Chapter 2: Q2.6 - 13E (page 76)

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.
$\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$

Short Answer

Expert verified

Homogeneous equation for the given equation is $\sqrt{1 + \frac{x^{2}}{t^{2}}} = \ln |t| + C$

Step by step solution

General form of Homogeneous equation

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

Evaluate the given equation

Given, $\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$

Evaluate it by dividing $t^{2}$ by both numerator and denominator.

$\frac{dx}{dt} = \frac{\frac{x^{2}}{t^{2}} + \sqrt{1 + \frac{x^{2}}{t^{2}}}}{\frac{x}{t}} \frac{dx}{dt} = \frac{{(\frac{x}{t})}^{2} + \sqrt{1 + {(\frac{x}{t})}^{2}}}{(\frac{x}{t})}$

Substitution method

Let us take $v = \frac{y}{x}$

Then $y = v x$

By Differentiating,

$\begin{array}{rcl} \frac{dx}{dt} & = & v + t \frac{dv}{dt} \\ (\frac{v^{2} + \sqrt{1 + v^{2}}}{v}) & = & v + t \frac{dv}{dt} \end{array}$

$\begin{array}{rcl} (v + \frac{\sqrt{1 + v^{2}}}{v}) & = & v + t \frac{dv}{dt} \\ \frac{\sqrt{1 + v^{2}}}{v} & = & t \frac{dv}{dt} \\ \frac{\sqrt{1 + v^{2}}}{v} \frac{1}{dv} & = & t \frac{1}{dt} \\ \frac{v}{\sqrt{1 + v^{2}}} dv & = & \frac{1}{t} dt \end{array}$

Integrate the equation

Now, integrate on both sides.

$\int \frac{v}{\sqrt{1 + v^{2}}} dv = \int \frac{1}{t} dt \int \frac{v}{\sqrt{1 + v^{2}}} dv = \ln |t| + C$

Integrate $\int \frac{v}{\sqrt{1 + v^{2}}} dv$ separately.

Let us take $w = 1 + v^{2}$

Then, $\frac{dw}{dv} = 2 v dv = \frac{1}{2 v} dw$

Now,

$\begin{array}{rcl} \int \frac{v}{\sqrt{w}} \frac{1}{2 v} dw & = & \frac{1}{2} \int \frac{1}{\sqrt{w}} dw \\ = & \frac{1}{2} (2) \sqrt{w} \\ = & \sqrt{w} \end{array}$

Substitute $w = 1 + v^{2}$

$\begin{array}{rcl} \int \frac{v}{\sqrt{1 + v^{2}}} dv & = & \sqrt{w} \\ = & \sqrt{1 + v^{2}} \end{array}$

Then,

$\sqrt{1 + v^{2}} = \ln |t| + C$

Substitute $v = \frac{y}{x}$

$\sqrt{1 + {(\frac{x}{t})}^{2}} = \ln |t| + C \sqrt{1 + \frac{x^{2}}{t^{2}}} = \ln |t| + C$

Therefore, Homogeneous equation for the given equation is $\sqrt{1 + \frac{x^{2}}{t^{2}}} = \ln |t| + C$

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Use the method discussed under “Homogeneous Equations” to solve problems 9-16.
$\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$

Short Answer

Step by step solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

Integrate the equation

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Use the method discussed under “Homogeneous Equations” to solve problems 9-16. dxdt=x2+tt2+x2tx

Short Answer

Step by step solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

Integrate the equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Statistics

Discrete Mathematics

Pure Maths

Logic and Functions

Geometry

Study anywhere. Anytime. Across all devices.

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.
$\frac{dx}{dt} = \frac{x^{2} + t \sqrt{t^{2} + x^{2}}}{tx}$