Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ " to solve problems 17-20.
$\frac{dy}{dx} = \sqrt{x + y} - 1$

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

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General form of dydx=Gax+by

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

Evaluate the given equation

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

Let us take

$z = x + y y = z - x$

Differentiate with respect to.

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

Integrate the equation

Now, integrate on both sides.

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

Substitute $z = x + y$

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

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

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ " to solve problems 17-20.
$\frac{dy}{dx} = \sqrt{x + y} - 1$

Step by Step Solution

General form of dydx=Gax+by

Evaluate the given equation

Integrate the equation

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the method discussed under “Equations of the Form dydx=Gax+by" to solve problems 17-20. dydx=x+y-1

Step by Step Solution

General form of dydx=Gax+by

Evaluate the given equation

Integrate the equation

Most popular questions for Math Textbooks

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Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ " to solve problems 17-20.
$\frac{dy}{dx} = \sqrt{x + y} - 1$