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

[khkwang]

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)

 

[pbandjay]

Does

$$\displaystyle\int \frac{dv}{a + bf(v)} = \displaystyle\int dx$$

not work?

 

[khkwang]

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

$$\displaystyle\int \frac{dv}{3 - 4\sqrt{v}} = \displaystyle\int dx$$

 

[pbandjay]

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.

 

[khkwang]

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

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

$$\displaystyle\frac{-3*\ln{|9-16v|}}{16} + \frac{\sqrt{v}}{2} + c = x$$

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

Is this good?

 

[pbandjay]

That method seems a little messy.

$$x = \int \frac{dv}{3-4\sqrt{v}} = \int \frac{3+4\sqrt{v}}{9-16v}dv$$

$$x = \int \frac{3}{9-16v}dv + 4 \int \frac{\sqrt{v}}{9-16v}dv$$

$$x = - \frac{3}{16}\ln|9-16v| + 4 \int \frac{\sqrt{v}}{9-16v}dv$$

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

$$x = \int \frac{2u}{3-4u}du$$

which is the same as

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

I think. Might want to double check the division.

 

[khkwang]

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?

 

[pbandjay]

It's long polynomial division. It's usually the first thing I try when seeing two polynomials in the integrand in the form u(x)/v(x). There's a nice example outlined here:

http://en.wikipedia.org/wiki/Polynomial_long_division

 

[khkwang]

Cheers