# What is the solution of the differential equation  dy/dx = 2y(5-3y)  with y(0)=2?

Dec 21, 2017

$y = \frac{10 {e}^{10 x}}{6 {e}^{10 x} - 1}$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \left(5 - 3 y\right)$

This is a First Order Separable ODE, so we can "separate the variables" to get

$\int \setminus \frac{1}{y \left(5 - 3 y\right)} \setminus \mathrm{dy} = \int \setminus 2 \setminus \mathrm{dx}$

The RHS is trivial, and the LHS can be integrated by decomposing the integrand into partial fractions:

$\frac{1}{y \left(5 - 3 y\right)} \equiv \frac{A}{y} + \frac{B}{5 - 3 y}$
$\text{ } = \frac{A \left(5 - 3 y\right) + B y}{y \left(5 - 3 y\right)}$

$1 \equiv A \left(5 - 3 y\right) + B y$

Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $y = 0 \implies 1 = 5 A \implies A = \frac{1}{5}$
Put $x = \frac{5}{3} \implies 1 = \frac{5 B}{3} \implies B = \frac{3}{5}$

So we can now write:

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \int \setminus \frac{\frac{1}{5}}{y} + \frac{\frac{3}{5}}{5 - 3 y} \setminus \mathrm{dy} = \int \setminus 2 \setminus \mathrm{dx}$

$\therefore \frac{1}{5} \int \setminus \frac{1}{y} - \frac{3}{3 y - 5} \setminus \mathrm{dy} = \int \setminus 2 \setminus \mathrm{dx}$

And integrating we get:

$\frac{1}{5} \left\{\ln | y | - \ln | 3 y - 5 |\right\} = 2 x + C$

Using the initial condition $y \left(0\right) = 2$ then:

$\frac{1}{5} \left\{\ln 2 - \ln 1\right\} = C \implies C = \frac{1}{5} \ln 2$

So we have:

$\frac{1}{5} \left\{\ln | y | - \ln | 3 y - 5 |\right\} = 2 x + \frac{1}{5} \ln 2$

$\therefore \ln | y | - \ln | 3 y - 5 | = 10 x + \ln 2$

$\therefore \ln | \frac{y}{3 y - 5} | = 10 x + \ln 2$

$\therefore \frac{y}{3 y - 5} = {e}^{10 x + \ln 2}$

$\therefore y = \left(3 y - 5\right) 2 {e}^{10 x}$

$\therefore y = 6 y {e}^{10 x} - 10 {e}^{10 x}$

$\therefore 6 y {e}^{10 x} - y = 10 {e}^{10 x}$

$\therefore \left(6 {e}^{10 x} - 1\right) y = 10 {e}^{10 x}$

$\therefore y = \frac{10 {e}^{10 x}}{6 {e}^{10 x} - 1}$

Dec 21, 2017

$y \left(x\right) = \frac{10}{6 - {e}^{- 10 x}}$

#### Explanation:

This is a separable differential equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \left(5 - 3 y\right)$

$\frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \mathrm{dx}$

$\left(1\right) \text{ } \int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \int \mathrm{dx}$

Solve the integral in $y$ using partial fractions:

$\frac{1}{2 y \left(5 - 3 y\right)} = \frac{A}{2 y} + \frac{B}{5 - 3 y}$

$\frac{1}{2 y \left(5 - 3 y\right)} = \frac{A \left(5 - 3 y\right) + 2 B y}{2 y \left(5 - 3 y\right)}$

$y \left(2 B - 3 A\right) + 5 A = 1$

$\left\{\begin{matrix}2 B - 3 A = 0 \\ 5 A = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{5} \\ B = \frac{3}{10}\end{matrix}\right.$

$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \frac{1}{10} \int \frac{\mathrm{dy}}{y} + \frac{3}{10} \int \frac{\mathrm{dy}}{5 - 3 y}$

$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \frac{1}{10} \left(\int \frac{\mathrm{dy}}{y} - \int \frac{d \left(5 - 3 y\right)}{5 - 3 y}\right)$

$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = \frac{1}{10} \left(\ln \left\mid y \right\mid - \ln \left\mid 5 - 3 y \right\mid\right) + C$

$\int \frac{\mathrm{dy}}{2 y \left(5 - 3 y\right)} = - \frac{1}{10} \ln \left\mid \frac{5}{y} - 3 \right\mid + C$

Substituting in $\left(1\right)$:

$x = - \frac{1}{10} \ln \left\mid \frac{5}{y} - 3 \right\mid + C$

$- 10 x + C = \ln \left\mid \frac{5}{y} - 3 \right\mid$

$c {e}^{- 10 x} = \frac{5}{y} - 3$

$\frac{5}{y} = 3 + c {e}^{- 10 x}$

$y = \frac{5}{3 + c {e}^{- 10 x}}$

For $x = 0$ the initial condition is $y \left(0\right) = 2$, so:

$2 = \frac{5}{3 + c}$

$c = - \frac{1}{2}$

The required solution is then:

$y \left(x\right) = \frac{10}{6 - {e}^{- 10 x}}$