Question

In Problems 1 and 2 Fill in the blank and then write this result as a linear first-order differential equation that is free of the symbol ${\mathit{c}}_{\mathbf{1}}$and has the form $\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}\mathbf{=}\mathit{f}\left(x,y\right)$. The symbol ${\mathit{c}}_{\mathbf{1}}$represents a constant.

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(5+c_1{e}^{-2x}\right)\mathbf{=}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}$

Short answer

Answer:

The answer is$\frac{dy}{dx}=-2y+10$.

Step-by-step solution

Define  chain rule

The chain rule is a formula in calculus that represents the derivative of a combination of two discrete functions f and g in terms of f and g 's derivatives.

Let the formula of the chain rule be, $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{u}}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}$. Here, $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$is a derivative of y with respect to x, $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{u}}$is a derivative of y with respect to u, $\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}$is a derivative of u with respect to x.

Determine the derivative of the function

Let the derivative of the given function be $\frac{d}{dx}\left(5+c_1{e}^{-2x}\right)=-2c_1{e}^{-2x}$.,

Rewrite the above equation as shown below.

$$\begin{gathered} \frac{d}{dx}\left(5+c_1{e}^{-2x}\right)=-2c_1{e}^{-2x}+5-5 \\ \frac{d}{dx}\left(5+c_1{e}^{-2x}\right)=-2\left(c_1{e}^{-2x}+5\right)-5 \end{gathered}$$

Substitute $y=5+c_1{e}^{-2x}$in the above differential equation.

$$\begin{gathered} \frac{dy}{dx}=-2\left(y-5\right) \\ \frac{dy}{dx}=-2y+10 \end{gathered}$$

Thus, the answer is$\frac{dy}{dx}=-2y+10$.