Homogeneous Differential Equations
A Differential Equation is an equation with a function and one or more of its derivatives:
Example: an equation with the function y and its
derivative \frac{dy}{dx}
Here we look at a special method for solving "Homogeneous Differential Equations"
Homogeneous Differential Equations
A first order Differential Equation is Homogeneous when it can be in this form:
\frac{dy}{dx} = F( \frac{y}{x} )
We can solve it using Separation of Variables but first we create a new variable v = y x
Using y = vx and dy dx = v + x dv dx we can solve the Differential Equation.
An example will show how it is all done:
Example: Solve \frac{dy}{dx} = \frac{x^{2} + y^{2}}{xy}
Can we get it in F( \frac{x}{y} ) style?
Start with: | \frac{x^{2} + y^{2}}{xy} | |
Separate terms: | \frac{x^{2}}{xy} + \frac{y^{2}}{xy} | |
Simplify: | \frac{x}{y} + \frac{y}{x} | |
Reciprocal of first term: | ( \frac{y}{x} )^{-1} + \frac{y}{x} |
Yes! So let's go:
Start with: | \frac{dy}{dx} = ( \frac{y}{x} )^{-1} + \frac{y}{x} | |
y = vx and dy dx = v + x dv dx | v + x \frac{dv}{dx} = v^{-1} + v | |
Subtract v from both sides: | x \frac{dv}{dx} = v^{-1} |
Now use Separation of Variables:
Separate the variables: | v dv = \frac{1}{x} dx | |
Put the integral sign in front: | ∫v dv = ∫\frac{1}{x} dx | |
Integrate: | \frac{v^{2}}{2} = ln(x) + C | |
Then we make C = ln(k): | \frac{v^{2}}{2} = ln(x) + ln(k) | |
Combine ln: | \frac{v^{2}}{2} = ln(kx) | |
Simplify: | v = ±√(2 ln(kx)) |
Now substitute back v = \frac{y}{x}
Substitute v = \frac{y}{x}: | \frac{y}{x} = ±√(2 ln(kx)) | |
Simplify: | y = ±x √(2 ln(kx)) |
And we have the solution.
Another example:
Example: Solve \frac{dy}{dx} = \frac{y(x−y)}{x^{2}}
Can we get it in F( \frac{x}{y} ) style?
Start with: | \frac{y(x−y)}{x^{2}} | |
Separate terms: | \frac{xy}{x^{2}} − \frac{y^{2}}{x^{2}} | |
Simplify: | \frac{y}{x} − ( \frac{y}{x} )^{2} |
Yes! So let's go:
Start with: | \frac{dy}{dx} = \frac{y}{x} − ( \frac{y}{x} )^{2} | |
y = vx and dy dx = v + x dv dx | v + x \frac{dv}{dx} = v − v^{2} | |
Subtract v from both sides: | x \frac{dv}{dx} = −v^{2} |
Now use Separation of Variables:
Separate the variables: | − \frac{1}{v^{2}} dv = \frac{1}{x} dx | |
Put the integral sign in front: | ∫− \frac{1}{v^{2}} dv = ∫\frac{1}{x} dx | |
Integrate: | \frac{1}{v} = ln(x) + C | |
Then we make C = ln(k): | \frac{1}{v} = ln(x) + ln(k) | |
Combine ln: | \frac{1}{v} = ln(kx) | |
Simplify: | v = \frac{1}{ln(kx)} |
Now substitute back v = \frac{y}{x}
Substitute v = \frac{y}{x}: | \frac{y}{x} = \frac{1}{ln(kx)} | |
Simplify: | y = \frac{x}{ln(kx)} |
And we have the solution.
And one last example:
Example: Solve \frac{dy}{dx} = \frac{x−y}{x+y}
Can we get it in F( \frac{x}{y} ) style?
Start with: | \frac{x−y}{x+y} | |
Divide through by x: | \frac{x/x−y/x}{x/x+y/x} | |
Simplify: | \frac{1−y/x}{1+y/x} |
Yes! So let's go:
Start with: | \frac{dy}{dx} = \frac{1−y/x}{1+y/x} | |
y = vx and dy dx = v + x dv dx | v + x \frac{dv}{dx} = \frac{1−v}{1+v} | |
Subtract v from both sides: | x \frac{dv}{dx} = \frac{1−v}{1+v} − v | |
Then: | x \frac{dv}{dx} = \frac{1−v}{1+v} − \frac{v+v^{2}}{1+v} | |
Simplify: | x \frac{dv}{dx} = \frac{1−2v−v^{2}}{1+v} |
Now use Separation of Variables:
Separate the variables: | \frac{1+v}{1−2v−v^{2}} dv = \frac{1}{x} dx | |
Put the integral sign in front: | ∫\frac{1+v}{1−2v−v^{2}} dv = ∫\frac{1}{x} dx | |
Integrate: | −\frac{1}{2} ln(1−2v−v^{2}) = ln(x) + C | |
Then we make C = ln(k): | −\frac{1}{2} ln(1−2v−v^{2}) = ln(x) + ln(k) | |
Combine ln: | (1−2v−v^{2})^{−½} = kx | |
Square and Reciprocal: | 1−2v−v^{2} = \frac{1}{k^{2}x^{2}} |
Now substitute back v = \frac{y}{x}
Substitute v = \frac{y}{x}: | 1−2( \frac{y}{x} )−( \frac{y}{x} )^{2} = \frac{1}{k^{2}x^{2}} | |
Multiply through by x^{2}: | x^{2}−2xy−y^{2} = \frac{1}{k^{2}} |
We are nearly there ... it is nice to separate out y though!
We can try to factor x^{2}−2xy−y^{2} but we must do some rearranging first:
Change signs: | y^{2}+2xy−x^{2} = − \frac{1}{k^{2}} | |
Replace − \frac{1}{k^{2}} by c: | y^{2}+2xy−x^{2} = c | |
Add 2x^{2} to both sides: | y^{2}+2xy+x^{2 }= 2x^{2}+c | |
Factor: | (y+x)^{2} = 2x^{2}+c | |
Square root: | y+x = ±√(2x^{2}+c) | |
Subtract x from both sides: | y = ±√(2x^{2}+c) − x |
And we have the solution.