# How do you use the differential equation dy/dx=(2x)/sqrt(2x^2-1) to find the equation of the function given point (5,4)?

Mar 4, 2017

The solution is $y = \sqrt{2 {x}^{2} - 1} - 3$.

#### Explanation:

This is a separable differential equation.

$\mathrm{dy} = \frac{2 x}{\sqrt{2 {x}^{2} - 1}} \mathrm{dx}$

Integrate both sides.

$\int \mathrm{dy} = \int \frac{2 x}{\sqrt{2 {x}^{2} - 1}} \mathrm{dx}$

It's true that trig substitution could be used to solve this integral, but a substitution would be easier.

$\therefore$ Let $u = 2 {x}^{2} - 1$. Then $\mathrm{du} = 4 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{4 x}$.

$\int \mathrm{dy} = \int \frac{2 x}{\sqrt{u}} \cdot \frac{\mathrm{du}}{4 x}$

$\int \mathrm{dy} = \frac{1}{2} \int \frac{1}{\sqrt{u}}$

$\int \mathrm{dy} = \frac{1}{2} \int {u}^{- \frac{1}{2}}$

$y = \frac{1}{2} \left(2 {u}^{\frac{1}{2}}\right) + C$

$y = {u}^{\frac{1}{2}} + C$

$y = {\left(2 {x}^{2} - 1\right)}^{\frac{1}{2}} + C$

We now solve for $C$ using the information given. We know that when $x = 5$, $y = 4$, so we can say the following:

$4 = \sqrt{2 {\left(5\right)}^{2} - 1} + C$

$4 = \sqrt{49} + C$

$4 - 7 = C$

$C = - 3$

The solution is therefore $y = \sqrt{2 {x}^{2} - 1} - 3$.

Hopefully this helps!