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
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:
𝑑𝑦
𝑑𝑥
𝑦
= F( )
𝑥
𝑦
We can solve it using Separation of Variables but first we create a new variable v = 𝑥
𝑦
v=𝑥
And
is also y = vx
dy
=
dx
𝑑(𝑣𝑥)
𝑑𝑥
𝑑𝑥
= v 𝑑𝑥+ x
dv
dx
Which can be simplified to
Using y = vx and
(by the Product Rule)
dy
dv
= v + x dx
dx
𝐝𝐲
𝑑𝑣
= v + x 𝑑𝑥 we can solve the Differential Equation.
𝐝𝐱
An example will show how it is all done:
𝐝𝐲
Example: Solve 𝐝𝐱=
𝐱 𝟐 +𝐲 𝟐
𝐱𝐲
𝑥
Can we get it in F( 𝑦 ) style?
Start with:
𝐱𝟐 + 𝐲𝟐
𝐱𝐲
Separate terms:
Simplify:
Reciprocal of first term:
𝑥 2 𝐲𝟐
+
𝑥𝑦 𝑥𝑦
𝑥
𝑦
+𝑥
𝑦
𝑦
𝑦
(𝑥 )-1 + 𝑥
Yes! So let's go:
Start with:
y = vx and
𝐝𝐲
𝐝𝐲
= v + x 𝐝𝐱
𝐝𝐱
Subtract v from both sides:
Now use Separation of Variables:
𝑑𝑦
𝑑𝑥
𝑦
𝑦
𝑥
𝑥
=( )-1 +
dv
v + x dx= v-1 + v
x
dv
dx
= v-1
Separate the variables:
Put the integral sign in front:
Integrate:
Then we make C = ln(k):
Combine ln:
Simplify:
v dv =
1
x
dx
1
∫v dv = ∫ 𝑥 dx
𝒗𝟐
𝟐
= ln(x) + C
v2/ 2 = ln(x) + ln(k)
v2/ 2 = ln(kx)
v = ±√(2 ln(kx))
y
Now substitute back v = x
Substitute v = y/ x :
Simplify:
y/ x = ±√(2 ln(kx))
y = ±x√(2 ln(kx))
And we have the solution.
Another example:
𝐝𝐲
Example: Solve 𝐝𝐱 =
𝐲(𝐱−𝐲)
𝐱𝟐
𝑥
Can we get it in F( 𝑦 ) style?
Start with:
Separate terms:
Simplify:
Yes! So let's go:
𝐲(𝐱−𝐲)
x2
xy
x2
𝑦
−
y2
x2
𝑦
− (𝑥 )2
𝑥
𝑑𝑦
y = vx and
𝐝𝐲
𝐝𝐱
𝑦
𝑦
= − (𝑥 )2
𝑑𝑥 𝑥
Start with:
𝑑𝑣
v+x
= v + x 𝑑𝑥
Subtract v from both sides:
x
𝑑𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
= v − v2
= −v2
Now use Separation of Variables:
Separate the variables:
−
Put the integral sign in front:
∫−
Combine ln:
v2
𝑣
1
dx
x
1
v
x
𝑣
1
dv =
1
1
Integrate:
Then we make C = ln(k):
1
2 dv = ∫ dx
= ln(x) + C
= ln(x) + ln(k)
1
𝑣
= ln(kx)
Simplify:
1
v = ln(kx)
𝑦
Now substitute back v = 𝑥
Substitute v = y/ x :
Simplify:
y/ x = 1 /ln(kx)
y = x /ln(kx)
And we have the solution.
And one last example:
Example
: Solve
𝐝𝐲
𝐝𝐱
𝒙−𝒚
= 𝒙+𝒚
𝑥
Can we get it in F( 𝑦 ) style?
Start with :
𝒙−𝒚
𝒙+𝒚
Divide through by x
:
Simplify:
𝒙−𝒚
𝑥
𝒙+𝒚
𝑥
𝑦
𝑥
𝑦
1+
𝑥
1−
Yes! So let's go:
𝑦
𝑑𝑦 1−𝑥
=
𝑦
𝑑𝑥 1+
Start with:
𝑥
y = vx and
𝒅𝒚
𝑑𝑣
𝒅𝒙
𝑑𝑥
=v+x
v+x
Subtract v from both sides:
Then:
x
𝑥
𝑑𝑣
𝑑𝑥
𝑑𝑣
𝑑𝑥
=
𝑑𝑣 1−𝑣
=
𝑑𝑥 1+𝑣
=
1−𝑣
1+𝑣
1−𝑣
1+𝑣
−v
−
𝑣+𝑣 2
1+𝑣
Simplify:
𝑑𝑣
𝑥 𝑑𝑥 =
1−2𝑣−𝑣 2
1+𝑣
Now use Separation of Variables:
Separate the variables:
1+𝑣
1
𝑑𝑣 = 𝑑𝑥
2
1 − 2𝑣 − 𝑣
𝑥
Put the integral sign in front:
Integrate:
Then we make C = ln(k):
∫
1+𝑣
1
𝑑𝑣 = ∫ 𝑑𝑥
2
1 − 2𝑣 − 𝑣
𝑥
−1
ln(1
2
−1
ln(1
2
− 2v − v2) = ln(x) + C
− 2v − v2) = ln(x)+ln(k)
−1
Combine ln: (1 − 2𝑣 − 𝑣 2 ) 2 = 𝑘𝑥
2
Square and Reciprocal: 1 − 2𝑣 − 𝑣 =
1
2
𝑘 𝑥2
Now substitute back v =
y
x
Substitute
v=
𝑦
𝑥
𝑦
𝑦 2
:1 − 2 ( ) − ( ) =
𝑥
𝑥
Multiply through by x2:
1
𝑘2𝑥 2
x2−2xy−y2 = 1/ k2
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:
Replace − 1 k2 by c:
Add 2x2 to both sides:
Factor:
y2+2xy−x2 = − 1 k2
y2+2xy−x2 = c
y2+2xy+x2 = 2x2+c
(y+x)2 = 2x2+c
Square root:
y+x = ± √(2x 2 + c)
Subtract x from both sides:
y = ± √(2x 2 + c) − x
And we have the solution.