Is It Linear in the Indicated Dependent Variable?

[find_the_fun]

I'm unclear how to do these kinds of questions. Here's an example

Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the equation $$a_1(x)\frac{dy}{dx} = + a_o(x)y=g(x)$$

[math](y^2-1)dx+xdy=0[/math]; in y; in x

The answer is it's linear in x but nonlinear in y.

First off I don't understand "indicated dependent variable", is it saying first consider y as a function of x and then consider x as a function of y? What difference does this have on the solution?

Here's what I tried.

[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)+x\frac{dy}{dx}=0[/math]
[math]x\frac{dy}{dx}+y^2=1[/math]
Therefore [math]a_1(x)=x[/math], [math]a_o(x)=invalid[/math] and [math]g(x)=1[/math] Since one is invalid the equation is non linear in y.

For it being linear in x
[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)\frac{dx}{dy}+x=0[/math]

 

[MarkFL]

Re: linear in x but not linear in y

I'm unclear how to do these kinds of questions. Here's an example

Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the equation $$a_1(x)\frac{dy}{dx} = + a_o(x)y=g(x)$$

[math](y^2-1)dx+xdy=0[/math]; in y; in x

The answer is it's linear in x but nonlinear in y.

First off I don't understand "indicated dependent variable", is it saying first consider y as a function of x and then consider x as a function of y? What difference does this have on the solution?

Here's what I tried.

[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)+x\frac{dy}{dx}=0[/math]
[math]x\frac{dy}{dx}+y^2=1[/math]
Therefore [math]a_1(x)=x[/math], [math]a_o(x)=invalid[/math] and [math]g(x)=1[/math] Since one is invalid the equation is non linear in y.

For it being linear in x
[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)\frac{dx}{dy}+x=0[/math]

I tend to like to write linear equations in the form:

$$\tag{1}\frac{dy}{dx}+P(x)y=Q(x)$$

Okay, we are given the ODE:

$$\left(y^2-1 \right)\,dx+x\,dy=0$$

To check for linearity in $y$, we try to express the ODE in the form given by (1). I find the form, which is equivalent to what you found:

$$\frac{dy}{dx}+\frac{1}{x}y^2=\frac{1}{x}$$

The fact that $y$ is squared means the ODE is non-linear in $y$.

To check for linearity in $x$, we want to express the ODE in the form:

$$\tag{2}\frac{dx}{dy}+P(y)x=Q(y)$$

We find the form equivalent to what you found:

$$\frac{dx}{dy}+\frac{1}{y^2-1}x=0$$

So the equation is linear in $x$.

In practice linear equations are much easier to solve that non-linear equations, and so if we can write an ODE in linear form, then we will be able to more easily solve it. The difference in the solution is we will get $x(y)$ as the solution rather than $y(x)$.