Question

What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)^2 + 1 = 0?

(a) Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zero

(b) Because for any real function, the left-hand side of the equation becomes zero

(c) Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zero

(d) Because for any real function, the left-hand side of the equation becomes infinity

The question was asked in exam.

This is a very interesting question from Solution of Second Order P.D.E. topic in division Partial Differential Equations of Engineering Mathematics

Answer 1

Correct answer is (c) Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zero

Easiest explanation: Given: (y’)^2 + 1 = 0

Consider if y = 2x, then y’ = 2 and hence the left-hand side of the equation becomes 3 which is greater than 1. Therefore, the left-hand side of the equation will always be greater than, or equal to one and thus cannot be zero and hence the differential equation is not satisfied.