MHB Why is the interval solution this?

[find_the_fun]

The question is solve, give transient term and interval of solution for [math]x \frac{dy}{dx}-y = x^2 \sin{x}[/math] and the answer key has [math]y=cx-x\cos{x}[/math] and [math](0, \infty)[/math]. Why wouldn't the interval be [math](-\infty, \infty)[/math]?

 

[MarkFL]

I can see why $x=0$ is eliminated in order to write the ODE in the form:

$$\frac{d}{dx}\left(\frac{y}{x}\right)=\sin(x)$$

But I don't know why they have restricted $x$ only to positive values. :D

 

[find_the_fun]

I can see why $x=0$ is eliminated in order to write the ODE in the form:

$$\frac{d}{dx}\left(\frac{y}{x}\right)=\sin(x)$$

But I don't know why they have restricted $x$ only to positive values. :D

Isn't it because interval has to be continuous i.e. can't have a break at 0?

 

[MarkFL]

Isn't it because interval has to be continuous i.e. can't have a break at 0?

What I mean is I don't see why the interval $(-\infty,0)$ couldn't be chosen either. Not both, but one or the other. :D

 

[Ackbach]

Was there an initial condition? If so, was it in $(-\infty,0)$ or $(0,\infty)$?