1. ## Seperable Differential Equation

Solve the separable differential equation

8x-3ysqrt(x^2+1)dy/dx=0

subject to initial condition y(0) = 1

This is what I have done so far....

-3ysqrt(x^2+1)dy/dx = -8x
-3ydy = -8x/sqrt(x^2+1)dx
-3/2y^2 = ?

2. Originally Posted by derekjonathon
Solve the separable differential equation

8x-3ysqrt(x^2+1)dy/dx=0

subject to initial condition y(0) = 1

This is what I have done so far....

-3ysqrt(x^2+1)dy/dx = -8x
-3ydy = -8x/sqrt(x^2+1)dx
-3/2y^2 = ?
$\int3y\,dy=\int\frac{8x}{\sqrt{x^2+1}}\,dx$

For the right integral, let $u=x^2+1$ so $du=2x\,dx$. Now it's $\int\frac{8}{2\sqrt{u}}\,du=8\sqrt{u}=8\sqrt{x^2+1 }$ so...

3. Originally Posted by derekjonathon
Solve the separable differential equation

8x-3ysqrt(x^2+1)dy/dx=0

subject to initial condition y(0) = 1

This is what I have done so far....

-3ysqrt(x^2+1)dy/dx = -8x
-3ydy = -8x/sqrt(x^2+1)dx
-3/2y^2 = ?
$8x - 3y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$

$3y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 8x$

$y\,\frac{dy}{dx} = \frac{8}{3}\,\frac{x}{\sqrt{x^2 + 1}}$

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

$\int{y\,dy} = \frac{8}{3}(x^2 + 1)^{\frac{1}{2}} + C_1$

$\ln{|y|} + C_2 = \frac{8}{3}\sqrt{x^2 + 1} + C_1$

$\ln{|y|} = \frac{8}{3}\sqrt{x^2 + 1} + C$ where $C = C_1 - C_2$

$|y| = e^{\frac{8}{3}\sqrt{x^2 + 1} + C}$

$|y| = e^Ce^{\frac{8}{3}\sqrt{x^2 + 1}}$

$y = \pm e^Ce^{\frac{8}{3}\sqrt{x^2 + 1}}$

$y = Ae^{\frac{8}{3}\sqrt{x^2 + 1}}$

Now you know $y(0) = 1$

$1 = Ae^{\frac{8}{3}\sqrt{0^2 + 1}}$

$1 = Ae^{\frac{8}{3}}$

$A = e^{-\frac{8}{3}}$.

Therefore $y = e^{-\frac{8}{3}}e^{\frac{8}{3}\sqrt{x^2 + 1}}$

$y = e^{\frac{8}{3}\sqrt{x^2 + 1} - \frac{8}{3}}$.

4. Never mind, you fixed it.

5. Originally Posted by redsoxfan325
I think you lost a square root sign going from step 2 to step 3.
Yes I did. Edited.