View source for Boundary Value Problems/Review: ODEs

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For more of an introduction to ODEs you are referred to [[../BVP-Ordinary-Differential-Equations]] and for examples please look at [[w:Examples_of_differential_equations|Examples of ordinary differential equations]]

== First order differential equation: ==
Let <math> t \in [a,b] </math> the equation
 
<p style="text-align:center;"><math>\frac{dy}{dt}=k(t)y+f(t) </math></p>

is a linear ordinary (single independent variable) differential equation of first order (first derivative). If <math> f(t)=0 </math> on <math> [a,b] </math> the equation is called "homogeneous" differential equation. Otherwise it is called nonhomogeneous otherwise.

Examples of homogeneous and non-homogeneous:

<math>  (Homogeneous) \frac{d}{dt} y = 10 y </math> where <math> k(t)=10 </math> is a constant coefficient.

<math> (Homogeneous) \frac{d}{dt} y = t y </math>  where <math> k(t)= t </math> is a coefficient that increases as <math> t 
</math> increases.

<math> (Nonhomgeneous) \frac{d}{dt} y = 10 y + sin(t) </math>  where <math>f(t)=sin(t)</math>.

== Solving homogeneous ODE with constant coefficients: Separable case ==

Solving <math>  \frac{d}{dt} y = 10 y </math>

<p style="text-align:center;"><math>  \frac{1}{y} \frac{d}{dt} y = 10  </math></p>

<p style="text-align:center;"><math>  \int \frac{1}{y} \frac{dy}{dt} dt= \int 10 dt </math>
</p>
<p style="text-align:center;"><math>  ln(|y|) = 10t + C_1 </math> </p>where <math>C_1</math> is the integration constant.

<math>  e^{ln(|y|)} = e^{10t + C_1} </math>

<math>  y(t) = e^{C_1}e^{10t} </math>

<math>  y(t) = C_2e^{10t} </math>

After integration there is a constant, an unknown. This will happen each time you integrate. To determine this unknown for a particular application you will need another piece of information. Typically it is another equation that requires <math> y(0)= intial value </math>. This condition with the above differential equation defines an '''Initial Value Problem'''. For this problem the value of <math> C_2 </math >is determined by using an intial condition such as <math> y(0)= 5 </math>. Then <math> y(0)=C_2 e^{0}=C_2 = 5 </math> and making the substitution provides the solution <math> y(t) = 5 e^{10t} </math> 
 
== Student work ==

Solve the following IVPs: 

#Differential eq:  <math>  \frac{dx}{dt} = 5x </math> where <math>x(t)</math> for all <math> \infty < t <\infty </math>. Initial value: <math> x(0)=-2 </math>

Click here for solution: [[../BVP-solution-1]]


#Differential eq:  <math>  \frac{dx}{dt} = 5x + 2 </math> where <math>x(t)</math> for all <math> \infty < t <\infty </math>. Initial value: <math> x(3)=-2 </math>

Click here for solution: [[../BVP-solution-2]]




===Pre-class work: ===
Complete each of the following:
* Read the following:
** Pages 1-10 of Powers
** Lecture notes (web page)
** (View) Associated Lecture Video
* Work the following problems [[../BVP-Exercise 1-1:Problems and Solutions]]

== Second order Differential Equations ==

== Cauchy Euler ==

== Bessel Equation ==

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