Question

what is the method of solving homogeneous equation?​

Answer 1

Solving Homogeneous Differential Equations

A homogeneous equation can be solved by substitution y=ux, which leads to a separable differential equation.

Answer 2

A Differential Equation is an equation with a function and one or more of its derivatives:

differential equation dy/dx = 5xy
Example: an equation with the function y and its derivative dydx

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:

dydx = F( yx )

We can solve it using Separation of Variables but first we create a new variable v = yx

v = yx which is also y = vx
And dydx = d (vx)dx = v dxdx + x dvdx (by the Product Rule)
Which can be simplified to dydx = v + x dvdx
Using y = vx and dydx = v + x dvdx we can solve the Differential Equation.

An example will show how it is all done:

Example: Solve dydx = x2 + y2xy

Can we get it in F( yx ) style?

Start with: x2 + y2xy
Separate terms: x2xy + y2xy
Simplify: xy + yx
Reciprocal of first term: ( yx )−1 + yx
Yes, we have a function of (y/x).

So let's go:

Start with: dydx = ( yx )−1 + yx
y = vx and dydx = v + x dvdx : v + x dvdx = v−1 + v
Subtract v from both sides: x dvdx = v−1
Now use Separation of Variables:

Separate the variables: v dv = 1x dx
Put the integral sign in front: ∫v dv = ∫ 1x dx
Integrate: v22 = ln(x) + C
Then we make C = ln(k): v22 = ln(x) + ln(k)
Combine ln: v22 = ln(kx)
Simplify: v = ±√(2 ln(kx))
Now substitute back v = yx

Substitute v = yx : yx = ±√(2 ln(kx))
Simplify: y = ±x √(2 ln(kx))
And we have the solution.

The positive portion looks like this:

y = x sqrt(2 ln(kx))



Another example:

Example: Solve dydx = y(x−y)x2

Can we get it in F( yx ) style?

Start with: y(x−y)x2
Separate terms: xyx2 − y2x2
Simplify: yx − ( yx )2
Yes! So let's go:

Start with: dydx = yx − ( yx )2
y = vx and dy dx = v + x dvdx v + x dvdx = v − v2
Subtract v from both sides: x dvdx = −v2
Now use Separation of Variables:

Separate the variables: − 1v2 dv = 1x dx
Put the integral sign in front: ∫− 1v2 dv = ∫ 1x dx
Integrate: 1v = ln(x) + C
Then we make C = ln(k): 1v = ln(x) + ln(k)
Combine ln: 1v = ln(kx)
Simplify: v = 1ln(kx)
Now substitute back v = yx

Substitute v = yx : yx = 1ln(kx)
Simplify: y = xln(kx)
And we have the solution.

Here are some sample k values:

y = x / ln(kx)

And one last example:

Example: Solve dydx = x−yx+y

Can we get it in F( yx ) style?

Start with: x−yx+y
Divide through by x: x/x−y/xx/x+y/x
Simplify: 1−y/x1+y/x
Yes! So let's go:

Start with: dydx = 1−y/x1+y/x
y = vx and dy dx = v + x dvdx v + x dvdx = 1−v1+v
Subtract v from both sides: x dvdx = 1−v1+v − v
Then: x dvdx = 1−v1+v − v+v21+v
Simplify: x dvdx = 1−2v−v21+v
Now use Separation of Variables:

Separate the variables: 1+v1−2v−v2 dv = 1x dx
Put the integral sign in front: ∫ 1+v1−2v−v2 dv = ∫ 1x dx
Integrate: − 12 ln(1−2v−v2) = ln(x) + C
Then we make C = ln(k): − 12 ln(1−2v−v2) = ln(x) + ln(k)
Combine ln: (1−2v−v2)−½ = kx
Square and Reciprocal: 1−2v−v2 = 1k2x2
Now substitute back v = yx

Substitute v = yx : 1−2( yx )−( yx )2 = 1k2x2
Multiply through by x2: x2−2xy−y2 = 1k2
We are nearly there ... it is nice to separate out y though!
We can try to factor x2−2xy−y2 but we must do some rearranging first:

Change signs: y2+2xy−x2 = − 1k2
Replace − 1k2 by c: y2+2xy−x2 = c
Add 2x2 to both sides: y2+2xy+x2 = 2x2+c
Factor: (y+x)2 = 2x2+c
Square root: y+x = ±√(2x2+c)
Subtract x from both sides: y = ±√(2x2+c) − x
And we have the solution.

The positive portion looks like this:

y = sqrt(2x^2+c) - x

Hope this helps you
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