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Sunday, 19 July 2015

Fundamentals of calculus Chapter 3: differential equations of functions involving one independent variable.


Fundamentals of calculus
Chapter 3: differential equations of functions involving one independent variable.


“Everything big is built out of smaller units; and if the smaller units fail to cooperate the bigger ones ultimately fall”

“A good learner never accepts the educative information he comes across unless he knows for sure, that there is no practical and logical way to contradict it”


Calculus as a whole evolved to support something new, a differential equation. One marvellous fact about differential equations are they don’t have any constants to describe the graphical structure they represent. In the previous article ‘chapter 2’, we saw how differentiation eliminates a constant. This fact can be taken advantage of to frame differential equations that elaborate the actual rate of change of different quantities. For example if, $z = ax^2 + bx +c$ then $\frac{dz}{dx} = 2ax + b$ and $\frac{d^2z}{dx^2} = 2a$ and $\frac{d^3z}{dx^3} = 0$.



Seeing the above expressions, the last one that has its value equated to zero best describes the differential equation; it does not have any constants involved. So what does this show? An example of a normal function would be $z = x^2 + 2x + 3$; this represents a particular behaviour of the equation. Now a family of equations would be $z = ax^2 + bx + c$. This does not have a specified shape and orientation but expresses a family of such equations. Unless the constants are specified, the rate of change of the equation, i.e., the differential, remains ambiguous; or we may say, depends on the terms $a$ and $b$. The value $c$ specifies the orientation of the function in the graph w.r.t (with respect to) the X- axis.


Let’s say that we have an equation $z = x^2 + 2x +c$. It is again a second degree equation, but has only one constant. We may eliminate that constant by writing $\frac {dz} {dx} = 2x + 2$. So now this serves as our differential equation. Now let’s write $z = ax^2 + 2x + 3$, differentiating $z$ w.r.t $x$ twice we get, $\frac {d^2z}{dx^2} = 2a$ and we may write $a = \frac{d^2z}{2dx^2} = a$. Substituting this in our original expression for $z$ we get, $z = \frac{d^2z}{2dx^2}x^2+2x + 3$ as our differential equation. So let’s see what does this mean by words. It means, $z$ is a function of $x$ for which the rate of change of rate of change of $z$ w.r.t is unknown, but the rate of change and the orientation of the function w.r.t the base axis is known.  


The benefit of this kind of mathematical expression will not seem clear unless we give a real example for this. Let an object move along a single axis with varying velocity, depending upon many unknown factors. But by observing it for long enough, let’s plot the change in position w.r.t the time taken. Time is taken in the X axis and the positional displacement w.r.t a specific point is denoted by the Z axis. So we may write $z = f(t)$. Here $z$ being some function of $t$ shows the position of the moving point at each point in time. If the mass of the object is $m$ then the force it provides along the Z axis for each unit time can be represented by the differential equation $F = m\frac {d^2z}{dt^2} $.


Let’s consider another differential equation $z = \frac{1}{6} \times \frac {d^2z}{dx^2}x^2$. What could this function be? Putting $z = ax^3$, we get $z = 6ax^3\times \frac{1}{6}$. Now let’s get this more challenging. $z = x^n$ now try finding the differential equation of this function, if $n$ is an unknown constant. Differentiating this for $n$ times w.r.t $x$ gives a differential equation. But this does not eliminate the constant $n$ because, we would be writing $\frac{d^nz}{dx^n} = 0$ and this still has the constant $n$.


 Example 1:
Consider the expression $z = x^n$ where $z$ is a function of $x$ with $n$ being the unknown constant. Express this function as a differential equation.

Solution:
In the examples elaborated in the above explanation, the unknown constant could be easily eliminated by differentiating and obtaining different functions which can be used to eliminate the constant value. Applying this here we get $\frac{dz}{dx} = nx^{n-1}$ which is $n\frac{z}{x} = \frac{dz}{dx} $ and can be further written as, $z = x^{\frac{x}{z}\times \frac{dz}{dx}}$.

Alternate:
A simpler solution will be to take log on both sides (note unless stated log to the base $e$ is taken).

$log z = n log x$
$\frac{dz}{dx}\times \frac{1}{z} = \frac{n}{x}$
$log z = x\frac{dz}{dx}\frac{1}{z}log x$
We have obviously eliminated $n$. In this case the solution involves logarithmic functions.       




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