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

Feb 22, 2017

$\left(\frac{7}{4} , 0\right)$.

#### Explanation:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{2 x - 3} ^ 2$

$\mathrm{dy} = \frac{6}{2 x - 3} ^ 2 \mathrm{dx}$

Integrate both sides.

$\int \mathrm{dy} = \int \frac{6}{2 x - 3} ^ 2 \mathrm{dx}$

The right-hand side can be integrated by a u-substitution. Let $u = 2 x - 3$. Then $\mathrm{du} = 2 \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{2}$.

$\int \mathrm{dy} = \int \frac{6}{u} ^ 2 \cdot \frac{\mathrm{du}}{2}$

$\int \mathrm{dy} = \int \frac{3}{u} ^ 2 \mathrm{du}$

$\int \mathrm{dy} = 3 {u}^{-} 2$

$y = - 3 {u}^{-} 1 + C$

$y = - \frac{3}{u} + C$

$y = - \frac{3}{2 x - 3} + C$

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

$5 = - \frac{3}{2 \left(3\right) - 3} + C$

$5 = - \frac{3}{3} + C$

$5 = C - 1$

$C = 6$

Therefore, the initial function is $y = - \frac{3}{2 x - 3} + 6$. The curve will cross the x-axis when $y = 0$.

$0 = - \frac{3}{2 x - 3} + 6$

$- 6 = - \frac{3}{2 x - 3}$

$- 6 \left(2 x - 3\right) = - 3$

$- 12 x + 18 = - 3$

$- 12 x = - 21$

$x = \frac{21}{12}$

$x = \frac{7}{4}$

Hopefully this helps!