Question

What is the general solution of the differential equation? : `(x^3 + y^3)=3xy^2 dy/dx`

Answer 1

See below.

Explanation:

Making `y = lambda x` then

`dy=x dlambda + lambda dx` so

`x dlambda + lambda dx =( (1+lambda^3)/(3lambda^2))dx` or

`(( (1+lambda^3)/(3lambda^2))-lambda)dx = x dlambda`

This is a separable differential equation

`(dlambda)/(( (1+lambda^3)/(3lambda^2))-lambda)=dx/x` and after integration

`-1/2 Log(1 - 2 lambda^3)=logx + C` and then

`1/sqrt(1-2lambda^3) = C_1 x` then

`1-2(y/x)^3=1/(C_1x)^2`

Answer 2

`y^3 = x^3/2+Cx`

Explanation:

We have:

`(x^3 + y^3)=3xy^2 dy/dx`

We can rearrange this Differential Equation as follows:

`3 \ dy/dx = (x^3 + y^3)/(xy^2)`
`" " = x^3/(xy^2) + y^3/(xy^2)`
`" " = x^2/y^2 + y/x`
`" " = (x/y)^2 + y/x`

This would lead us to try a substitution, Let:

`v = y/x => y=vx`

Then:

`dy/dx = v + x(dv)/dx`

And substituting into the above DE, to eliminate `y`:

`3(v + x(dv)/dx) = (1/v)^2+v`

`:. 3v + 3x(dv)/dx = (1/v)^2+v`
`:. 3x(dv)/dx = 1/v^2-2v`
`:. 3x(dv)/dx = (1-2v^3)/v^2`

`:. v^2/(1-2v^3)(dv)/dx = 1/(3x)`

This is now a separable DIfferential Equation, and so "separating the variables" gives us:

`int \ v^2/(1-2v^3) \ dv = int \ 1/(3x) \ dx`

This is now a trivial integration problem, thus:

`-1/6ln(1-2v^3) = 1/3lnx + ln A`

`:. ln(1-2v^3) = -2lnx - 2ln A`
`:. ln(1-2v^3) = ln(B/x^2)`

`:. 1-2v^3 = B/x^2`
`:. 2v^3 = 1-B/x^2`
`:. v^3 = 1/2-B/(2x^2)`

And restoring the substitution we get:

`:. (y/x)^3 = 1/2-B/(2x^2)`

`:. y^3 = x^3(1/2-B/(2x^2))`
`:. " " = x^3/2+Cx`