Chapter 2: Q2.6 - 18E (page 76)

Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ ” to solve problems 17-20.
$\frac{dy}{dx} = {(x + y + 2)}^{2}$

Short Answer

Expert verified

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

Step by step solution

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, $\frac{dy}{dx} = {(x + y + 2)}^{2}$

Let us take $z = x + y y = z - x$

Differentiate with respect to x.

$\begin{array}{rcl} \frac{dy}{dx} & = & \frac{dz}{dx} - 1 \\ {(z + 2)}^{2} & = & \frac{dz}{dx} - 1 \\ {(z + 2)}^{2} + 1 & = & \frac{dz}{dx} \\ \frac{1}{{(z + 2)}^{2} + 1} dz & = & dx \end{array}$

Integrate the equation

Now, integrate on both sides.

$\int \frac{1}{{(z + 2)}^{2} + 1} dz = dx \int \frac{1}{{(z + 2)}^{2} + 1} dz = x + C$

Evaluate $\int \frac{1}{{(z + 2)}^{2} + 1} dz$ separately.

Let; $\begin{array}{rcl} v & = & z + 2 \\ \frac{dv}{dz} & = & 1 \\ dz & = & dv \end{array}$

$\begin{array}{rcl} \int \frac{1}{v^{2} + 1} dv & = & x + C \\ \tan^{- 1} v & = & x + C \end{array}$

Substitute $v = z + 2$ and $z = x + y$

$\begin{array}{rcl} \tan^{- 1} (z + 2) & = & x + C \\ \tan^{- 1} (x + y + 2) & = & x + C \\ x + y + 2 & = & \tan (x + C) \\ y & = & \tan (x + C) - x - 2 \end{array}$

Therefore, Equation of the form of $\frac{dy}{dx} = G (ax + by)$ for the given equation isrole="math" localid="1655207486825" $y = \tan (x + C) - x - 2$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ ” to solve problems 17-20.
$\frac{dy}{dx} = {(x + y + 2)}^{2}$

Short Answer

Step by step solution

General form of dydx=Gax+by

Evaluate the given equation

Integrate the equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Mechanics Maths

Applied Mathematics

Statistics

Geometry

Pure Maths

Study anywhere. Anytime. Across all devices.

Use the method discussed under “Equations of the Form dydx=Gax+by” to solve problems 17-20.dydx=x+y+22

Short Answer

Step by step solution

General form of dydx=Gax+by

Evaluate the given equation

Integrate the equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Mechanics Maths

Applied Mathematics

Statistics

Geometry

Pure Maths

Study anywhere. Anytime. Across all devices.

Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ ” to solve problems 17-20.
$\frac{dy}{dx} = {(x + y + 2)}^{2}$