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Q2.6 - 17E

Expert-verified
Found in: Page 76

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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

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 ${\mathbf{y}}{\mathbf{=}}{\mathbf{-}}{\mathbf{x}}$.

See the 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

$\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{=}\mathbf{z}\mathbf{-}\mathbf{x}$

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 ${\mathbf{y}}{\mathbf{=}}{\mathbf{-}}{\mathbf{x}}$.