# First Order Separable Differential equation

• Feb 24th 2009, 10:01 PM
ibanez270dx
First Order Separable Differential equation
Hello,
I'm in Differential Equations and I have my midterm tomorrow... I can do most everything on my review sheet, but I'm a bit confused when it comes to a very simple initial value problem:

y' = x*y^3, y(0)=1

Here is what I'm doing:
I figure I can use separation of variables like so:

dy/y^3 = x*dx

..and after integrating both sides:

-1/(2*y^2) = (x^2)/2

solving for y gives:

y = 1/x

...which the initial condition y(0) = 1 does not hold.

Can somebody point out what I'm doing wrong? I'm quite confused and I'm running out of time...

- Jeff
• Feb 24th 2009, 10:22 PM
Soroban
Hello, ibanez270dx!

You forgot the "plus C" . . .

Quote:

$y' \:=\: xy^3,\quad y(0)\,=\,1$

Here is what I'm doing: . $\frac{dy}{y^3} \:=\: x\,dx$

Integrating: . $-\frac{1}{2y^2} \:= \:\frac{x^2}{2} \:{\color{red}+ \;C}$

Multiply by -2: . $\frac{1}{y^2} \:=\:-x^2 + C$

Invert: . $y^2 \:=\:\frac{1}{C - x^2}$

Now plug in the initial value . . .

• Feb 24th 2009, 10:28 PM
ibanez270dx
thanks! wow, that totally makes a lot more sense. I knew it had to be something simple.

Thanks again, your help is much appreciated!!
- Jeff