91Ó°ÊÓ

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

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

$$\begin{gathered} ∫\frac{1}{0.4y-100}dy=∫dx \\ \frac{1}{0.4}\ln |0.4y-100|=x+C_1 \\ \ln |0.4y-100|=0.4x+C(0.4C_1=C) \\ 0.4y=100+e^{0.4x+C} \end{gathered}$$

Simplify the above expression further

$$\begin{gathered} y=\frac{1}{0.4}\left[100+e^{0.4x+C}\right] \\ =\left[250+A{e}^{0.4x}\right]\left(\frac{1}{0.4}{e}^C=A\right) \end{gathered}$$

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

$$\begin{gathered} 0=250+A \\ A=-250 \end{gathered}$$

Substitute this value of the constant A in the solution of the differential equation and obtain the solution of the initial-value problem$\frac{dy}{dx}=-2\left(1-3y\right),y\left(0\right)=2asy\left(x\right)=250\left[1-e^{0.4x}\right]$