Question

In problems 1 and 2, $\mathit{y}\mathbf{=}\mathbf{1}\mathbf{/}\left(1+c_1{e}^{-x}\right)$is a one-parameter family of solutions of the first-order DE$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{y}\mathbf{-}{\mathit{y}}^{\mathbf{2}}$. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition.

$\mathit{y}\left(0\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}$

Short answer

The solution of the differential equation is $y=\frac{1}{1-4e^{-x}}$.

Step-by-step solution

Definition of Initial Value Problems

The unknown function $y\left(x\right)$and its derivatives at a number $x_0$. On some interval I containing $x_0$the problem of solving an nth-order differential equation subject to n side conditions specified at:

Solve: $\frac{d^{n}y}{d{x}^n}=f(x,y,y\text{'},...,y^{(n-1)})$

Subject to: $y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$

Where $y_0,y_1,...,y_{n-1}$are arbitrary constants, is called n-thorder Initial Value Problem (IVP). The values of $y\left(x\right)$and its first n-1 derivatives at $x_0$ $y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}$ are called Initial Conditions.

Compute the value of c1

Given $y=\frac{1}{1+c_1{e}^{-x}}$

Substitute the initial condition

$$\begin{gathered} y\left(0\right)=-\frac{1}{3} \\ -\frac{1}{3}=\frac{1}{1+c_1{e}^0}=\frac{1}{1+c_1} \\ 1+c_1=-3 \\ {c}_1=-4 \end{gathered}$$

Substitute the above value, we get

$y=\frac{1}{1-4e^{-x}}$.

Hence,the solution of the differential equation is $y=\frac{1}{1-4e^{-x}}$.