Q. 47 - Calculus

Question

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{dy}{dx}=3y(1-y),y\left(0\right)=1$

Step-by-Step Solution

Answer

The solution for the given initial value problem$\frac{dy}{dx}=3y(1-y),y\left(0\right)=1$ as $y\left(x\right)=\frac{2e^{3x}}{1+2e^{3x}}$

Step 1. Given information

The given initial value problem $\frac{dy}{dx}=3y(1-y),y\left(0\right)=1..............\left(1\right)$

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value

$y=1$Note that the differential equation in (1) does not contain the independent variable at all, so technically the variables have already been separated. Hence, the differential equation can be solved by antidifferentiating. Thus, the solution of the differential equation involved in the initialvalue problem is given by

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

Simplify the integral on the left hand side by resolving the fraction in to the partial fractions.So, first obtain the partial fractions of the integrated by applying cover up rule

$$\begin{gathered} \frac{1}{y(1-y)}=\frac{A}{y}+\frac{B}{1-y} \\ A=1 \\ B=1 \\ \frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y} \end{gathered}$$

Substitute these partial fractions on the left hand side of equations (2),and integrate

$$\begin{gathered} \int \left[\frac{1}{y}+\frac{1}{1-y}\right]dy=3x+C \\ \ln \left|y\right|-\ln |1-y|=3x+C \\ \ln \left|\frac{y}{1-y}\right|=3x+C \\ \frac{y}{1-y}=e^{3x+C} \end{gathered}$$

Simplify the above expression further

$$\begin{gathered} \frac{1-y}{y}=e^{-3x-C} \\ =A{e}^{-3x} \left(e^{-C}=A\right) \\ \frac{1}{y}=1+A{e}^{-3x} \\ y=\frac{1}{1+A{e}^{-3x}} \end{gathered}$$

Now, use the given initial condition $y\left(0\right)=\frac{2}{3}$, that is take$x=0,y=\frac{2}{3}$ in the above result and evaluate the constant A

$$\begin{gathered} \frac{2}{3}=\frac{1}{1+A} \\ 1+A=\frac{3}{2} \\ A=\frac{1}{2} \end{gathered}$$

Substitute this value of the constant A in the solution of the differential equation and obtain the solution of the initial-value problem $\frac{dy}{dx}=3y(1-y),y\left(0\right)=1$as

$$\begin{gathered} y\left(x\right)=\frac{1}{1+{\displaystyle \frac{1}{2}}{e}^{-3x}} \\ y\left(x\right)=\frac{2}{2+e^{-3x}} \\ y\left(x\right)=\frac{2e^{3x}}{1+2e^{3x}} \end{gathered}$$

[Note: The initial condition $y\left(0\right)=1$ is not valid. In fact the carrying capacity of the problem is 1 . So, graphically both $y=0$ and  are asymptotes to the graph of the curve. Hence, the value of y remains between 0 and 1 . Keeping this in view the initial condition has been changed to $y\left(0\right)=\frac{2}{3}$ and problem solved with this initial condition.]