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 "hom*ogeneous Differential Equations"
A first order Differential Equation is hom*ogeneous when it can be in this form:
dy dx = F( y x )
We can solve it using Separation of Variables but first we create a new variable v = y x
v = y x which is also y= vx
And dy dx = d (vx) dx = v dx dx + x dv dx (by the Product Rule)
Which can be simplified to dy dx = v + x dv dx
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 dy dx = x2 + y2xy
Can we get it in F( y x ) style?
Start with: x2 + y2 xy
Separate terms: x2 xy + y2 xy
Simplify: x y + y x
Reciprocal of first term:( y x )-1 + y x
Yes, we have a function of yx.
So let's go:
Start with: dy dx = ( y x )-1 + y x
y = vx and dydx = v + x dvdx:v + x dv dx = v-1 + v
Subtract v from both sides:x dv dx = v-1
Now use Separation of Variables:
Separate the variables:v dv = 1 x dx
Put the integral sign in front:∫v dv = ∫ 1 x dx
Integrate: v2 2 = ln(x) + C
Then we make C = ln(k): v2 2 = ln(x) + ln(k)
Combine ln: v2 2 = ln(kx)
Simplify:v = ±√(2 ln(kx))
Now substitute back v = y x
Substitute v = y x : y x = ±√(2 ln(kx))
Simplify:y = ±x √(2 ln(kx))
And we have the solution.
The positive portion looks like this:
Another example:
Example: Solve dy dx = y(x−y) x2
Can we get it in F( y x ) style?
Start with: y(x−y) x2
Separate terms: xy x2 − y2 x2
Simplify: y x − ( y x )2
Yes! So let's go:
Start with: dy dx = y x − ( y x )2
y = vx and dy dx = v + x dvdx v + x dv dx = v − v2
Subtract v from both sides:x dv dx = −v2
Now use Separation of Variables:
Separate the variables:− 1 v2 dv = 1 x dx
Put the integral sign in front:∫− 1 v2 dv = ∫ 1 x dx
Integrate: 1 v = ln(x) + C
Then we make C = ln(k): 1 v = ln(x) + ln(k)
Combine ln: 1 v = ln(kx)
Simplify:v = 1 ln(kx)
Now substitute back v = y x
Substitute v = y x : y x = 1 ln(kx)
Simplify:y = x ln(kx)
And we have the solution.
Here are some sample k values:
And one last example:
Example: Solve dy dx = x−y x+y
Can we get it in F( y x ) style?
Start with: x−y x+y
Divide through by x: x/x−y/x x/x+y/x
Simplify: 1−y/x 1+y/x
Yes! So let's go:
Start with: dy dx = 1−y/x 1+y/x
y = vx and dy dx = v + x dvdx v + x dv dx = 1−v 1+v
Subtract v from both sides:x dv dx = 1−v 1+v − v
Then:x dv dx = 1−v 1+v − v+v2 1+v
Simplify:x dv dx = 1−2v−v2 1+v
Now use Separation of Variables:
Separate the variables: 1+v 1−2v−v2 dv = 1 x dx
Put the integral sign in front:∫ 1+v 1−2v−v2 dv = ∫ 1 x dx
Integrate:− 1 2 ln(1−2v−v2) = ln(x) + C
Then we make C = ln(k):− 1 2 ln(1−2v−v2) = ln(x) + ln(k)
Combine ln:(1−2v−v2)-½ = kx
Square and Reciprocal:1−2v−v2 = 1 k2x2
Now substitute back v = y x
Substitute v = y x :1−2( y x )−( y x )2 = 1 k2x2
Multiply through by x2: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:y2+2xy−x2 = − 1 k2
Replace − 1 k2 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:
9419, 9420, 9421, 9422, 9423, 9424, 9425, 9426, 9427, 9428
hom*ogeneous Functions Differential Equation Calculus Index
FAQs
A hom*ogeneous differential equation can often be solved by making the substitution v(x)=yx v ( x ) = y x , where v=v(x) v = v ( x ) is a function of x.
Is a hom*ogeneous equation always correct? ›
A hom*ogenous equation can be written in the form Ax = 0, where A is an mxn matrix and 0 is the zero vector in R". Such a system Ax = 0 always has at least one nontrivial solution. Thus a hom*ogenous equation is always consistent.
How many cases are there to solve linear hom*ogeneous difference equations? ›
There are still the three main cases: real distinct roots, repeated roots and complex roots (although these can now also be repeated as we'll see). In 2nd order differential equations each differential equation could only involve one of these cases.
What is the rule for hom*ogeneous differential equations? ›
The general form of the hom*ogeneous differential equation is of the form f(x, y). dy + g(x, y). dx = 0. The hom*ogeneous differential equation has the same degree for the variables x, y within the equation.
Can a hom*ogeneous equation be inconsistent? ›
Since a hom*ogeneous system always has a solution (the trivial solution), it can never be inconsistent. Thus a hom*ogeneous system of equations always either has a unique solution or an infinite number of solutions. 1. So the trivial solution (x1,x2,x3) = (0,0,0) is the only solution.
What are the cases where an equation can be hom*ogeneous but incorrect? ›
How do we give a hom*ogenous equation but an incorrect equation at the same time? Removing constants from correct equations make them hom*ogeneous but incorrect. For example, the correct formula for kinetic energy is K.E.=(1/2) m v2. If we remove the constant 1/2 we get K.E=m v2, which is hom*ogenous but incorrect.
Are all hom*ogeneous equations physically correct? ›
Being hom*ogeneous does not necessarily mean the equation will be true, since it does not take into account numerical factors. For example, E = mv2 could be or could not be the correct formula for the energy of a particle of mass m traveling at speed v, and one cannot know if hc/λ should be divided or multiplied by 2π.
Do hom*ogeneous equations have infinite solutions? ›
Every hom*ogeneous system has either exactly one solution or infinitely many solutions. If a hom*ogeneous system has more unknowns than equations, then it has infinitely many solutions.
How do you tell if a differential equation is hom*ogeneous or inhom*ogeneous? ›
where Fi(x) F i ( x ) and G(x) are functions of x, the differential equation is said to be hom*ogeneous if G(x)=0 G ( x ) = 0 and non-hom*ogeneous otherwise.
What makes a PDE hom*ogeneous? ›
An ODE/PDE is hom*ogeneous if u = 0 is a solution of the ODE/PDE. An equation which is not hom*ogeneous is said to be inhom*ogeneous. For example, du. dt + 3u = 2 is inhom*ogeneous because u = 0 is not a valid solution.
In hom*ogeneous differential equation, the equation has zero in R.H.S while in nonhom*ogeneous differential equation, it contains functions on the right-hand side of the equation.
Can a nonlinear differential equation be hom*ogeneous? ›
Answer and Explanation:
Yes, the concept of linearity and hom*ogeneity are two concepts that do not exclude each other. For example, the following equation y ″ + y 2 = 0 is hom*ogeneous and it is not linear.
Do hom*ogeneous differential equations have to be linear? ›
No, the word is only used in the context of linear equations. It could be said hom*ogeneous if it is expressed in the form:p(y′,y)=0 where p is a hom*ogeneous polynomial of two variables. hom*ogeneity is a notion related to polynomials of several variables.
How to prove a hom*ogeneous differential equation? ›
A first‐order differential equation is said to be hom*ogeneous if M( x,y) and N( x,y) are both hom*ogeneous functions of the same degree. is hom*ogeneous because both M( x,y) = x 2 – y 2 and N( x,y) = xy are hom*ogeneous functions of the same degree (namely, 2).
Can a hom*ogeneous differential equation have a constant? ›
Definition. A linear hom*ogeneous second order ODE with constant coefficients is an ordinary differential equation in the form: ad2ydx2+bdydx+cy=0, where a , b and c are constants.
Can a differential equation be hom*ogeneous and exact? ›
Consider the equation (5y−2x)(dydx)−2y=0 This equation is Exact and Homgeneous differential equation.
Can a non-linear differential equation be hom*ogeneous? ›
Answer and Explanation:
Yes, the concept of linearity and hom*ogeneity are two concepts that do not exclude each other. For example, the following equation y ″ + y 2 = 0 is hom*ogeneous and it is not linear.
