## Monday, June 19, 2017

### Unizor - Ordinary Differential Equations - Homogeneous Equations

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

Checking for homogeneousness.
Substitute x with λ·x and y with λ·y:
λ·x·y' = λ·x·sin(λ·y/(λ·x)) + λ·y
Obviously, λ cancels out completely, which proves homogeneous character of the equation.
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 xz 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 = ey

Checking for homogeneousness.
Substitute x with λ·x and y with λ·y:
[(λy − λx·y')/λx]λx = eλy
Cancel λ in the ratio, getting:
[(y − x·y')/x]λx = eλy
This can be written as
{[(y − x·y')/x]x}λ = [ey]λ
Raising both sides to power 1/λ (or, which is the same, extracting a root of power λ) we come to the original equation, which proves homogeneous character of the equation.
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 xz 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 = ez·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] = ez
Cancel x:
z − (z'·x+z) = ez
Cancel z:
−z'·x = ez
This equation is separable, let's separate x from x, getting
−e−z·dz = dx/x
−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

Solution must be checked.
It's easier, instead of checking the original equation
[(y − x·y')/x]x = ey
to check the equality of logarithms from both sides:
x·ln[(y − x·y')/x] = y
or, simpler,
ln(y/x − y') = y/x
where we should substitute
y = −x·ln(ln(x)) + C
and
y' = −ln(ln(x))−x·(1/ln(x))·(1/x)
or, simpler,
y' = −ln(ln(x)) − 1/ln(x)
Let's disregard constant C in this checking to make manipulations simpler.
Then, since
y/x = −ln(ln(x))
we will have to check that ln(−ln(ln(x)) + ln(ln(x)) + 1/ln(x)) = −ln(ln(x))
Canceling opposite positive and negative members under logarithm on the left, we come to an obvious equality
ln(1/ln(x)) = −ln(ln(x))
which proves the correctness of our solution.

Example 3

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

Checking for homogeneousness.
Substitute x with λ·x and y with λ·y:
λx·λy·y' = (λx+λy)²
λ²x·y·y' = λ²(x+y)²
Obviously, λ cancels out, and we get the same original equation.
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 xz 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.
dz/(1+2z) = dx/x
Integrating the left side of this equation:
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.

Solution must be checked.
If this equality that includes function y(x) is correct, derivatives of both parts are also equal. Let's differentiate them both.
−2y/x² + 2y'/x − (1/(1+2y/x))·(−2y/x²+2y'/x) = 4/x
Simplifying by multiplying by :
−2y + 2xy' − x·(−2y+2xy')/(x+2y) = 4x
Multiplying by x+2y:
−2xy−4y²+2x²y'+4xyy'+2xy−2x²y' = 4x²+8xy
After cancellation of mutually opposing by sign members and dividing by 4 we get:
−y²+xyy' = x²+2xy
which easily transforms into
xyy' = (x+y)²
that corresponds to original differential equation.
This proves the correctness of the answer (A) as an equation that includes x and y(x) without derivatives that we obtained above.