Using v substitution for first order homogenous DE and constraining solution

[Mayhem]

TL;DR

Understanding this substitution for a first order homogenous equation as well as a solution to an example of such an equation.

My considers a type of differential equation $$\frac{\mathrm{d} y}{\mathrm{d} x} = f\left(\frac{y}{x} \right )$$ and proposes that it can be solved by letting $v(x) = \frac{y}{x}$ which is equivalent to $y = xv(x)$. Then it says $$\frac{\mathrm{d} y}{\mathrm{d} x} = v + x\frac{\mathrm{d} v}{\mathrm{d} x}$$ without any further explanation. Is it correctly understand that it just let's $\frac{\mathrm{d} y}{\mathrm{d} x} = y' = \left(xv(x) \right)' = v + x \frac{\mathrm{d} v}{\mathrm{d} x}$ ?

Next, I would like to ask if I'm understanding this right. For the DE $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{x^2+xy+y^2}{x^2}$$ the solution is $$y = x\tan{\left(\ln(|x|) + C \right)}$$ (I got the same answer as the book). But shouldn't it be noted that this solution is only true for $-\pi/2 < y/x < \pi/2$? The book mentions nothing of the sort in the answer key.

 

[pasmith]

Summary:: Understanding this substitution for a first order homogenous equation as well as a solution to an example of such an equation.

My considers a type of differential equation $$\frac{\mathrm{d} y}{\mathrm{d} x} = f\left(\frac{y}{x} \right )$$ and proposes that it can be solved by letting $v(x) = \frac{y}{x}$ which is equivalent to $y = xv(x)$. Then it says $$\frac{\mathrm{d} y}{\mathrm{d} x} = v + x\frac{\mathrm{d} v}{\mathrm{d} x}$$ without any further explanation. Is it correctly understand that it just let's $\frac{\mathrm{d} y}{\mathrm{d} x} = y' = \left(xv(x) \right)' = v + x \frac{\mathrm{d} v}{\mathrm{d} x}$ ?

Yes.

Next, I would like to ask if I'm understanding this right. For the DE $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{x^2+xy+y^2}{x^2}$$ the solution is $$y = x\tan{\left(\ln(|x|) + C \right)}$$ (I got the same answer as the book). But shouldn't it be noted that this solution is only true for $-\pi/2 < y/x < \pi/2$? The book mentions nothing of the sort in the answer key.

To determine C, you need to apply an initial condition y(x_0) = y_0 where x_0 \neq 0. Using a more convenient definition of C the general solution is <br /> y = x\tan \left( \ln \left| \frac{x}{x_0}\right| + C\right) where the aboluste value signs are not strictly necessary as x and x_0 will have the same sign so x/x_0 &gt; 0 and we obtain <br /> \frac{y_0}{x_0} = \tan C. Now the range of \tan is the whole of \mathbb{R} so such a C always exists, and is equal to \arctan(y_0/x_0) \in (-\frac 12\pi, \frac12\pi).

The solution is valid for -\frac{\pi}{2} &lt; \ln \left|\frac{x}{x_0}\right| + \arctan\left(\frac{y_0}{x_0}\right) &lt; \frac{\pi}{2}..

 

[Mayhem]

Yes.To determine C, you need to apply an initial condition y(x_0) = y_0 where x_0 \neq 0. Using a more convenient definition of C the general solution is <br /> y = x\tan \left( \ln \left| \frac{x}{x_0}\right| + C\right) where the aboluste value signs are not strictly necessary as x and x_0 will have the same sign so x/x_0 &gt; 0 and we obtain <br /> \frac{y_0}{x_0} = \tan C. Now the range of \tan is the whole of \mathbb{R} so such a C always exists, and is equal to \arctan(y_0/x_0) \in (-\frac 12\pi, \frac12\pi).

The solution is valid for -\frac{\pi}{2} &lt; \ln \left|\frac{x}{x_0}\right| + \arctan\left(\frac{y_0}{x_0}\right) &lt; \frac{\pi}{2}..

How do I learn to think like this? I don't think this line of reasoning would have come to me no matter how long I looked at the problem.

 

[Mark44]

How do I learn to think like this?

Experience.
Solving differential equations requires a bag of tricks that you learn as you go along.

 

[Mayhem]

Experience.
Solving differential equations requires a bag of tricks that you learn as you go along.

Seems reasonable. I think the same is true for integrals.