Chapter 2, Problem 2.1P

To determine

(a)

Whether the given differential equation is linear or non-linear with supportive reason.

Answer to Problem 2.1P

$y\ddot{y}+5\dot{y}+y=0$ is a non-linear differential equation.

As, here, derivative of dependent variable is multiplied with itself.

Explanation of Solution

Given:

$y\ddot{y}+5\dot{y}+y=0$.

Concept Used:

An ordinary differential for $y=y\left(t\right)$ is said to be linear if it can be written in the form

$a_n\left(t\right)y^n+a_{n-1}\left(t\right)y^{n-1}+\mathrm{...}+a_2\left(t\right)y^2+a_1\left(t\right)y^{\prime }+a_0\left(t\right)y=g\left(t\right)$

where the ‘coefficient’ functions $a_n\left(t\right),a_{n-1}\left(t\right),\mathrm{...},a_2\left(t\right),a_1\left(t\right),a_0\left(t\right),g\left(t\right)$ can be any functions of t, (including the zero function).

In other words a differential equation is said to be linear if and only if the derivative of dependent variable should not be multiplied with dependent variable itself, otherwise it should be non-linear.

Calculation:

Given differential equation is

$y\ddot{y}+5\dot{y}+y=0$

Which is a non-linear differential equation.

As, here, derivative of dependent variable is multiplied with itself.

To determine

(b)

Whether the given differential equation is linear or non-linear with supportive reason.

Answer to Problem 2.1P

$\dot{y}+\sin y=0$ is a linear differential equation.

As, here, derivative of dependent variable is not multiplied with itself.

Explanation of Solution

Given:

$\dot{y}+\sin y=0$.

Concept Used:

An ordinary differential for $y=y\left(t\right)$ is said to be linear if it can be written in the form

$a_n\left(t\right)y^n+a_{n-1}\left(t\right)y^{n-1}+\mathrm{...}+a_2\left(t\right)y^2+a_1\left(t\right)y^{\prime }+a_0\left(t\right)y=g\left(t\right)$

where the ‘coefficient’ functions $a_n\left(t\right),a_{n-1}\left(t\right),\mathrm{...},a_2\left(t\right),a_1\left(t\right),a_0\left(t\right),g\left(t\right)$ can be any functions of t, (including the zero function).

In other words a differential equation is said to be linear if and only if the derivative of dependent variable should not be multiplied with dependent variable itself, otherwise it should be non-linear.

Calculation:

Given differential equation is

$\dot{y}+\sin y=0$ is a linear differential equation. As, here, derivative of dependent variable is not multiplied with itself.

To determine

(c)

Whether the given differential equation is linear or non-linear with supportive reason.

Answer to Problem 2.1P

$\dot{y}+\sqrt{y}=0$ is a linear differential equation.

As, here, derivative of dependent variable is not multiplied with itself.

Explanation of Solution

Given:

$\dot{y}+\sqrt{y}=0$.

Concept Used:

An ordinary differential for $y=y\left(t\right)$ is said to be linear if it can be written in the form

$a_n\left(t\right)y^n+a_{n-1}\left(t\right)y^{n-1}+\mathrm{...}+a_2\left(t\right)y^2+a_1\left(t\right)y^{\prime }+a_0\left(t\right)y=g\left(t\right)$

where the ‘coefficient’ functions $a_n\left(t\right),a_{n-1}\left(t\right),\mathrm{...},a_2\left(t\right),a_1\left(t\right),a_0\left(t\right),g\left(t\right)$ can be any functions of t, (including the zero function).

In other words a differential equation is said to be linear if and only if the derivative of dependent variable should not be multiplied with dependent variable itself, otherwise it should be non-linear.

Calculation:

Given differential equation is

$\dot{y}+\sqrt{y}=0$ is a linear differential equation.

As, here, derivative of dependent variable is not multiplied with itself.

To determine

(d)

Whether the given differential equation is linear or non-linear with supportive reason.

Answer to Problem 2.1P

$y\ddot{y}+5t^2\dot{y}+3y=0$

is a non-linear differential equation.

As, here, derivative of dependent variable is multiplied with itself.

Explanation of Solution

Given:

$y\ddot{y}+5t^2\dot{y}+3y=0$.

Concept Used:

An ordinary differential for $y=y\left(t\right)$ is said to be linear if it can be written in the form

$a_n\left(t\right)y^n+a_{n-1}\left(t\right)y^{n-1}+\mathrm{...}+a_2\left(t\right)y^2+a_1\left(t\right)y^{\prime }+a_0\left(t\right)y=g\left(t\right)$

where the ‘coefficient’ functions $a_n\left(t\right),a_{n-1}\left(t\right),\mathrm{...},a_2\left(t\right),a_1\left(t\right),a_0\left(t\right),g\left(t\right)$ can be any functions of t, (including the zero function).

In other words a differential equation is said to be linear if and only if the derivative of dependent variable should not be multiplied with dependent variable itself, otherwise it should be non-linear.

Calculation:

Given differential equation is

$y\ddot{y}+5t^2\dot{y}+3y=0$

Which is a non-linear differential equation.

As, here, derivative of dependent variable is multiplied with itself.

To determine

(e)

Whether the given differential equation is linear or non-linear with supportive reason.

Answer to Problem 2.1P

$\ddot{y}+3t^2\sin y=0$ is a non-linear differential equation.

As, here, derivative of dependent variable is multiplied with independent variable.

Explanation of Solution

Given:

$\ddot{y}+3t^2\sin y=0$.

Concept Used:

An ordinary differential for $y=y\left(t\right)$ is said to be linear if it can be written in the form

$a_n\left(t\right)y^n+a_{n-1}\left(t\right)y^{n-1}+\mathrm{...}+a_2\left(t\right)y^2+a_1\left(t\right)y^{\prime }+a_0\left(t\right)y=g\left(t\right)$

where the ‘coefficient’ functions $a_n\left(t\right),a_{n-1}\left(t\right),\mathrm{...},a_2\left(t\right),a_1\left(t\right),a_0\left(t\right),g\left(t\right)$ can be any functions of t, (including the zero function).

In other words a differential equation is said to be linear if and only if the derivative of dependent variable should not be multiplied with dependent variable itself, otherwise it should be non-linear.

Calculation:

Given differential equation is

$\ddot{y}+3t^2\sin y=0$

Which is a non-linear differential equation.

As, here, function of dependent variable is multiplied with independent variable.

To determine

(f)

Whether the given differential equation is linear or non-linear with supportive reason.

Answer to Problem 2.1P

$\dot{y}+e^{t}y=0$ which is a linear differential equation.

As, here, function of dependent variable is not multiplied with independent variable.

Explanation of Solution

Given:

$\dot{y}+e^{t}y=0$.

Concept Used:

An ordinary differential for $y=y\left(t\right)$ is said to be linear if it can be written in the form

$a_n\left(t\right)y^n+a_{n-1}\left(t\right)y^{n-1}+\mathrm{...}+a_2\left(t\right)y^2+a_1\left(t\right)y^{\prime }+a_0\left(t\right)y=g\left(t\right)$

where the ‘coefficient’ functions $a_n\left(t\right),a_{n-1}\left(t\right),\mathrm{...},a_2\left(t\right),a_1\left(t\right),a_0\left(t\right),g\left(t\right)$ can be any functions of t, (including the zero function).

In other words a differential equation is said to be linear if and only if the derivative of dependent variable should not be multiplied with dependent variable itself, otherwise it should be non-linear.

Calculation:

Given differential equation is

$\dot{y}+e^{t}y=0$

which is a linear differential equation.

As, here, function of dependent variable is not multiplied with independent variable.