-\ln(1-x^{2})=-1/2y^{-2}+c$Now since you didn't provide an initial condition, there is no way to find an explicit solution. Only a general solution can be found because there is no initial condition to solve for the constant c. So, we simplify to get y alone. $1/(2y^{2})=\ln(1-x^{2})+c$$1/2(\ln(1-x^{2})+c)=y^{2}$$\pm \sqrt{1/(\ln((1-x^{2})^{2})+2c)}=y$ There is no way to tell which sign on the square root is correct since no initial condition was given to determine interval of validity.