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

$\frac{dy}{dx}=-2y(1-3y),y\left(0\right)=2$

The given initial value problem $\frac{dy}{dx}=-2y(1-3y),y\left(0\right)=2.............\left(1\right)$

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

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

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

$$\begin{gathered} \frac{1}{y(1-3y)}=\frac{A}{y}+\frac{B}{1-3y} \\ A=1 \\ B=3 \\ \frac{1}{y(1-3y)}=\frac{1}{y}+\frac{3}{1-3y} \end{gathered}$$

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

$$\begin{gathered} \int \left[\frac{1}{y}+\frac{3}{1-3y}\right]dy=-2x+C \\ \ln \left|y\right|-\ln |1-3y|=-2x+C \\ \ln \left|\frac{y}{1-3y}\right|=-2x+C \\ \frac{y}{1-3y}=e^{-2x+C} \end{gathered}$$

Simplify the above expression further

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

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

$$\begin{gathered} 2=\frac{1}{3+A} \\ 3+A=\frac{1}{2} \\ A=-\frac{5}{2} \end{gathered}$$

Substitute this value of the constant A in the solution of the differential equation and obtain the solution of the initial-value problem$\frac{dy}{dx}=-2y(1-3y),y\left(0\right)=2asy\left(x\right)=\frac{2}{6-5e^{2x}}$