# Differential equations/checking an answer - help, please!

1. Apr 8, 2005

### DivGradCurl

Problem from my textbook - Solve:

$$\frac{dy}{dx} = \frac{4y - 3x}{2x - y}$$

Answer provided:

$$\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.} \mbox{ Also, } y = -3x \mbox{.}$$

My answer is pretty close to that one, but it seems like I missed some step. Anyhow, here's what's I've got:

Left-hand side:

$$\frac{dy}{dx} = \frac{d}{dx} \left( xv \right)$$

$$\frac{dy}{dx} = xv^{\prime} + v$$

$$\frac{dy}{dx} = x\frac{dv}{dx} + v$$

Right-hand side:

$$\frac{dy}{dx} = \frac{4y - 3(y/v)}{2(y/v) - y}$$

$$\frac{dy}{dx} = -\left( \frac{4v-3}{v-2} \right)$$

Hence, this equation is homogeneous. As a result, we find

$$x\frac{dv}{dx} + v = -\left( \frac{4v-3}{v-2} \right)$$

$$x\frac{dv}{dx} = \frac{-v^2 -2v + 3}{v-2} \mbox{,}$$

which is separable. Thus, we obtain

$$\int \frac{v-2}{-v^2 -2v + 3} \: dv = \int \frac{dx}{x}$$

$$\frac{1}{4} \ln \left| v - 1 \right| - \frac{5}{4} \ln \left| v + 3 \right| = \ln \left| x \right| + \mathrm{C}$$

$$\ln \left| v - 1 \right| - 5 \ln \left| v + 3 \right| = 4 \ln \left| x \right| + \mathrm{C}$$

$$\ln \frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \ln \left( x^4 \right) + \mathrm{C}$$

$$\frac{\left| v - 1 \right|}{\left| v + 3 \right| ^5} = \mathrm{C} x^4$$

$$\frac{\left| (y/x) - 1 \right|}{\left| (y/x) + 3 \right| ^5} = \mathrm{C} x^4$$

$$\frac{\left| y - x \right|}{\left| y + 3x \right| ^5} = \mathrm{C} x^4$$

$$\left| y - x \right| = \mathrm{C} x^4 \left| y + 3x \right| ^5 \mbox{,}$$

which is different from

$$\left| y - x \right| = \mathrm{C} \left| y + 3x \right| ^5 \mbox{.}$$

Again, it seems like I missed some step. Can anyone help me out with that?

By the way, I can't see why $$y=-3x$$ also is an answer. Could you please clarify it?

Any help is highly appreciated.

Last edited: Apr 8, 2005
2. Apr 8, 2005

### jdavel

thiago,

Don't you have a sign error in your "Right-hand side" equation (second line)? I'm not sure if that will get rid of the x^4 factor or not.

3. Apr 8, 2005

### stunner5000pt

here is your problem
$$\frac{\left| (y/x) - 1 \right|}{\left| (y/x) + 3 \right| ^5} = \mathrm{C} x^4$$

will turn into $$\frac{\frac{y-x}{x}}{\frac{(y+3)^5}{x^5}} = Cx^4$$ which cancels out the x^4

4. Apr 8, 2005

### DivGradCurl

Oh... sure!!! I certainly skipped that step. About the y=-3x, don't bother explaining. I've just figured it out.

Thank you so much.

5. Apr 9, 2005

### DivGradCurl

One more question!!!

I've just noticed that $$y=x$$ also seems to work. Is it really possible?

Thanks

6. Apr 9, 2005

### stunner5000pt

what do you mean y=x seems to work? Do you mean for the substitution part? Or do you mean for the equilibrium points?

7. Apr 9, 2005

### DivGradCurl

As an equilibrium solution, maybe. I'm not sure. Probably, something like y=-3x is.

8. Apr 9, 2005

### stunner5000pt

isnt hte equilibrium solution obtained when dy/dx = 0?? in which case y = 3/4x?

9. Apr 9, 2005

### AKG

Yes, y=x is a solution, but it doesn't need to be stated in addition, since it is simply the solution you get when C = 0. Well, actually, C is of the form exp(A^4) where A is the constant of integration, and this will be a positive number (i.e. C cannot be 0). So perhaps y=x does need to be written separately.

10. Apr 9, 2005

### DivGradCurl

That makes sense. There still is a question I'd like to ask, though. I need to check something. My procedure with respect to $$y=-3x$$ is to plug into the given DE:

$$\frac{dy}{dx} = \frac{4y - 3x}{2x - y}$$

which gives $$-3$$, which is the expected result. Thus, it has been verified as an additional solution. A mistake that I made previously was to do the same thing with $$y=x$$, because it also implies that $$\mathrm{C}=0$$, which is false.

My question is: does $$y=-3x$$ work because it satisfies the procedure above and is not dependent on the value of the constant $$\mathrm{C}$$?

Thanks

I literally was dozing off when I suggested otherwise. Sorry about that. :rofl: Thanks for reminding me of the definition.

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