Difference between revisions of "Homogeneous Differential Equations"

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:<math> 2 x^2 \frac{d^2y}{dx^2} - 3 x \frac{dy}{dx} + y = 2 \,. </math>
 
:<math> 2 x^2 \frac{d^2y}{dx^2} - 3 x \frac{dy}{dx} + y = 2 \,. </math>
 
The existence of a constant term is a sufficient condition for an equation to be inhomogeneous, as in the above example.
 
The existence of a constant term is a sufficient condition for an equation to be inhomogeneous, as in the above example.
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===Characteristic equation===
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If the equation is linear homogeneous and further <math>p(t),q(t)</math> are constant, then the equation is referred to as a '''constant-coefficients''' equation:
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<math display="block">ay''+by'+cy=0</math>
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and we can apply the method of characteristic equations to solve such an equation. Note that <math>a</math> is assumed to be non-zero since we are working with a second order equation.
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===Method formal steps===
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# We assume that the solution is of the form <math>y(t)=e^{rt}</math> (this is called making an ansatz). This gives <math display="block">(ar^{2}+br+c)e^{rt}=0\Longrightarrow\,ar^{2}+br+c=0,</math> which equation is called the '''characteristic equation'''.
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# So to solve the above ODE, it suffices to find the two roots <math>r_{1},r_{2}</math>.
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#Then the general solution is of the form:<math display="block"> y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}. </math>
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===Example-presenting the method===
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Consider a mass <math>m</math> hanging at rest on the end of a vertical spring of length <math>l</math>, spring constant <math>k</math> and damping constant <math>\gamma</math>.
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Let <math>u\left(t\right)</math> denote the displacement, in units of feet, from the equilibrium position. Note that since <math>u(t)</math> represents the amount of displacement from the spring's equilibrium position (the position obtained when the downward force of gravity is matched by the will of the spring to not allow the mass to stretch the spring further) then <math>u(t)</math> should increase downward. Then by Newton's Third Law one can obtain the equation
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<math display="block">mu'' (t )+\gamma u' (t )+ku (t )=F (t),</math>
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where <math>F\left (t\right )</math> is any external force, which for simplicity we will assume to be zero.
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#First we obtain the characteristic equation:<math display="block">mr^{2}+\gamma r+k=0.</math>
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#Suppose that <math>m=1\text{lb},\gamma=5 \text{lb}/\text{ft}/\text{s}</math> and <math>k=6 \text{lb}/\text{ft}</math> then we obtain the roots <math>r_{1}=-2</math>, <math>r_{2}=-3</math>.
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#Therefore, the general solution will be<math display="block">u (t )=c_{1}e^{-2t}+c_{2}e^{-3t}.</math>
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#Further if <math>u(0)=0,\,u'(0)=1</math> we obtain <math>c_{1}=1,c_{2}=-1</math>:<math display="block">u (t )=e^{-2t}-e^{-3t}.</math>
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====Examples====
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*Consider the IVP
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<math display="block">4y''-y=0,\,\,y (-2 )=1,\,\,y' (-2 )=-1.</math>
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#We obtain the characteristic equation <math>4r^{2}-1=0\Rightarrow r=\pm \frac{1}{2}</math> and so the general solution will be<math display="block">y (t )=c_{1}e^{\frac{t}{2}}+c_{2}e^{-\frac{t}{2}}.  </math>
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#Using the initial conditions we obtain:<math display="block">1=c_{1}e^{-1}+c_{2}e\,\text{ and }\,-1=\frac{1}{2} (c_{1}e^{-1}-c_{2}e ).</math>
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#Solving these two equations gives: <math>c_{1}=\frac{-1}{2}e,c_{2}=\frac{3}{2}e^{-1}</math> and so the solution for our IVP is:<math display="block"> y\left (t\right )=-\frac{1}{2}e^{1+\frac{t}{2}}+\frac{3}{2}e^{-\frac{t}{2}-1}.  </math>
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#Therefore, as <math>t\to +\infty</math> we obtain <math>y\to -\infty</math>.
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*Consider the IVP
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<math display="block"> y''+5y'+6y=0,\,\,y\left (0\right )=2,\,\,y'\left (0\right )=\beta </math>
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#The characteristic equation is <math>r^{2}+5r+6=0\Rightarrow r=-2,-3</math> and so the general solution will be:<math display="block">y (t )=c_{1}e^{-2t}+c_{2}e^{-3t}</math>
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#Using the initial conditions we obtain:<math display="block">2=c_{1}+c_{2}\,\text{ and } \,\beta=-2c_{1}-3c_{2}.</math>
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#Solving these two equations gives: <math>c_{1}= (6+\beta),\,c_{2}=-\left (4+\beta\right )</math> and so the solution for our IVP is:<math display="block"> y (t )= (6+\beta )e^{-2t}- (4+\beta )e^{-3t}.  </math>
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#Therefore, as <math>t\to +\infty</math> we obtain <math>y\to 0</math>.
  
 
== Licensing ==  
 
== Licensing ==  
 
Content obtained and/or adapted from:
 
Content obtained and/or adapted from:
 
* [https://en.wikipedia.org/wiki/Homogeneous_differential_equation Homogeneous differential equation, Wikipedia] under a CC BY-SA license
 
* [https://en.wikipedia.org/wiki/Homogeneous_differential_equation Homogeneous differential equation, Wikipedia] under a CC BY-SA license

Revision as of 09:13, 22 October 2021

A differential equation can be homogeneous in either of two respects.

A first order differential equation is said to be homogeneous if it may be written

where f and g are homogeneous functions of the same degree of x and y.[1] In this case, the change of variable y = ux leads to an equation of the form

which is easy to solve by integration of the two members.

Otherwise, a differential equation is homogeneous if it is a homogeneous function of the unknown function and its derivatives. In the case of linear differential equations, this means that there are no constant terms. The solutions of any linear ordinary differential equation of any order may be deduced by integration from the solution of the homogeneous equation obtained by removing the constant term.

Homogeneous first-order differential equations

Template:Differential equations

A first-order ordinary differential equation in the form:

is a homogeneous type if both functions M(x, y) and N(x, y) are homogeneous functions of the same degree n.[2] That is, multiplying each variable by a parameter λ, we find

Thus,

Solution method

In the quotient , we can let t = Template:Sfrac to simplify this quotient to a function f of the single variable Template:Sfrac:

That is

Introduce the change of variables y = ux; differentiate using the product rule:

This transforms the original differential equation into the separable form

or

which can now be integrated directly: ln x equals the antiderivative of the right-hand side (see ordinary differential equation).

Special case

A first order differential equation of the form (a, b, c, e, f, g are all constants)

where afbe can be transformed into a homogeneous type by a linear transformation of both variables (α and β are constants):

Homogeneous linear differential equations

Template:See also A linear differential equation is homogeneous if it is a homogeneous linear equation in the unknown function and its derivatives. It follows that, if φ(x) is a solution, so is (x), for any (non-zero) constant c. In order for this condition to hold, each nonzero term of the linear differential equation must depend on the unknown function or any derivative of it. A linear differential equation that fails this condition is called inhomogeneous.

A linear differential equation can be represented as a linear operator acting on y(x) where x is usually the independent variable and y is the dependent variable. Therefore, the general form of a linear homogeneous differential equation is

where L is a differential operator, a sum of derivatives (defining the "0th derivative" as the original, non-differentiated function), each multiplied by a function fi of x:

where fi may be constants, but not all fi may be zero.

For example, the following linear differential equation is homogeneous:

whereas the following two are inhomogeneous:

The existence of a constant term is a sufficient condition for an equation to be inhomogeneous, as in the above example.

Characteristic equation

If the equation is linear homogeneous and further are constant, then the equation is referred to as a constant-coefficients equation:

and we can apply the method of characteristic equations to solve such an equation. Note that is assumed to be non-zero since we are working with a second order equation.

Method formal steps

  1. We assume that the solution is of the form (this is called making an ansatz). This gives
    which equation is called the characteristic equation.
  2. So to solve the above ODE, it suffices to find the two roots .
  3. Then the general solution is of the form:

Example-presenting the method

Consider a mass hanging at rest on the end of a vertical spring of length , spring constant and damping constant . Let denote the displacement, in units of feet, from the equilibrium position. Note that since represents the amount of displacement from the spring's equilibrium position (the position obtained when the downward force of gravity is matched by the will of the spring to not allow the mass to stretch the spring further) then should increase downward. Then by Newton's Third Law one can obtain the equation

where is any external force, which for simplicity we will assume to be zero.

  1. First we obtain the characteristic equation:
  2. Suppose that and then we obtain the roots , .
  3. Therefore, the general solution will be
  4. Further if we obtain :


Examples

  • Consider the IVP

  1. We obtain the characteristic equation and so the general solution will be
  2. Using the initial conditions we obtain:
  3. Solving these two equations gives: and so the solution for our IVP is:
  4. Therefore, as we obtain .
  • Consider the IVP

  1. The characteristic equation is and so the general solution will be:
  2. Using the initial conditions we obtain:
  3. Solving these two equations gives: and so the solution for our IVP is:
  4. Therefore, as we obtain .

Licensing

Content obtained and/or adapted from:

  • Template:Cite book
  • Template:Harvnb