# Math Help - Separable D.E. problem

1. ## Separable D.E. problem

So here's my problem, I'm having trouble starting out:

Solve the separable differential equation:
$9x-5y\sqrt{x^2+1}\frac{dy}{dx}=0$
Subject to the initial condition: $y(0)=2$

So I have this to start out with.

$-5y dy=-9x+\frac{1}{\sqrt{x^2+1}} dx$

Am I headed in the right direction? Any help is greatly appreciated!

2. Originally Posted by ascendancy523
So here's my problem, I'm having trouble starting out:

Solve the separable differential equation:
$9x-5y\sqrt{x^2+1}\frac{dy}{dx}=0$
Subject to the initial condition: $y(0)=2$

So I have this to start out with.

$-5y dy=-9x+\frac{1}{\sqrt{x^2+1}} dx$

Am I headed in the right direction? Any help is greatly appreciated!
$9x - 5y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$

$\frac{9x}{\sqrt{x^2 + 1}} - 5y\,\frac{dy}{dx} = 0$

$5y\,\frac{dy}{dx} = \frac{9x}{\sqrt{x^2 + 1}}$

$\int{5y\,\frac{dy}{dx}\,dx} = \int{\frac{9x}{\sqrt{x^2 + 1}}\,dx}$

$\int{5y\,dy} = \int{9x(x^2 + 1)^{-\frac{1}{2}}\,dx}$.

You should be able to continue from here. You will need to use a $u$ substitution to evaluate the RHS.

3. Where would I use the u substitution?

4. Originally Posted by ascendancy523
Where would I use the u substitution?
Taking a wild guess, I'd say in the integral that requires it ....

5. Originally Posted by ascendancy523
Where would I use the u substitution?

Let $u = x^2 + 1$

Then $\frac{du}{dx} = 2x$

Substituting that into your original integral:

$\frac{9}{2} \int \frac{1}{\sqrt{u}} \ du$