Chapter 2: Q2.6-1E (page 76)

In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ . $2 t \times dx + (t^{2} - x^{2}) dt = 0$

Short Answer

Expert verified

The given equation is the form of both homogeneous and Bernoulli.

Step by step solution

General form of homogeneous, Bernoulli, linear coefficients of the form of y'=Gax+by

Homogeneous equation

If the right-hand side of the equation $\frac{d y}{d x} = f (x, y)$ can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

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

Bernoulli’s equation

A first-order equation that can be written in the form $\frac{d y}{d x} + P (x) y = Q (x) y^{n}$ , where $P (x)$ and $Q (x)$ are continuous on interval $(a, b)$ and localid="1654934017689" $n$ is a real number, is called a Bernoulli equation.

Equation of Linear coefficients

We have used various substitutions for $y$ to transform the original equation into a new equation that we could solve. In some cases, we must transform both $x$ and $y$ into new variables, say $u$ and $v$ . This is the situation for equations with linear coefficients-that is, equations of the form $(a_{1} x + b_{1} y + c_{1}) d x + (a_{2} x + b_{2} y + c_{2}) d y = 0$

Evaluate the given equation

Given, $2 t x d x + (t^{2} - x^{2}) d t = 0$

By Evaluation,

role="math" localid="1654934940793" $\begin{array}{rcl} 2 t x d x + (t^{2} - x^{2}) d t & = & 0 \\ 2 t x d x & = & - (t^{2} - x^{2}) d t \\ \frac{d x}{d t} & = & \frac{t^{2} - x^{2}}{2 t x} \\ = & - \frac{1}{2} (\frac{t}{x} - \frac{x}{t}) \\ = & \frac{1}{2} (\frac{x}{t} - \frac{1}{\frac{x}{t}}) \end{array}$ .........(1)

Substitution method

Let us consider $v = \frac{x}{t}$

From Equation (1),

role="math" localid="1654934950153" $\begin{array}{rcl} \frac{x^{2} - t^{2}}{2 t x} & = & G (\frac{x}{t}) \\ = & \frac{1}{2} (\frac{x}{t} - \frac{1}{\frac{x}{t}}) \\ G (v) & = & \frac{1}{2} (v - \frac{1}{v}) \end{array}$

So, the given equation is homogeneous.

It can be rewritten as,

$\begin{array}{rcl} \frac{d x}{d t} & = & \frac{x^{2} - t^{2}}{2 t x} \\ = & \frac{x}{2 t} - \frac{t}{2 x} \\ \frac{d x}{d t} + (- \frac{1}{2 t}) x & = & (- \frac{t}{2}) x^{- 1} \end{array}$

Comparing the found equation with Bernoulli general form of equation. It seems that the given equation is Bernoulli equation as well.

Therefore, the given equation is the form of both homogeneous and Bernoulli.

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In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ . $2 t \times dx + (t^{2} - x^{2}) dt = 0$

Short Answer

Step by step solution

General form of homogeneous, Bernoulli, linear coefficients of the form of y'=Gax+by

Evaluate the given equation

Substitution method

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Most popular questions from this chapter

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In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form y'=G(ax+by).2t×dx+(t2-x2)dt=0

Short Answer

Step by step solution

General form of homogeneous, Bernoulli, linear coefficients of the form of y'=Gax+by

Evaluate the given equation

Substitution method

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Discrete Mathematics

Theoretical and Mathematical Physics

Pure Maths

Logic and Functions

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ . $2 t \times dx + (t^{2} - x^{2}) dt = 0$