# Boundary Value Problems/Review: ODEs

For more of an introduction to ODEs you are referred to Boundary Value Problems/BVP-Ordinary-Differential-Equations and for examples please look at Examples of ordinary differential equations

## First order differential equation:

Let ${\displaystyle t\in [a,b]}$ the equation

${\displaystyle {\frac {dy}{dt}}=k(t)y+f(t)}$

is a linear ordinary (single independent variable) differential equation of first order (first derivative). If ${\displaystyle f(t)=0}$ on ${\displaystyle [a,b]}$ the equation is called "homogeneous" differential equation. Otherwise it is called nonhomogeneous otherwise.

Examples of homogeneous and non-homogeneous:

${\displaystyle (Homogeneous){\frac {d}{dt}}y=10y}$ where ${\displaystyle k(t)=10}$ is a constant coefficient.

${\displaystyle (Homogeneous){\frac {d}{dt}}y=ty}$ where ${\displaystyle k(t)=t}$ is a coefficient that increases as ${\displaystyle t}$ increases.

${\displaystyle (Nonhomgeneous){\frac {d}{dt}}y=10y+sin(t)}$ where ${\displaystyle f(t)=sin(t)}$.

## Solving homogeneous ODE with constant coefficients: Separable case

Solving ${\displaystyle {\frac {d}{dt}}y=10y}$

${\displaystyle {\frac {1}{y}}{\frac {d}{dt}}y=10}$

${\displaystyle \int {\frac {1}{y}}{\frac {dy}{dt}}dt=\int 10dt}$

${\displaystyle ln(|y|)=10t+C_{1}}$

where ${\displaystyle C_{1}}$ is the integration constant.

${\displaystyle e^{ln(|y|)}=e^{10t+C_{1}}}$

${\displaystyle y(t)=e^{C_{1}}e^{10t}}$

${\displaystyle y(t)=C_{2}e^{10t}}$

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 ${\displaystyle y(0)=intialvalue}$. This condition with the above differential equation defines an Initial Value Problem. For this problem the value of ${\displaystyle C_{2}}$is determined by using an intial condition such as ${\displaystyle y(0)=5}$. Then ${\displaystyle y(0)=C_{2}e^{0}=C_{2}=5}$ and making the substitution provides the solution ${\displaystyle y(t)=5e^{10t}}$

## Student work

Solve the following IVPs:

1. Differential eq: ${\displaystyle {\frac {dx}{dt}}=5x}$ where ${\displaystyle x(t)}$ for all ${\displaystyle \infty . Initial value: ${\displaystyle x(0)=-2}$

1. Differential eq: ${\displaystyle {\frac {dx}{dt}}=5x+2}$ where ${\displaystyle x(t)}$ for all ${\displaystyle \infty . Initial value: ${\displaystyle x(3)=-2}$