In summary: Yes, y=-3x does work because it satisfies the procedure above and is not dependent on the value of the constant \mathrm{C}.
Problem from my textbook - Solve:

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

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

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:
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.

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

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

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.

One more question!

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

Thanks

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

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

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

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.

AKG said:
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.

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

stunner5000pt said:
isnt hte equilibrium solution obtained when dy/dx = 0?? in which case y = 3/4x?

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

## 1. What are differential equations?

Differential equations are mathematical equations that describe how a physical quantity changes over time. They are used to model many natural phenomena, such as the growth of populations, the flow of fluids, and the motion of objects.

## 2. How do I check if my answer to a differential equation is correct?

The best way to check your answer is to plug it back into the original differential equation and see if it satisfies the equation. You can also use a graphing calculator to graph both the original equation and your solution to visually compare them.

## 3. What are the different types of differential equations?

There are several types of differential equations, including ordinary differential equations (ODEs), partial differential equations (PDEs), and stochastic differential equations (SDEs). ODEs involve a single independent variable, PDEs involve multiple independent variables, and SDEs involve random variables.

## 4. Do I need to know calculus to understand and solve differential equations?

Yes, a strong understanding of calculus is necessary to understand and solve differential equations. Differential equations involve derivatives and integrals, which are fundamental concepts in calculus.

## 5. Are there any common techniques for solving differential equations?

Yes, there are several common techniques for solving differential equations, including separation of variables, integrating factors, and substitution. It is important to understand which technique is most appropriate for a given differential equation.

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