Chapter 3.4, Problem 31PS

To determine

(a)

All functions that satisfy $\frac{dy}{dx}=0$

$y\left(x\right)=\text{Constant}C$.

Given:

An equation, $\frac{dy}{dx}=0$.

Calculation:

We have

$\frac{dy}{dx}=0$

Integrate both sides of the equation,

$\begin{array}{c}y\left(x\right)=0+C\\ y\left(x\right)=C\end{array}$

Where C is constant.

Therefore, the function that satisfies $\frac{dy}{dx}=0$is $y\left(x\right)=\text{Constant}C$.

(b)

A particular function that satisfy $\frac{dy}{dx}=3$

$y_p\left(x\right)=3x$.

Given:

An equation, $\frac{dy}{dx}=3$.

Calculation:

It is noted that

$\frac{dy}{dx}=3$

$\Rightarrow y_p\left(x\right)=3x$ is a particular function which solves the equation $\frac{dy}{dx}=3$.

(c)

All functions that satisfy $\frac{dy}{dx}=3$

$y\left(x\right)=3x+C$.

Given:

An equation, $\frac{dy}{dx}=3$.

Calculation:

Suppose $y\left(x\right)$is a solution tom:math>dydx=3.

Then,

$\begin{array}{c}\frac{d}{dx}\left(y\left(x\right)-y_p\left(x\right)\right)=\frac{dy}{dx}-\frac{d{y}_p}{dx}\\ =3-3\\ =0\end{array}$

Thus, $\left(y\left(x\right)-y_p\left(x\right)\right)$is a solution to equation $\frac{dy}{dx}=0$,

However, we know that $\left(y\left(x\right)-y_p\left(x\right)\right)=C$, for some constant.

$\begin{array}{c}y\left(x\right)=y_p\left(x\right)+C\\ y\left(x\right)=3x+C\end{array}$

Therefore, $y\left(x\right)=3x+C$satisfy $\frac{dy}{dx}=3$.

To determine

(b)

A particular function that satisfy $\frac{dy}{dx}=3$

$y_p\left(x\right)=3x$.

Given:

An equation, $\frac{dy}{dx}=3$.

Calculation:

It is noted that

$\frac{dy}{dx}=3$

$\Rightarrow y_p\left(x\right)=3x$ is a particular function which solves the equation $\frac{dy}{dx}=3$.

(c)

All functions that satisfy $\frac{dy}{dx}=3$

$y\left(x\right)=3x+C$.

Given:

An equation, $\frac{dy}{dx}=3$.

Calculation:

Suppose $y\left(x\right)$is a solution tom:math>dydx=3.

Then,

$\begin{array}{c}\frac{d}{dx}\left(y\left(x\right)-y_p\left(x\right)\right)=\frac{dy}{dx}-\frac{d{y}_p}{dx}\\ =3-3\\ =0\end{array}$

Thus, $\left(y\left(x\right)-y_p\left(x\right)\right)$is a solution to equation $\frac{dy}{dx}=0$,

However, we know that $\left(y\left(x\right)-y_p\left(x\right)\right)=C$, for some constant.

$\begin{array}{c}y\left(x\right)=y_p\left(x\right)+C\\ y\left(x\right)=3x+C\end{array}$

Therefore, $y\left(x\right)=3x+C$satisfy $\frac{dy}{dx}=3$.

To determine

(c)

All functions that satisfy $\frac{dy}{dx}=3$

$y\left(x\right)=3x+C$.

Given:

An equation, $\frac{dy}{dx}=3$.

Calculation:

Suppose $y\left(x\right)$is a solution tom:math>dydx=3.

Then,

$\begin{array}{c}\frac{d}{dx}\left(y\left(x\right)-y_p\left(x\right)\right)=\frac{dy}{dx}-\frac{d{y}_p}{dx}\\ =3-3\\ =0\end{array}$

Thus, $\left(y\left(x\right)-y_p\left(x\right)\right)$is a solution to equation $\frac{dy}{dx}=0$,

However, we know that $\left(y\left(x\right)-y_p\left(x\right)\right)=C$, for some constant.

$\begin{array}{c}y\left(x\right)=y_p\left(x\right)+C\\ y\left(x\right)=3x+C\end{array}$

Therefore, $y\left(x\right)=3x+C$satisfy $\frac{dy}{dx}=3$.