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

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Answer 1 · 2017-02-22T16:37:10

$(7/4, 0)$ .

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!

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