# What is the solution of the Differential Equation dy/dx=(x-3)y^2/x^3?

Jun 12, 2017

$y = \frac{2 {x}^{2}}{2 x - 3 + A {x}^{2}}$

#### Explanation:

We have:

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

This is a first Order non-linear Separable Differential Equation, we can collect terms by rearranging the equation as follows

$\frac{1}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - 3}{x} ^ 3$

And now we can "separate the variables" to get

$\int \setminus \frac{1}{y} ^ 2 \setminus \mathrm{dy} = \int \setminus \frac{x - 3}{x} ^ 3 \setminus \mathrm{dx}$

$\therefore \int \setminus \frac{1}{y} ^ 2 \setminus \mathrm{dy} = \int \setminus \frac{1}{x} ^ 2 - \frac{3}{x} ^ 3 \setminus \mathrm{dx}$

And integrating gives us:

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

$\therefore - \frac{1}{y} = - \frac{1}{x} + \frac{3}{2 {x}^{2}} + {C}_{1}$

$\therefore - \frac{1}{y} = \frac{3 + 2 {C}_{1} {x}^{2} - 2 x}{2 {x}^{2}}$

$\therefore \frac{1}{y} = \frac{2 x - 3 + A {x}^{2}}{2 {x}^{2}}$

$y = \frac{2 {x}^{2}}{2 x - 3 + A {x}^{2}}$

Jun 12, 2017

$- \frac{1}{y} = - \frac{1}{x} + \frac{3}{2 {x}^{2}} + C$

#### Explanation:

Use separation of variables, that is put the $y$ terms in the left, and $x$ terms in the right.

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

$\left(\frac{1}{y} ^ 2\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - 3}{x} ^ 3$

$\left(\frac{1}{y} ^ 2\right) \mathrm{dy} = \frac{x - 3}{x} ^ 3 \mathrm{dx}$

Integrate both sides:

$\int \left(\frac{1}{y} ^ 2\right) \mathrm{dy} = \int \frac{x - 3}{x} ^ 3 \mathrm{dx}$

$- \frac{1}{y} = \int \frac{1}{x} ^ 2 - \frac{3}{x} ^ 3 \mathrm{dx}$

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

$- \frac{1}{y} = - \frac{1}{x} + \frac{3}{2 {x}^{2}} + C$