### Differential equations: Introduction to Differential equations

### Solutions of differential equations

The exponential function \[y(t)=\e^t\] is equal to its own derivative and so satisfies the differential equation \(y'=y\). In other words, it is a **solution** of the ODE. The equation \(y'=y\) has other solutions, for example: \[y(t)= 2\e^t,\quad y(t)= -\e^t,\quad y(t)= -\tfrac{1}{3}\e^t \] These solutions are all of the form #y(t)=C\cdot \e^t# for a certain constant \(C\). Every solution is of this form:

Solution of the exponential growth equation

Let #\lambda# be a real number. The general solution of the differential equation \[\frac{\dd y}{\dd t}=\lambda\cdot y\] is \[y(t)=C\cdot \e^{\lambda\cdot t}\] where \(C\) is a constant.

The special case where #\lambda=0# is already known: the only functions having derivatives #0#, are the constant functions.

In general, a differential equation can have multiple solutions, but by using constants, often called **integration constants**, we can sometimes describe the **general solution** of a differential equation by means of a function rule containing integration constants.

Initial Value Problem

The existence of global solutions shows that an ODE may have a whole host of solutions. To pinpoint a given solution additional conditions are required. When these conditions are all related to the value of the solution #y# or a derivative thereof for a given value of the independent variable (think of the state #y(0)# at time \(0\) or the speed #y'(0)# at time \(0\)), then we speak of an **initial value problem.**

If there is a unique solution of the ODE satisfying the **initial condition** (by this we mean the condition(s) of the initial value problem), we call it the **specific solution** of the initial value problem (or associated with the initial condition).

If all the additional requirements are related to the boundary of an interval on which the function is defined (think of #y(2)=7# and #y(9)=13# for the function #y# on the interval #\ivcc{2}{9}#), then we also speak of a **boundary value problem.** We also talk about the **specific solution** of the boundary value problem.

The general solution of this differential equation of exponential growth equals \[y= C\cdot \e^t\] We use the condition \(y(2)=4\) by substituting #t=2# and #y=4# into it. This gives the equation \(4=C\cdot \e^{2} \), so #C = 4\cdot \e^ {- 2 }#. Substituting this value of #C# into the general solution, we find the special solution of the initial value problem:

\[ y = 4\cdot \e^ {- 2 } \cdot \e ^ t =4\cdot \e^{t-2} \]

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