Question

`y = c^2 + c/x` is solution of ______.

विकल्प

• `x^4 ((dy)/(dx))^2 - x (dy)/(dx) - y = 0`

• `x^2 ((dy)/(dx))^2 + y = 0`

• `x^3 ((d^2y)/(dx^2)) - x (dy)/(dx) + y = 0`

• `x (d^2y)/(dx^2) = 4y`

Answer 1

`y = c^2 + c/x` is solution of `bbunderline(x^4 ((dy)/(dx))^2 - x (dy)/(dx) - y = 0)`.

Explanation:

Let’s find the first and second derivatives of y with respect to x:

Step 1: First derivative:

`(dy)/(dx) = d/(dx) (c^2 + c/x) = 0 - c/x^2 = -c/x^2`

Step 2: Second derivative:

`(d^2y)/(dx^2) = d/dx (-c/x^2)= (2c)/(x^3)`

Multiplying both sides of the second derivative by x³:

`x^3 · (d^2y)/(dx^2) = x^3 · (2c)/(x^3) = 2c`

`y = c^2 + c/x => xy = c^2x + c => xy - c^2x = c`

Multiply both sides by 2: 

2(xy − c²x) = 2c ⇒ 2xy − 2c²x = 2c

Try Option (3):

`x^3 ((d^2y)/(dx^2)) - x (dy)/(dx) + y = 0`

Substitute `(dy)/(dx) = -c/x^2`

`(d^2y)/(dx^2) = (2c)/(x^3)`

`y = c^2 + c/x`

`x^3 · (2c)/(x^3) - x · (-c/x^2) + (c^2 + c/x) = 0 => 2c + c/x + c^2 + c/x = 0 => 2c + c^2 + (2c)/x ≠ 0`

This is not zero, so this option is incorrect.

Try Option (1):

`x^4 ((dy)/(dx))^2 - x(dy)/(dx) - y = 0`

Use `(dy)/(dx) = -c/x^2`

`y = c^2 + c/x`

Then `x^4 (-c/x^2)^2 - x(-c/x^2)-(c^2 + c/x) = x^4 · (c^2)/(x^4) + c/x - c^2 - c/x = c^2 + c/x - c^2 - c/x = 0`

LHS = 0, so this equation is satisfied.