Chapter 8, Problem 1RE

Classify the following first-order differential equations as separable, linear, both, or neither.

$\begin{array}{l}\text{(a)}\frac{dy}{dx}-3y=\sin x\\ \left(\text{b}\right)\frac{dy}{dx}+xy=x\\ \left(\text{c}\right)y\frac{dy}{dx}-x=1\\ \text{(d)}\frac{dy}{dx}+x{y}^2=\sin (xy)\end{array}$

To determine

The given differential equation, $\frac{dy}{dx}-3y=\sin x$ , is separable, linear, both, or neither.

Answer to Problem 1RE

The given equation is linear.

Explanation of Solution

For an equation to be separable, it should be of the form, $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ .

Since the equation, $\frac{dy}{dx}-3y=\sin x$ , is not of the form, $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ , it is non separable.

For an equation to be linear, it should be of the form, $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$ .

Since the equation, $\frac{dy}{dx}-3y=\sin x$ , is of the form, $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$ , it is linear.

Hence, the given equation is linear.

To determine

The given differential equation, $\frac{dy}{dx}+xy=x$ , is separable, linear, both, or neither.

Answer to Problem 1RE

The given equation is both linear and separable.

Explanation of Solution

For an equation to be separable, it should be of the form, $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ .

Consider the given equation.

$\begin{array}{c}\frac{dy}{dx}+xy=x\\ \frac{dy}{dx}=x-xy\\ \frac{dy}{dx}=x\left(1-y\right)\\ \frac{1}{\left(1-y\right)}\frac{dy}{dx}=x\end{array}$

So, the given equation follows $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ , it is separable.

For an equation to be linear, it should be of the form, $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$ .

Since the equation, $\frac{dy}{dx}+xy=x$ , is of the form, $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$ , it is linear.

Hence, the given equation is both linear and separable.

To determine

The given differential equation, $y\frac{dy}{dx}-x=1$ , is separable, linear, both, or neither.

Answer to Problem 1RE

The given equation is separable.

Explanation of Solution

For an equation to be separable, it should be of the form, $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ .

Consider the given equation.

$\begin{array}{c}y\frac{dy}{dx}-x=1\\ y\frac{dy}{dx}=\left(1+x\right)\end{array}$

So, the given equation is of the form, $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ , it is separable.

It can’t be linear as it cannot be written in the form of $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$ .

Hence, the given equation is separable.

To determine

The given differential equation, $\frac{dy}{dx}+x{y}^2=\sin \left(xy\right)$ , is separable, linear, both, or neither.

Answer to Problem 1RE

The given equation is neither separable nor linear.

Explanation of Solution

For an equation to be separable, it should be of the form, $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ .

Consider the given equation.

$\begin{array}{c}\frac{dy}{dx}+x{y}^2=\sin \left(xy\right)\\ \frac{dy}{dx}+x{y}^2=2\sin x\cos y\\ \end{array}$

Since the equation, $\frac{dy}{dx}+x{y}^2=\sin (xy)$ , is not of the form, $h\left(y\right)\frac{dy}{dx}=g\left(x\right)$ , the equation is non separable.

It can’t be linear as it not of the form, $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$ .

Hence, the given equation is neither separable nor linear.