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Simplicity is the ultimate sophistication.” — Leonardo da Vinci
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Sunday, 2 August 2015

Fundamentals of calculus chapter 4: More on differential equations

Fundamentals of calculus chapter 4: More on differential equations

Differential equations do not have immediate means to solve them. The reason is, because differential equations are defined from differentiation and so is its integral, it’s easier to find the differential equation rather than it’s solution. If our motive is purely to find the solution of differential equations, then our best approach will be to try and find how the expressions “behave” while trying to differentiate them. Then we may back trace this sequence of steps to find the original expression.

Many textbooks that talk differential equation’s solutions fail to explain the exact way to obtain the solutions. Most times, solutions written down in academic sources fail to explain exactly how the unbelievable solution to a differential equation was brought about. Generally textbooks will come up with an “unbelievable” or “brilliant” substitution to a problem and hence obtain the final solution and when this subject is thought in class (either schools or universities) the mathematics teacher would first give the students some “time” to try and solve the problem themselves and eventually when the entire class fails to come up with a solution the mathematics teacher would “do his little magic substitution” and show the entire class that the problem can actually be solved. But this is not the case in reality. The solution was actually discovered by a simpler method. And that’s probably the only universal way to solve differential equations, unless they are not too complicated enough.

I am not going to give you an easier method to solve the problems, because it’s something I can’t do too. But I wish to throw light upon how these magical substitutions ended up in your textbook. Could it be that someone really thought about that amazing substitution to solve the differential equation? That’s possible. Is there a better way to understand how it’s all done or how the substitution works? Yes, the only way to do it is to find its original source.

There is no standard way to find the solution to a differential equation but there is a standard algorithm to find the differential of an equation. Let’s take advantage of this. Let there be an equation of form, $ax^2+bx+c = z$. Let $c$ be an unknown constant and $a, b$ be known constants. Then the differential equation is $\frac{dz}{dx} = 2ax + b$. Now let $b$ be the unknown constant and $a, c$ be the known constants. Then we have,
$ax^2 + (\frac{dz}{dx}-2ax)x+c = z …(1)$.    

Here it’s obvious that $b = \frac{dz}{dx}-2ax $. Let’s try to solve this part. $\frac{dz}{dx} = b + 2ax$ and integrating w.r.t $x$ we get
$z = bx + ax^2 + c…(2)$.
 Comparing the coefficients of (1) and (2) we get $b = (\frac{dz}{dx}-2ax)$. We got back our result. But is this method universal? Let’s test is out. We have $\frac{dz}{dx} = 2ax + b$ with $a, b$ known constants and $c$ as unknown constant, now if we follow the same algorithm, we get $z = ax^2 + bx +c$. Now let’s compare the coefficients. We would be getting, $b = 2a$, $c = b$ and $a = 0$. We cannot consider this result because the term with the highest degree has a coefficient of zero and this reduces the equation’s degree. We cannot set a finite value to the variables $a, b, c$. Though $a$ and $b$ are known, we defined it to be arbitrary. We must have our algorithm fix values for neither of these variables. Now what do we do to make our algorithm hold good for both these cases?

One possibility will be to assign another rule. The rule should prevent us from comparing the coefficients for the second case. So with a little thought about this we can arrive at this: We must stop proceeding once we have all the coefficients we need. While solving (1) we considered the value of $\frac{dz}{dx} – 2ax$ to be a constant. But assume we didn’t obtain (1) from the expression $z = ax^2 + bx + c$ and we got to figure this out ourselves. So could $\frac{dz}{dx} -2ax$ be a function of $x$?  If it was, then our rule (the one we just created) should be confined even further to a smaller subset of problems.  

We need to eliminate the constant to frame the differential equation. To eliminate one constant we need two equations. But to eliminate two constants we need two equations. The second order differential is required to obtain the second equation. Next we solve them algebraically eliminating the unknown constant. But we must be careful not to eliminate the known constants, if any. Because, if we do eliminate the known constant, then it is certain that the differential equation we get becomes independent of them. It’s not right to treat a known constant as an unknown constant, as the information it represents is lost. Instead becomes ambiguous and satisfies for all possible values in place of that constant.

Framing differential equation’s solution
Let’s summarise the rules for framing a differential equation’s solution. We first need to write down all the possible variations involving certain functions, constants and unknown constants. Next we try to frame the general differential equation of each of these cases. We then study the nature of how we obtained our differential equation form the original equation and back-trace to find the solution, which is the original equation.  We study this nature with many other cases and frame a simplified rule for finding the solution.  





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