Question

.Verify that $y\left(x\right)=\sqrt{x+c}$is a solution of the differential equation $\frac{dy}{dx}=\frac{1}{2y}$for every constant C.

Short answer

The solution of the differential equation has been verified.

Step-by-step solution

Step 1. Given information

Solution to the differential equation:

$\frac{dy}{dx}=\frac{1}{2y}$is $y\left(x\right)=\sqrt{x+C}$, where $C$is a constant.

Step 2. Verify the solution by differentiating given solution.

$y=\sqrt{x+C}$

Let $z=x+C$

So,

$$\begin{gathered} \frac{dz}{dx}=1+0 \\ \frac{dz}{dx}=1 \end{gathered}$$

$$\begin{gathered} y=\sqrt{z} \\ y={\left(z\right)}^{\frac{1}{2}} \end{gathered}$$

$$\begin{gathered} \frac{dy}{dz}=\frac{1}{2}{z}^{\left(\frac{1}{2}-1\right)} \\ =\frac{1}{2}{z}^{-\frac{1}{2}} \\ =\frac{1}{2\sqrt{z}} \end{gathered}$$

Differentiate this with respect to x using chain rule.

localid="1652095282669" $$\begin{gathered} \frac{dy}{dx}=\frac{dy}{dz}\times \frac{dz}{dx} \\ \frac{dy}{dx}=\frac{1}{2\sqrt{z}}\times 1 \\ \frac{dy}{dx}=\frac{1}{2\sqrt{x+C}} \\ \frac{dy}{dx}=\frac{1}{2y} \end{gathered}$$

Thus, the solution has been verified.