Unizor - Ordinary Differential Equations - Homogeneous Equations
Channel: Zor Shekhtman on UNIZOR Education
Date: 2017-06-19

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Homogeneous Ordinary Differential Equations

We have defined homogeneous ordinary differential equations of the first order as an equation
F(x, y, y')=0
which does not change if we replace x with λ·x and y with λ·y, where λ - any real number not equal to zero.
In other words,
F(x, y, y') = F(λ·x, λ·y, y')
Examples:
F(x, y, y') = y'+y/x
F(x, y, y') = 3y'+x·y/(x²+y²)
etc.

The recommended technique to solve these equations is to substitute function y(x) with x·z(x) and solve the equation for z(x), after which determine y(x)=x·z(x).

Let's solve a few equations of this kind.

Example 1

Check for homogeneousness and solve the following equation:
x·y' = x·sin(y/x) + y

Now let's solve this equation using the substitution z(x)=y(x)/x, which results in y(x)=z(x)·x, and express the initial equation in terms of x, z and z'.
x·(z'·x + z) = x·sin(z) + z·x
Simplifying:
z'·x + z = sin(z) + z
z'·x = sin(z)
dz/sin(z) = dx/x
∫dz/sin(z) = ∫dx/x
The right side is easy, the integral equals to ln(|x|)+C.
The left side is more involved.
∫dz/sin(z) =
= ∫dz/(2sin(z/2)·cos(z/2)) =
= ∫d(z/2)/(sin(z/2)·cos(z/2))
Substitute u=z/2, getting
∫du/(sin(u)·cos(u)) =
= ∫cos(u)d(u)/(sin(u)·cos²(u)) =
= ∫d(sin(u))/(sin(u)·cos²(u))
Substitute t=sin(u), getting
∫dt/[t·(1−t²)]
The polynomial in the denominator is
t·(1−t²) = t·(1−t)·(1+t)
Its inverse can be represented as
1/t − 1/[2(1+t)] + 1/[2(1−t)]
which makes our integral equal to
∫{[2/t−1/(1+t)+1/(1−t)]/2}dt
The last expression can be represented as a sum of three integrals, the result of integration is:
ln(|t|) − ln(|1+t|)/2 − ln(|1−t|)/2
where t=sin(z/2)
This leads us to a final solution of our differential equation.
ln(|sin(z/2)|) −
− ln(1+sin(z/2))/2 −
− ln(1−sin(z/2))/2 = ln(|x|)+C
and then we should substitute z=y/x to get the final expression
ln(|sin(y/2x)|) −
− ln(1+sin(y/2x))/2 −
− ln(1−sin(y/2x))/2 = ln(|x|)+C
Using this as an exponent, we come up with an expression without logarithms
|sin(y/2x)| / [(1+sin(y/2x))·(1−sin(y/2x))] = C·|x|
A simplification in the denominator results in
|sin(y/2x)| / cos²(y/2x) = C·|x|
We leave it "as is" without resolving for y(x).

Example 2

Check for homogeneousness and solve the following equation:
[(y − x·y')/x]^x = e^y

Now we will solve it using the recommended technique.
Substitute z(x)=y(x)/x, which results in y(x)=z(x)·x and express the initial equation in terms of x, z and z'.
The expression for a derivative y' is:
y' = (z·x)' = z'·x+z
New equation is, therefore,
[(z·x − x·(z'·x+z))/x]^x = e^(z·x)
Simplifying it by raising to power 1/x both sides (or, equivalently, extracting a root of power x):
[(z·x − x·(z'·x+z))/x] = e^z
Cancel x:
z − (z'·x+z) = e^z
Cancel z:
−z'·x = e^z
This equation is separable, let's separate x from x, getting
−e^(−z)·dz = dx/x
Ready to integrate:
∫−e^(−z)·dz = ∫dx/x
e^(-z) = ln(x)+C
(assuming for simplicity positive only sign for x, so integral on the right is ln(x) instead of ln(|x|))
From the last equation we derive:
−z = ln(ln(x)) + C
z = −ln(ln(x)) + C
Now we can use it to find an expression for y:
y = −x·ln(ln(x)) + C

Example 3

Check for homogeneousness and solve the following equation:
x·y·y' = (x+y)²

Now let's solve it by substituting z(x)=y(x)/x, which results in y(x)=z(x)·x and express the initial equation in terms of x, z and z'.
The expression for a derivative y' is:
y' = (z·x)' = z'·x+z
So, our equation looks like
x·(z·x)·(z'·x+z) = (x+z·x)²
Simplifying by opening all parenthesis, we get
x²·(x·z·z'+z²) = x²·(1+z)²
x·z·z'+z² = (1+z)²
x·z·z' = 1+2z
This equation can be solved using the method of separation.
z·dz/(1+2z) = dx/x
Integrating the left side of this equation:
∫z·dz/(1+2z) =
= (1/2)∫(1+2z−1)·dz/(1+2z) =
= (1/2)[∫dz − ∫dz/(1+2z)] =
= (1/2)[z−(1/2)ln(1+2z)] + C
Integrating the right side of the equation:
∫dx/x = ln(x) + C
Since integral of both sides are equal,
(1/2)[z−(1/2)ln(1+2z)] =
= ln(x)+ C
which can be simplified
2z − ln(1+2z) = 4ln(x) + C
Though this equation for z(x) cannot be easily solve for z, it allows to replace the original differential equation for y with purely algebraic one, replacing z with y/x:
(A) 2y/x − ln(1+2y/x) =
= 4ln(x) + C
This is the final algebraic answer to our differential equation. Though it's not resolved for y(x), it's still the best solution we can come up with.