Question

The differential form of the equation (y - a)² = 2bx is:

1. 2x\(\rm d^2y\over dx^2\) + \(\rm {dy\over dx}\)= 0
2. 2\(\rm d^2y\over dx^2\) + x\(\rm {dy\over dx}\)= 0
3. \(\rm d^2y\over dx^2\) + 2x \(\rm {dy\over dx}\)= 0
4. x\(\rm d^2y\over dx^2\) + \(\rm {dy\over dx}\)= 0

Answer 1

Correct Answer - Option 1 : 2x\(\rm d^2y\over dx^2\) + \(\rm {dy\over dx}\)= 0

Concept:

To form the differential equation of the given equation

• Differentiate the equation, the number of times as many as the constants are there.
• Find out the constants in terms of the variables.
• Substitute the variables in the original equation.

Calculation:

Given equation is (y - a)2 = 2bx               ....(1)

There are two constants a and b so differentiate two times

2(y - a)\(\rm dy\over dx\) = 2b

b = (y - a)\(\rm dy\over dx\)                                            ....(2)

Differentiating one more time w.r.t x

0 = \(\rm \left(dy\over dx\right)^2\) + (y - a)\(\rm d^2y\over dx^2\)

Multiply (y - a) on both sides, we get

(y - a)\(\rm \left(dy\over dx\right)^2\) + (y - a)²\(\rm d^2y\over dx^2\) = 0

From equation (1) and (2), (y - a)2 = 2bx  and (y - a)\(\rm dy\over dx\) = b

(y - a)\(\rm {dy\over dx} \times {dy\over dx}\) + (2bx) \(\rm d^2y\over dx^2\) = 0              

b\(\rm {dy\over dx}\) + (2bx) \(\rm d^2y\over dx^2\) = 0

2x\(\rm d^2y\over dx^2\) + \(\rm {dy\over dx}\)= 0