How do you find all solutions of the differential equation #dy/dx=xy^2#?

Question

3666 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-23T18:23:09

# y = -2/(x^2 + C)#

We have:

#dy/dx = xy^2#

which is a first order separable Differential Equation, so we can just rearrange and separate the variables as follows;

# 1/y^2dy/dx = x#
# :. int \ 1/y^2 \ dy = int \ x \ dx#
# :. int \ y^(-2) \ dy = int \ x \ dx#

We can now integrate to get;

# y^(-1)/(-1) = x^2/2 + A#
# -1/y = (x^2 + 2A)/2#
# -y = 2/(x^2 + 2A)#

And so the General Solution is:

# y = -2/(x^2 + C)#

Answer 2 · 2017-01-23T18:30:59

#y= 2/(C-x^2)#

This is a separable differential equation, so we proceed by separating the variables and integrating:

#(dy)/(dx) = xy^2#

#dy/y^2 = xdx#

#int dy/y^2 = int xdx#

#-1/y = x^2/2+C#

#y= -1/(x^2/2-C) = 2/(C-x^2)#

and we can verify that:

#(dy)/(dx) = (4x)/((C-x^2)^2) = x (2/(C-x^2))^2 = xy^2#