Differentiation Particular Solutions
This chapter explores particular solutions to differential equations. The chapter covers finding the particular solution to a first order differential equations by suing given conditions. Before attempting this chapter you must have prior knowledge of solving first order differential equations by separating the variables.
Introduction
In previous lessions we managed to find general solutions to the differential equations. We simply left tge constant of integration in the solution. This provided an infinite number of differential solutions for example.
Suppose we had to find the general solution of the differential equation
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…and find the particular solution for which y=1/2 when x=0.
First we separate the variables and integrate;
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The above is the general solution, usually we rearrange to make y the subject but we already know y to be 1/2 when x=0. Let us find C. We substitute y and x to get;
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We can conclude that the solution is;
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It is best to make y the subject. This would give the final particular solution as;
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By finding the particular solution we have managed to find the constant at the end. Generally we are provided actual values which you can substitute into the general solution to find the constant at the end. Below is an example;
Example
Suppose we had to find the particular solution of;
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…given that y=1 when x=3
First we write;
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Now we have;
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…we separate the variables…
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…we integrate…
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Now we can substitute in the known values in the equation, we know that y=1 when x=3 so;
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So the particular solutoin becomes;
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…we can simplify further…
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Example
Here is another example; Finding the particular solution of;
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…given that y=4 when x=1
First we separate the variables to get;
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Now we can integrate to get;
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Then we substitute in the known values to find the value of C.
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The solution becomes;
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…or we can simplify further to get;
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Real life example
The following is a real life example;
A colony of mould spores on a piece of breadm initially of size 400, increases at a rate proportional to it’s size.
- White this as a differential equation with N as the number of spores at time t hours
- If the size of the colony increases to 2000 in 10 hours find N in terms of t
- How many spores are there after 20 hours?
- How long does it take for the colony to increase to a million spores?
The rate of growth is given by;
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…this is proportial to N, we can write this as an equation;
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For the second equation that asks to find N in terms of t. First we separate the variables;
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because N>0
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…or…
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At the start t=0 and N = 400;
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…we have;
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…we know that after 10 hours N=2000;
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The solution becomes;
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The third question asks how many spores there are after 20 hours. We use the above formula;
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There are 10000 spores after 20 hours.
For the fourth question to find the time we replace N with 1000000, so we have;
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We take logs on both sides;
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After 48 hours and 37 minutes there will be 1000000 mould spores.
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At first, we sample $f(x)$ in the $N$ ($N$ is odd) equidistant points around $x^*$:
\[
f_k = f(x_k),\: x_k = x^*+kh,\: k=-\frac{N-1}{2},\dots,\frac{N-1}{2}
\]
where $h$ is some step.
Then we interpolate points $\{(x_k,f_k)\}$ by polynomial
\begin{equation} \label{eq:poly}
P_{N-1}(x)=\sum_{j=0}^{N-1}{a_jx^j}
\end{equation}
Its coefficients $\{a_j\}$ are found as a solution of system of linear equations:
\begin{equation} \label{eq:sys}
\left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2}
\end{equation}
Here are references to existing equations: (\ref{eq:poly}), (\ref{eq:sys}).
Here is reference to non-existing equation (\ref{eq:unknown}).