Patterns – Powerful tool to solve complex math problems

All of you must be familiar with word patterns, its there in all walks of life. In each period, fashion has some pattern related to dresses, hairstyle, Eyeglasses etc. Even we have a design patterns in architecture, automobile etc. So the point is same concept is applicable for for solving math problems too. There is a fixed and repetitive pattern even solving the maths problems too, its even applicable for India’s toughest exam IIT JEE. Let me prove my point by following set of examples.

I am giving below the set of questions asked in IIT previous year exams, i am focusing on differential equations as an example. When you go through it then you will realize that there is common way aka substitution or transformation trick to solve these problems.

Example 1 (IIT 2013): A curve passes through the point (1, pi /6). Let the slope of the curve at each point (x, y) be y/x+ sec(y/x) . Then the equation of the curve is
(A) sin y/x = log x +1/2  (B) cosec y/x = log x + 1/2 (C) sin 2y/x = log x + 1/2  (D) cos 2y/x = log x + 1/2

Solution: There are two concepts involved to solve above problem:

Concept 1 – slope of a curve is nothing but m = dy/dx hence

dy/dx = y/x+ sec(y/x)  – its given in the question

Concept 2 – above is posed as a differential equation (DE) to the candidate but solving it is not easy, why? The term sec(y/x) is an issue, we can not separate it out to solve the DE with known procedure. So the trick here use of the technique called substitution, just put y = px in both sides then DE will transform into following shape:

d(px)/dx = p+x dp/dx = p + sec(p)

x dp/dx = sec p

dp/sec (p) = dx/x

Integrating the both side will yield following solution:

sin(p) = log(x)+c   where c is a constant

Or sin (y/x) = log(x)+c   – reverse substitution of p = y/x

So the answer is A, we can even find the value of c by putting (1, pi /6) in above equation. Isn’t it an easy question to solve?

Example 2 (IIT 1996): Determine the equation of the curve passing through the
origin in the from y= f (x), which satisfies the differential equation
dy/dx = sin(10x+6y)

Solution: Again here the problem is same, its not possible to seperate the 10x+6y from Sin() so i guess by now you would have already seen the substitution pattern here. Lets substitute here 10x+6y = p in LHS & RHS, it gives

LHS = 10 + 6 dy/dx = dp/dx  or on rearranging dy/dx = 1/6[dp/dx -10]

RHS – sin (p)

=> 1/6[dp/dx -10]  = sin(p)

=> dp/dx = 6 sin(p) + 10

=> dp/[ 6 sin(p) + 10]  = dx

Now integrating both sides will solve for P and X and reverse substitution of p = 10x+6y will give the final answer.

Example 3 (IIT 2004): A curve C passing through (2,0) and the slope of tangent at point P(x,y) equals [(x+1)^2+(y-3)]/(x+1). Find the equation of the curve.

Solution: This question is similar to example 1 and a repeated pattern. As per explanation of example 1:

dy/dx =  [(x+1)^2+(y-3)]/(x+1)

Now this question turned from tangent to simple DE to solve. As evident from RHS side that seperating x and y and performing integration is a complex task. Lets use the power of substitution again to solve above DE.

Lets assume x+1 = p and y-3=q  this gives

LHS -dp/dq =  dy/dx  and RHS is p+q/p

dq/dp = p + q/p  rearranging it to below

dq/dp – (1/p)q = p

Above is easiest way to solve the DE using IF it gives

q/p = p + C     where C is a constant

now reverse substitution gives

(y-3)/(x-1) = x-1 +C   or

y-3 = (x-1)^2 + C (x-1)

Now its mentioned in the question that this curve is passing through the (2,0) hence

-3 = 1 + C so it gives C = -4

So the complete equation of the curve is  y-3 = (x-1)^2  -4 (x-1)

So by now you would have understood the power of patterns in solving the questions easily even from the India’s toughest exam IIT. Same patterns could be identified easily in all the topics and solution would be on your fingertip.