Bernoulli equation

**colerelm1** · February 11th 2013, 08:15 PM

I'm trying to figure out how to do this problem and we've just covered bernoulli equations in class so I know I have to use that. This is the problem:

Solve the equation
$y' = (X cos(t) + T)y - y^3$
where X and T are constants.

So far I have this:

$dy/dt - (X cos(t) + T)y = -y^3$

then using integrating factor:

$e^\int(Xcos(t) + T)) = e^(Xsin(t) + Tt + C) = e^(Tt)* e^(xsin(t)) * e^C$

But where do I go from here?

**Prove It** · February 11th 2013, 09:32 PM

You can't go straight to the integrating factor, because you haven't done the substitution required to turn the DE into something first order linear.

Let $\displaystyle \begin{align*} v = y^{1 - 3} = y^{-2} \end{align*}$ . Then we have $\displaystyle \begin{align*} y = v^{-\frac{1}{2}} \implies \frac{dy}{dt} = -\frac{1}{2}v^{-\frac{3}{2}}\, \frac{dv}{dt} \end{align*}$ . Substituting into the DE gives

$\displaystyle \begin{align*} \frac{dy}{dt} - \left( X \cos{(t)} + T \right) y &= -y^3 \\ -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dt} - \left( X \cos{(t)} + T \right) v^{-\frac{1}{2}} &= -v^{-\frac{3}{2}} \\ \frac{dv}{dt} + 2 \left( X \cos{(t)} + T \right) v &= 2 \end{align*}$

This is now first order linear, so now you can apply an integrating factor.

**colerelm1** · February 12th 2013, 09:21 AM

Alright, thanks a lot. One questions though: How did you get $-v^{-\frac{3}{2}}$ on the right side in the 2nd line?

**Prove It** · February 12th 2013, 03:35 PM

We know $\displaystyle \begin{align*} y = v^{-\frac{1}{2}} \end{align*}$ so $\displaystyle \begin{align*} y^3 = \left( v^{-\frac{1}{2}} \right)^3 = v^{-\frac{3}{2}} \end{align*}$ .

**colerelm1** · February 12th 2013, 04:24 PM

So for the integrating factor I get $e^{2Xsin(t) + Tt + C}$ , is that correct? Do I then multiply the entire equation by this? I've kind of forgotten how to do the rest of the steps

.

**Prove It** · February 12th 2013, 06:00 PM

Not quite. Your integrating factor is $\displaystyle \begin{align*} e^{\int{2 \left[ X\cos{(t)} + T \right] dt}} = e^{2 \left[ X\sin{(t)} + Tt \right]} \end{align*}$ . Note that any of the family of solutions to this integral will work, so for simplicity we disregard the +C. Anyway, yes we multiply both sides of the DE by this integrating factor to give

$\displaystyle \begin{align*} e^{2\left[ X\sin{(t)} + Tt \right]}\,\frac{dv}{dt} + 2e^{2\left[ X\cos{(t)} + Tt \right]} \left[X \cos{(t)} + T \right] v &= 2e^{2\left[ X\sin{(t)} + Tt \right]} \\ \frac{d}{dt} \left[ e^{2\left[ X\sin{(t)} + Tt \right] }\, v \right] &= 2e^{2\left[ X\sin{(t)} + Tt \right]} \\ e^{2\left[ X\sin{(t)} + Tt \right]} &= \int{2e^{2\left[ X\sin{(t)} + Tt \right]}\,dt} \end{align*}$

Can you go from here?

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