# Bernoulli Equation

1. Sep 22, 2005

### amcavoy

I cannot get the correct answer to this for some reason:

$$t^2y'+2ty-y^3=0$$

I use the substitution $v=y^{1-n}=y^{-2}\implies y=v^{-\frac{1}{2}}$ and come up with $y'=-\frac{1}{2}v^{-\frac{3}{2}}$ and $y^3=v^{-\frac{3}{2}}$.

$$-\frac{1}{2}t^2v^{-\frac{3}{2}}+2tv^{-\frac{1}{2}}-v^{-\frac{3}{2}}=0$$

Then multiplying through by $v^{-2}$ gives:

$$-\frac{1}{2}t^2v^{3}+2t-v^{3}=0=2t-\left(\frac{1}{2}t^2+1\right)v^3$$

For which I would say:

$$v=\left(\frac{2t}{\frac{1}{2}t^2+1}\right)^{\frac{1}{3}}$$

But according to the back of my book, that answer is incorrect. What did I do wrong?

2. Sep 23, 2005

### HallsofIvy

Staff Emeritus
If $y= v^{-\frac{1}{2}}$ then $y'= -\frac{1}{2}v^{-\frac{3}{2}}v'$.

You seem to have dropped v' throughout. Your differential equation became an algebraic equation!

3. Sep 23, 2005

### saltydog

And what happen to the constant of integration?

So you got:

$$t^2y^{'}+2ty-y^3=0$$

Divide by $t^2$ and then divide by $y^3$, leads to:

$$y^{-3}dy+\frac{2}{t}y^{-2}t=\frac{1}{t^2}dt$$

Noting that:

$$d(y^{-2})=-2y^{-3}dy$$

and that's what you have up there on the left, we make the substitution:

$$v=y^{-2}$$

and:

$$dv=-2y^{-3}dy$$

Plug that in up there, rearrange to a first-order ODE in terms of v and t, find the integrating factor, integrate and not forgetting about the constant of integration, solve for v(t), then converting back to y, you should get:

$$y(t)=\pm \sqrt{\frac{5t}{2+5ct^5}}$$

Now what?

4. Sep 23, 2005

### amcavoy

Ahhh, I can't believe I forgot the v'! Thanks for your help.