Question

Use the method discussed under “Equations of the Form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$" to solve problems 17-20.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\sqrt{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{-}\mathbf{1}$

Short answer

Equation of the form of $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{G}\left(\mathrm{ax}+\mathrm{by}\right)$for the given equation is $\mathbf{y}\mathbf{=}\frac{{\left(\mathbf{x}\mathbf{+}\mathbf{C}\right)}^{\mathbf{2}}}{\mathbf{4}}\mathbf{-}\mathbf{x}$ and$\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{x}$.

Step-by-step solution

General form of dydx=Gax+by

Equations of the form$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$ When the right-hand side of the equation$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{f}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)$ can be expressed as a function of the combination $\mathbf{ax}\mathbf{+}\mathbf{by}$, where a and b are constants, that is,$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$ then the substitution$\mathbf{z}\mathbf{=}\mathbf{ax}\mathbf{+}\mathbf{by}$ transforms the equation into a separable one.

Evaluate the given equation

Given, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\sqrt{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{-}\mathbf{1}$

Let us take

$$\begin{gathered} \mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y} \\ \mathbf{y}\mathbf{=}\mathbf{z}\mathbf{-}\mathbf{x} \end{gathered}$$

Differentiate with respect to.

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \frac{\mathbf{dz}}{\mathbf{dx}}\mathbf{-}\mathbf{1}\\ \sqrt{\mathbf{z}}\mathbf{-}\mathbf{1}& \mathbf{=}& \frac{\mathbf{dz}}{\mathbf{dx}}\mathbf{-}\mathbf{1}\\ \sqrt{\mathbf{z}}& \mathbf{=}& \frac{\mathbf{dz}}{\mathbf{dx}}\\ \frac{\mathbf{1}}{\sqrt{\mathbf{z}}}\mathbf{dz}& \mathbf{=}& \mathbf{dx}\end{array}$

Integrate the equation

Now, integrate on both sides.

$\begin{array}{rcl}\int \frac{\mathbf{1}}{\sqrt{\mathbf{z}}}\mathbf{dz}& \mathbf{=}& \int \mathbf{dx}\\ \mathbf{2}\sqrt{\mathbf{z}}& \mathbf{=}& \mathbf{x}\mathbf{+}\mathbf{C}\\ \sqrt{\mathbf{z}}& \mathbf{=}& \frac{\mathbf{x}\mathbf{+}\mathbf{C}}{\mathbf{2}}\\ & & \end{array}$

Substitute$\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}$

$\begin{array}{rcl}\sqrt{\mathbf{x}\mathbf{+}\mathbf{y}}& \mathbf{=}& \frac{\mathbf{x}\mathbf{+}\mathbf{C}}{\mathbf{2}}\\ \mathbf{x}\mathbf{+}\mathbf{y}& \mathbf{=}& {\left(\frac{\mathbf{x}\mathbf{+}\mathbf{C}}{\mathbf{2}}\right)}^{\mathbf{2}}\\ \mathbf{y}& \mathbf{=}& \frac{{\left(\mathbf{x}\mathbf{+}\mathbf{C}\right)}^{\mathbf{2}}}{\mathbf{4}}\mathbf{-}\mathbf{x}\\ & & \end{array}$

Therefore, Equation of the form of $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$for the given equation is $\mathbf{y}\mathbf{=}\frac{{\left(\mathbf{x}\mathbf{+}\mathbf{C}\right)}^{\mathbf{2}}}{\mathbf{4}}\mathbf{-}\mathbf{x}$and$\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{x}$.