Ordinary Differential Equations: Classification of ODEs
Classification of ODEs
Order
The order of an ODE is the order of the highest derivative appearing in the equation. For example, the following equation (Newton’s equation) is a second-order ODE:
      ![Rendered by QuickLaTeX.com \[m \frac{\mathrm{d}^2x}{\mathrm{d}t^2}=mg-kx\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-75138af5866e45c949aeb74da1d04c5e_l3.png)
while the beam equation is a fourth-order ODE:
      ![Rendered by QuickLaTeX.com \[EI\frac{\mathrm{d}^4y}{\mathrm{d}x^4}=q\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-e021488c0e90cbc02242eb278bebc111_l3.png)
Linear vs. Nonlinear
An ODE is linear if it can be written as the linear combination:
      ![Rendered by QuickLaTeX.com \[\frac{\mathrm{d}^ny}{\mathrm{d}x^n}+\sum_{i=0}^{n-1} a_i(x)\frac{\mathrm{d}^iy}{\mathrm{d}x^i}=f(x)\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-ad1fb46254c21f94953fb97f230698e6_l3.png)
where  and
 and  are functions of the independent variable
 are functions of the independent variable  . All the examples above are considered linear ODEs.
. All the examples above are considered linear ODEs.
The following are two examples of nonlinear ODEs with  being the dependent variable and
 being the dependent variable and  being the independent variable:
 being the independent variable:
      ![Rendered by QuickLaTeX.com \[\begin{split}\frac{\mathrm{d}x}{\mathrm{d}t}&=\frac{k}{x}\\\frac{\mathrm{d}^2x}{\mathrm{d}t^2}&=\sin{x}\end{split}\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-8d836b640042971e74dcec61d0a1a470_l3.png)
Homogeneous vs. Nonhomogeneous
A homogeneous ODE is an equation whose every term contains either the dependent variable or one of its derivatives. For example, Newton’s second law applied to a spring without the gravitational term is a linear homogeneous ODE:
      ![Rendered by QuickLaTeX.com \[m \frac{\mathrm{d}^2x}{\mathrm{d}t^2}=-kx\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-474911722e5cc65a4df0da8e2bea0f23_l3.png)
If one of the non-zero terms in the ODE contains neither the dependent variable nor any of its derivatives, the equation is nonhomogeneous. As an example, if the gravitational term is added to Newton’s second law, then, the equation becomes nonhomogeneous:
      ![Rendered by QuickLaTeX.com \[m \frac{\mathrm{d}^2x}{\mathrm{d}t^2}=mg-kx\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-75138af5866e45c949aeb74da1d04c5e_l3.png)
The free term (the term devoid of the dependent variables and any of its derivatives) is called the source term.
A linear homogeneous ODE has the following form:
      ![Rendered by QuickLaTeX.com \[\frac{\mathrm{d}^ny}{\mathrm{d}x^n}+\sum_{i=0}^{n-1} a_i(x)\frac{\mathrm{d}^iy}{\mathrm{d}x^i}=0\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-2572b269b15db53edebac2c15322d416_l3.png)
while a linear nonhomogeneous ODE has the form:
      ![Rendered by QuickLaTeX.com \[\frac{\mathrm{d}^ny}{\mathrm{d}x^n}+\sum_{i=0}^{n-1} a_i(x)\frac{\mathrm{d}^iy}{\mathrm{d}x^i}=f(x)\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-ad1fb46254c21f94953fb97f230698e6_l3.png)
The term  is called the source term. The solution to a linear homogeneous equation is called the complementary solution
 is called the source term. The solution to a linear homogeneous equation is called the complementary solution  while the solution when a source term appears in the equation is the sum of the complementary solution
 while the solution when a source term appears in the equation is the sum of the complementary solution  and the particular solution
 and the particular solution  (which is particular to the source term
 (which is particular to the source term  ). For example, the solution to Newton’s second law without the gravitational term has the form:
). For example, the solution to Newton’s second law without the gravitational term has the form:
      ![Rendered by QuickLaTeX.com \[x=C_1\cos\left(\frac{\sqrt{k}t}{\sqrt{m}}\right)+C_2\sin\left(\frac{\sqrt{k}t}{\sqrt{m}}\right)\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-81a05da9e6f634024df8c302161c6e76_l3.png)
while the solution when the gravitational term appears has the form:
      ![Rendered by QuickLaTeX.com \[x=\frac{mg}{k}+C_1\cos\left(\frac{\sqrt{k}t}{\sqrt{m}}\right)+C_2\sin\left(\frac{\sqrt{k}t}{\sqrt{m}}\right)\]](https://engcourses-uofa.ca/wp-content/ql-cache/quicklatex.com-286487137285e2f649a81de0cd3180a4_l3.png)
The particular solution is the term
 .
.
IVP vs. BVP
Depending on the boundary conditions, an ODE can be classified as either an Initial Value Problem (IVP) or a Boundary Value Problem (BVP). An initial value problem is an ODE given with initial conditions of the dependent variable and its derivative at a particular value of the independent variable. This usually applies to dynamic systems whose independent variable is time. For example, Newton’s second law of motion is an initial value problem because the initial value at  of the displacement
 of the displacement  and the velocity
 and the velocity  are required to reach a solution. The initial values lead to a particular path that is a function of time
 are required to reach a solution. The initial values lead to a particular path that is a function of time  .
.
A boundary value problem is an ODE given with boundary conditions at different points. This usually applies to static systems whose independent variable is position. For example, the Euler-Bernoulli beam deflection equation is a boundary value problem. The boundary conditions of the displacement  , the rotation
, the rotation  , and the third and fourth derivatives, are usually given at boundary points on the beam which will then dictate the equilibrium deflection of the beam as a function of position
, and the third and fourth derivatives, are usually given at boundary points on the beam which will then dictate the equilibrium deflection of the beam as a function of position  .
.
