How Do You Solve and Visualize a Homogeneous Differential Equation?

[aznkid310]

Homework Statement

dy/dx = (x + 3y)/(x - y)

A) Solve the Differential Eqn
B) Draw a Direction Field and some integral curves. Are they symmetric w/ respect to the origin?

Homework Equations

I believe i solved the equation correctly, but i don't know how to draw the direction fields and integral curves. I tried plotting y v. y' in order to create the resulting direction field and integral curves, but i don't know what it looks like.

*Also, how do I integrate the left side? I just used an online calculator to get the answer, but i would like to know how to solve this

The Attempt at a Solution

A) After dividing by x and substituting v = y/x:

(1 + 3v)/(1 - v) = xv' + v

v' = (1/x)( (1 + 3v)/(1 - v) - v )

* (1-3v)/(1+3v) - 1/v dv = dx/x

After integrating and solving for c:

C = (2/3)ln(3y/x + 1) - y/x -ln(y/x) - ln(x)
Also, y = -x

 

[benorin]

* Use long division: $$\frac{1-3v}{1+3v}=-1+\frac{2}{1+3v}$$ then

$$\int \left(\frac{1-3v}{1+3v}-\frac{1}{v}\right)\, dv = \int \left(-1+\frac{2}{1+3v}-\frac{1}{v}\right)\, dv = -v+{\textstyle\frac{2}{3}} \ln |1+3v| -\ln |v|+C$$​

A direction field (a.k.a http://mathworld.wolfram.com/SlopeField.html" , see link for a better definition) is a plot of unit vectors with slope determined by the DE (e.g., on our direction field at (5,1) we plot a unit vector with a slope of $y^{\prime}=\frac{5+3}{5-1}= 2$).

An http://mathworld.wolfram.com/IntegralCurve.html" is a particular solution to a differential equation corresponding to a specific value of the equation's free parameters.

 

[tiny-tim]

Hey guys!

Isn't it (1 - v)dv/(1 + v)² ?

 

[benorin]

Thanks tiny-tim.
Oops ...

$$x\frac{dv}{dx}=\frac{1+3v}{1-v}-v$$
$$\frac{1-v}{(1+v)^2}dv = \frac{dx}{x}$$