Using v substitution for first order homogenous DE and constraining solution

In summary: It's just that the bag of tricks for integrals is much more extensive than for DEs.Yes. In summary, the conversation discusses understanding a substitution for a first-order homogeneous equation and a solution to an example of such an equation. The solution is valid within a certain range which can be determined using an initial condition. Developing a bag of tricks through experience is important when solving differential equations, similar to solving integrals.
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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.
 
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Mayhem said:
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 [itex]C[/itex], you need to apply an initial condition [itex]y(x_0) = y_0[/itex] where [itex]x_0 \neq 0[/itex]. Using a more convenient definition of [itex]C[/itex] the general solution is [tex]
y = x\tan \left( \ln \left| \frac{x}{x_0}\right| + C\right)[/tex] where the aboluste value signs are not strictly necessary as [itex]x[/itex] and [itex]x_0[/itex] will have the same sign so [itex]x/x_0 > 0[/itex] and we obtain [tex]
\frac{y_0}{x_0} = \tan C.[/tex] Now the range of [itex]\tan[/itex] is the whole of [itex]\mathbb{R}[/itex] so such a [itex]C[/itex] always exists, and is equal to [itex]\arctan(y_0/x_0) \in (-\frac 12\pi, \frac12\pi)[/itex].

The solution is valid for [tex]-\frac{\pi}{2} < \ln \left|\frac{x}{x_0}\right| + \arctan\left(\frac{y_0}{x_0}\right) < \frac{\pi}{2}.[/tex].
 
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  • #3
pasmith said:
Yes.To determine [itex]C[/itex], you need to apply an initial condition [itex]y(x_0) = y_0[/itex] where [itex]x_0 \neq 0[/itex]. Using a more convenient definition of [itex]C[/itex] the general solution is [tex]
y = x\tan \left( \ln \left| \frac{x}{x_0}\right| + C\right)[/tex] where the aboluste value signs are not strictly necessary as [itex]x[/itex] and [itex]x_0[/itex] will have the same sign so [itex]x/x_0 > 0[/itex] and we obtain [tex]
\frac{y_0}{x_0} = \tan C.[/tex] Now the range of [itex]\tan[/itex] is the whole of [itex]\mathbb{R}[/itex] so such a [itex]C[/itex] always exists, and is equal to [itex]\arctan(y_0/x_0) \in (-\frac 12\pi, \frac12\pi)[/itex].

The solution is valid for [tex]-\frac{\pi}{2} < \ln \left|\frac{x}{x_0}\right| + \arctan\left(\frac{y_0}{x_0}\right) < \frac{\pi}{2}.[/tex].
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.
 
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Mayhem said:
How do I learn to think like this?
Experience.
Solving differential equations requires a bag of tricks that you learn as you go along.
 
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  • #5
Mark44 said:
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.
 

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