Chapter 1.RP, Problem 1RP

To determine

To classify:

The equation $5\frac{dx}{dt}+5x^2+3=0$

Answer to Problem 1RP

Solution:

In $5\frac{dx}{dt}+5x^2+3=0$,

$t$ is the independent variable and $x$ is the dependent variable

And also a nonlinear equation.

Explanation of Solution

Given:

$5\frac{dx}{dt}+5x^2+3=0$

Calculation:

Independent and dependent variables:

If an equation involves the derivative of one variable with respect to another, then the former is called a dependent variable and latter an independent variable.

In the given equation $t$ is the independent variable and $x$ is the dependent variable.

Linear or nonlinear equation:

A differential equation is linear if it has the format as below,

$a_n\left(x\right)\frac{d^{n}y}{d{x}^n}+a_{n-1}\left(x\right)\frac{d^{n-1}y}{d{x}^{n-1}}+\dots +a_{1n}\left(x\right)\frac{dy}{dx}+a_0\left(x\right)y=F(x)$

Where, $a_n\left(x\right),a_{n-1}\left(x\right)\dots .,a_0\left(x\right) and F(x)$ depends only on the independent variable $x$.

If the equation is not in the above format, then it is called nonlinear equation.

$5\frac{dx}{dt}+5x^2+3=0$ is the nonlinear equation, since it is not in the above format.

Hence, In $5\frac{dx}{dt}+5x^2+3=0$ is,

$t$ is the independent variable and $x$ is the dependent variable

And also a nonlinear equation.

Final statement:

In $5\frac{dx}{dt}+5x^2+3=0$,

$t$ is the independent variable and $x$ is the dependent variable

And also a nonlinear equation.