Let me start this post with a simple equation.

x(x − 2) = 3x

Let us solve it in a *particular* manner.

Dividing both sides by x we get

x − 2 = 3

Adding 2 to both sides we get

x = 5

Indeed, x = 5 is a solution to this equation. When we substitute x with 5 in the equation we get 15 on both sides. However, when we solved the equation in this manner, we lost one solution, x = 0. If we substitute x with 0, the equation still holds good as we get 0 on both sides.

Why did we lose this solution?

We lost this solution in the step where we divided both sides by x. In doing so, we made the wrong assumption that x ≠ 0. If the possibility of x = 0 is allowed, then we can not divide both sides by x since division by 0 is undefined in arithmetic.

We usually get around this problem by bringing all terms involving x to one side rather than dividing both sides by x.

x(x − 2) = 3x ⇔ x(x − 2) − 3x = 0 ⇔ x(x − 2 − 3) = 0 ⇔ x = 0 or x = 5

Often fallacies or spurious proofs of contradictions is formed by sneaking such a division by 0 into the proof. Here is one I could think of.

Let a = b. So,

a + b = 2a ⇔ a − b = 2a − 2b ⇔ (a − b) = 2(a − b)

Dividing both sides by a − b we get

1 = 2

We have learnt these things in our high school. However, I forgot this important lesson while solving differential equations and lost a solution. The differential equation I was trying to solve was

y' = x^{2}y^{2}

If we divide both sides by y^{2}, the variables separate and
solving it becomes a routine job.

y^{−2}y' = x^{2}
⇔ −y^{−1}y' =
^{x3}⁄_{3} + c' … (c' is
constant of integration)
⇔ y = ^{−3}⁄_{(x3 + c)}
… (c = 3c')

This is indeed a valid solution of the differential equation. However,
we lost one solution while dividing both sides by y^{2}. This
solution is y = 0. It can be verified that with y = 0, we get 0 on both
sides of the equation.

We lost this solution in the step where we divided both sides by
y^{2}. I couldn't find a way to get around this problem. So,
whenever we need to divide both sides of the equation by y to separate
the variables, we also need to perform an additional step of verifying
whether y = 0 is a solution of the equation.

In general, when we need to divide both sides of the equation by f(y) to separate the variables, we need to perform an additional step of verifying whether f(y) = 0 is a solution of the equation.

## 3 comments

## phani said:

good one.Keep it up.

## callezee said:

nice one,very hard to write a post on maths,u did it,keep it up..

## tariq said:

thanks! it is very helpful for me.