Question

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)?

Answer 1

The solution is `y = sqrt(2x^2 - 1) - 3`.

Explanation:

This is a separable differential equation.

`dy = (2x)/sqrt(2x^2 - 1) dx`

Integrate both sides.

`int dy = int (2x)/sqrt(2x^2 - 1) dx`

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

`:.` Let `u = 2x^2 - 1`. Then `du = 4x dx` and `dx = (du)/(4x)`.

`int dy = int (2x)/sqrt(u) * (du)/(4x)`

`int dy = 1/2int 1/sqrt(u)`

`int dy = 1/2int u^(-1/2)`

`y = 1/2(2u^(1/2)) + C`

`y = u^(1/2) + C`

`y = (2x^2 - 1)^(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(5)^2 - 1) + C`

`4 = sqrt(49) + C`

`4 - 7 = C`

`C = -3`

The solution is therefore `y = sqrt(2x^2 - 1) - 3`.

Hopefully this helps!