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

Jul 14, 2017

See below.

#### Explanation:

Making $y = \lambda x$ then

$\mathrm{dy} = x \mathrm{dl} a m b \mathrm{da} + \lambda \mathrm{dx}$ so

$x \mathrm{dl} a m b \mathrm{da} + \lambda \mathrm{dx} = \left(\frac{1 + {\lambda}^{3}}{3 {\lambda}^{2}}\right) \mathrm{dx}$ or

$\left(\left(\frac{1 + {\lambda}^{3}}{3 {\lambda}^{2}}\right) - \lambda\right) \mathrm{dx} = x \mathrm{dl} a m b \mathrm{da}$

This is a separable differential equation

$\frac{\mathrm{dl} a m b \mathrm{da}}{\left(\frac{1 + {\lambda}^{3}}{3 {\lambda}^{2}}\right) - \lambda} = \frac{\mathrm{dx}}{x}$ and after integration

$- \frac{1}{2} L o g \left(1 - 2 {\lambda}^{3}\right) = \log x + C$ and then

$\frac{1}{\sqrt{1 - 2 {\lambda}^{3}}} = {C}_{1} x$ then

$1 - 2 {\left(\frac{y}{x}\right)}^{3} = \frac{1}{{C}_{1} x} ^ 2$

Jul 14, 2017

${y}^{3} = {x}^{3} / 2 + C x$

#### Explanation:

We have:

$\left({x}^{3} + {y}^{3}\right) = 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

We can rearrange this Differential Equation as follows:

$3 \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{3} + {y}^{3}}{x {y}^{2}}$
$\text{ } = {x}^{3} / \left(x {y}^{2}\right) + {y}^{3} / \left(x {y}^{2}\right)$
$\text{ } = {x}^{2} / {y}^{2} + \frac{y}{x}$
$\text{ } = {\left(\frac{x}{y}\right)}^{2} + \frac{y}{x}$

This would lead us to try a substitution, Let:

$v = \frac{y}{x} \implies y = v x$

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = v + x \frac{\mathrm{dv}}{\mathrm{dx}}$

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

$3 \left(v + x \frac{\mathrm{dv}}{\mathrm{dx}}\right) = {\left(\frac{1}{v}\right)}^{2} + v$

$\therefore 3 v + 3 x \frac{\mathrm{dv}}{\mathrm{dx}} = {\left(\frac{1}{v}\right)}^{2} + v$
$\therefore 3 x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{v} ^ 2 - 2 v$
$\therefore 3 x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1 - 2 {v}^{3}}{v} ^ 2$

$\therefore {v}^{2} / \left(1 - 2 {v}^{3}\right) \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{3 x}$

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

$\int \setminus {v}^{2} / \left(1 - 2 {v}^{3}\right) \setminus \mathrm{dv} = \int \setminus \frac{1}{3 x} \setminus \mathrm{dx}$

This is now a trivial integration problem, thus:

$- \frac{1}{6} \ln \left(1 - 2 {v}^{3}\right) = \frac{1}{3} \ln x + \ln A$

$\therefore \ln \left(1 - 2 {v}^{3}\right) = - 2 \ln x - 2 \ln A$
$\therefore \ln \left(1 - 2 {v}^{3}\right) = \ln \left(\frac{B}{x} ^ 2\right)$

$\therefore 1 - 2 {v}^{3} = \frac{B}{x} ^ 2$
$\therefore 2 {v}^{3} = 1 - \frac{B}{x} ^ 2$
$\therefore {v}^{3} = \frac{1}{2} - \frac{B}{2 {x}^{2}}$

And restoring the substitution we get:

$\therefore {\left(\frac{y}{x}\right)}^{3} = \frac{1}{2} - \frac{B}{2 {x}^{2}}$

$\therefore {y}^{3} = {x}^{3} \left(\frac{1}{2} - \frac{B}{2 {x}^{2}}\right)$
$\therefore \text{ } = {x}^{3} / 2 + C x$