How do you solve #dy/dx = xy^2#?

Question

49216 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-23T02:55:12

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

Let's separate our variables, IE, have each side of the equation only in terms of one variable. This entails

#dy/y^2=xdx#

Integrate each side:

#intdy/y^2=intxdx#

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

Note that we would technically have constants of integration on both sides, but we moved them all over to the right and absorbed them into #C.#

Now, let's get an explicit solution with #y# as a function of #x:#

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

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

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

Let's get the fraction out of the denominator. It just looks messy.

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

Well, #2C=C,# as #2# multiplied by a constant just yields another constant.

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

Thus,

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

Answer 2 · 2018-03-23T02:55:52

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

where #C# is an arbitrary constant.

#dy/dx = xy^2#

We separate variables:

#dy/y^2 = xdx#

Now we integrate both sides:

#\int1/y^2dy = \intxdx#

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

where #C# is an arbitrary constant of integration.

Now we solve for #y#.

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

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

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

where #C# absorbed a factor of #2# since it is an arbitrary constant.