Question

In Problems $\mathbf{11}\mathbf{-}\mathbf{14}$verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

$\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{*}\mathbf{/}\mathbf{2}}$.

Short answer

Answer:

The indicated function is an explicit solution of the given differential equation in the interval $\mathrm{(}\mathrm{-}\mathrm{\infty }\mathrm{,}\mathrm{\infty }\mathrm{)}$.

Step-by-step solution

Step by Step solution:Step 1: Define an explicit function.

In explicit solution unknown functions (dependable variables) is indicated (expressed) in terms of undependable variable and constants. This is known as “explicit solution”.

Let the given function be $y=e^{-*/2}$.

Then, the first derivative of the function is $y\text{'}=-\frac{1}{2}{e}^{-x/2}$.

Determine it is nonlinear.

Substituteandin the left-hand side of the differential equation.

$$\begin{gathered} \mathbf{2}\left(-\frac{1}{2}{e}^{-x/2}\right)\mathbf{+}{\mathit{e}}^{\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{e}}^{\mathbf{x}\mathbf{/}\mathbf{2}}\mathbf{+}{\mathit{e}}^{\mathbf{x}\mathbf{/}\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

That is same as the right-hand side of the differential equation for every real number role="math" localid="1653645622995" $\mathbf{x}$. Thus, the indicated function is an explicit solution of the given differential equation in the interval $\left(-\infty ,\infty \right)$.