# Separable Differential Equation Problem I'm having

• Mar 21st 2010, 04:59 PM
s3a
Separable Differential Equation Problem I'm having
Question:
Solve the separable differential equation Subject to the initial condition: http://gauss.vaniercollege.qc.ca/web...0ef732ac51.png, y = ?

(My work is attached)

Any help would be greatly appreciated!
• Mar 21st 2010, 05:53 PM
Prove It
Quote:

Originally Posted by s3a
Question:

Solve the separable differential equation

Subject to the initial condition: http://gauss.vaniercollege.qc.ca/web...0ef732ac51.png, y = ?

(My work is attached)

Any help would be greatly appreciated!

$\displaystyle 3x - 4y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$

$\displaystyle \frac{3x}{\sqrt{x^2 + 1}} - 4y\,\frac{dy}{dx} = 0$

$\displaystyle \frac{3x}{\sqrt{x^2 + 1}} = 4y\,\frac{dy}{dx}$

$\displaystyle \int{\frac{3x}{\sqrt{x^2 + 1}}\,dx} = \int{4y\,\frac{dy}{dx}\,dx}$

$\displaystyle \int{3x(x^2 + 1)^{-\frac{1}{2}}\,dx} = \int{4y\,dy}$.

You should be able to evaluate both integrals now. To do the left hand side, you need to use the substitution $\displaystyle u = x^2 + 1$ so that $\displaystyle \frac{du}{dx} = 2x$.
• Mar 21st 2010, 07:24 PM
s3a
Did you check my work? (because that's exactly what I did)

Quote:

Originally Posted by Prove It
$\displaystyle 3x - 4y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$

$\displaystyle \frac{3x}{\sqrt{x^2 + 1}} - 4y\,\frac{dy}{dx} = 0$

$\displaystyle \frac{3x}{\sqrt{x^2 + 1}} = 4y\,\frac{dy}{dx}$

$\displaystyle \int{\frac{3x}{\sqrt{x^2 + 1}}\,dx} = \int{4y\,\frac{dy}{dx}\,dx}$

$\displaystyle \int{3x(x^2 + 1)^{-\frac{1}{2}}\,dx} = \int{4y\,dy}$.

You should be able to evaluate both integrals now. To do the left hand side, you need to use the substitution $\displaystyle u = x^2 + 1$ so that $\displaystyle \frac{du}{dx} = 2x$.

• Mar 22nd 2010, 03:10 PM
Prove It
1. Don't be so rude - most people on this forum do not open attachments.

2. When you have got it to the step:

$\displaystyle \frac{3}{2}\int{u^{-\frac{1}{2}}\,du} = \int{4y\,dy}$

$\displaystyle 3u^{\frac{1}{2}} + C_1 = 2y^2 + C_2$

$\displaystyle 3\sqrt{x^2 + 1} + C_1 - C_2 = 2y^2$

$\displaystyle \frac{3}{2}\sqrt{x^2 + 1} + C = y^2$, where $\displaystyle C = \frac{1}{2}(C_1 - C_2)$

$\displaystyle y = \pm \sqrt{\frac{3}{2}\sqrt{x^2 + 1} + C}$.

If $\displaystyle y(0) = 1$

$\displaystyle 1 = \pm \sqrt{\frac{3}{2}\sqrt{0^2 + 1} + C}$

$\displaystyle 1 = \frac{3}{2}\sqrt{1} + C$

$\displaystyle 1 = \frac{3}{2} + C$

$\displaystyle C = -\frac{1}{2}$.

So $\displaystyle y = \pm \sqrt{\frac{3}{2}\sqrt{x^2 + 1} - \frac{1}{2}}$

$\displaystyle y = \pm \sqrt{\frac{3\sqrt{x^2 + 1} - 1}{2}}$

$\displaystyle y = \pm \frac{\sqrt{3\sqrt{x^2 + 1} - 1}}{\sqrt{2}}$

$\displaystyle y = \pm \frac{\sqrt{6\sqrt{x^2 + 1} - 2}}{2}$.
• Mar 23rd 2010, 02:10 PM
s3a
I get it now thanks and I didn't try to sound rude, it just probably came that way to you because it's hard to see "emotion" (for the lack of a better word) through text.