Question

If `dy/dx = 6/(2x - 3)^2`, and `y(3) = 5`, what is the x-intercept of `y`?

Answer 1

`(7/4, 0)`.

Explanation:

We will have to solve a differential equation to find the initial function. This d.e. is separable.

`dy/dx = 6/(2x -3)^2`

`dy = 6/(2x - 3)^2dx`

Integrate both sides.

`intdy = int 6/(2x - 3)^2 dx`

The right-hand side can be integrated by a u-substitution. Let `u = 2x - 3`. Then `du = 2dx` and `dx = (du)/2`.

`intdy =int 6/u^2 * (du)/2`

`intdy = int 3/u^2 du`

`intdy = 3u^-2`

`y = -3u^-1+ C`

`y = -3/u + C`

`y = -3/(2x - 3) + C`

Now solve for `C`. We know that when `x = 3`, `y = 5`.

`5 = -3/(2(3) - 3) + C`

`5 = -3/3 + C`

`5 = C - 1`

`C = 6`

Therefore, the initial function is `y = -3/(2x - 3) + 6`. The curve will cross the x-axis when `y = 0`.

`0 = -3/(2x - 3) + 6`

`-6 = -3/(2x - 3)`

`-6(2x - 3) = -3`

`-12x + 18 = -3`

`-12x = -21`

`x = 21/12`

`x = 7/4`

Hopefully this helps!