Differential Equations of form y'(x)=f(ax+by+c)

  • Thread starter khkwang
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  • #1
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Homework Statement


My professor states that a differential equation of form y'(x)=f(ax+by+c) can be reduced to a separable equation by substituting in v=ax+by+c, but I don't see how.

Edit: more specifically: y'(x)= sqrt(3x -4y +2)


Homework Equations


y'(x)=f(ax+by+c)
v=ax+by+c

The Attempt at a Solution



If v=ax+by+c, then dv/dx = a + b*dy/dx
Then dv/dx = a + b*f(v)

But this isn't a separable differential equation... the constant a is in the way.

Edit: more specifically:
dv/dx = 3 - 4*sqrt(v)
 
Last edited:

Answers and Replies

  • #2
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Does

[tex]\displaystyle\int \frac{dv}{a + bf(v)} = \displaystyle\int dx[/tex]

not work?
 
  • #3
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yeah, i was about to post that i'm a moron for not realizing such a solution a few moments ago...

But now I'm stuck on how to solve

[tex]\displaystyle\int \frac{dv}{3 - 4\sqrt{v}} = \displaystyle\int dx[/tex]
 
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  • #4
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I just solved that particular integral by hand. I don't know if I should give away the method I used yet. It's an easy straightforward one, once you see what to do. No trigonometric substitution or anything like that, obviously.
 
  • #5
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I'll give it another shot. If i still can't figure it out in a while, I'll ask again?

Edit:

I think I solved it, but it's a little messy

[tex]
\displaystyle\int \frac{dv*(3 + 4\sqrt{v})}{9 - 16v} = \displaystyle\int dx
[/tex]

[tex]
\displaystyle\frac{-3*\ln{|9-16v|}}{16} + \frac{\sqrt{v}}{2} + c = x
[/tex]

[tex]
\displaystyle\ln{|64y - 48x -23|} - \frac{8\sqrt{3x - 4y + 2}}{3} - \frac{16x}{3} + c = 0
[/tex]

Is this good?
 
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  • #6
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That method seems a little messy.

[tex]x = \int \frac{dv}{3-4\sqrt{v}} = \int \frac{3+4\sqrt{v}}{9-16v}dv[/tex]

[tex]x = \int \frac{3}{9-16v}dv + 4 \int \frac{\sqrt{v}}{9-16v}dv[/tex]

[tex]x = - \frac{3}{16}\ln|9-16v| + 4 \int \frac{\sqrt{v}}{9-16v}dv[/tex]

I fear the second integral is not equal to sqrt(v)/2.

My suggestion would have been to use the substitution u = sqrt(v), then the integral becomes

[tex]x = \int \frac{2u}{3-4u}du[/tex]

which is the same as

[tex]x = \int \left( - \frac{1}{2} + \frac{3/2}{3-4u}\right) du[/tex]

I think. Might want to double check the division.
 
  • #7
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Ahh substitution. It seems obvious now lol thanks.

But can you tell me how you brought the 1/2 out of the last equation to get rid of the extra u? I can see that the last two equations are equivalent, but I don't understand the process. Did you just guess at taking -1/2 out? Or is there some basic rule I'm forgetting? On another problem right now i need to get the integral of (2u + 1)du/(5u + 5). Would the same steps apply there?

Edit: Scratch helping me solve (2u + 1)du/(5u +5), I figured it out using the same method you showed me. Still, could you tell me if that process has a name? I don't feel like I've ever been taught it before. Or is it just supposed to be obvious to the student?
 
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  • #9
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Cheers
 

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