Question

In problems 23–28 Find an explicit solution to the given initial-value problem.

$\frac{\text{dy}}{\text{dt}}\text{= 1 - 2y,}\text{y(0) =}\frac{5}{2}$

Short answer

$y=\frac{1+4e^{-2t}}{2}$

Step-by-step solution

Definition of separable equation

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Separate the variables

Rearrange and separate the variables.

$$\begin{gathered} \frac{dy}{dt}=1-2y \\ \frac{dy}{1-2y}=dt \end{gathered}$$

Integrate both sides.

$\int \frac{dy}{1-2y}=\int dt$

Integration

Do the integration.

$-\frac{1}{2}\ln (1-2y)=t+c_1$

$Relabeling$$e^{2c_1}=C$

$$\begin{gathered} (1-2y)=-2\left(t+c_1\right) \\ 1-2y=C{e}^{-2t} \\ y=\frac{1-C{e}^{-2t}}{2} \end{gathered}$$

Substituting initial condition $y\left(0\right)=5/2$

$$\begin{gathered} \frac{5}{2}=\frac{1-C{e}^{-2\left(0\right)}}{2} \\ 5=1-C \\ C=-4 \end{gathered}$$

Therefore, the solution is$\text{y =}\frac{{\text{1 + 4e}}^{\text{- 2t}}}{\text{2}}$