Checking a Bernoulli Equation

**Lyov** · February 21st 2011, 05:07 PM

The Equation is:

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

I have the solution, which thanks to Mathematica, I know to be correct, it is:

$\pm \sqrt{\frac{5t}{2+5t^5c}}$

I have to do a solution check, and keep getting terribly lost in the algebra. I am wondering if anyone has any suggestions or tricks on how to plug this mess back in without it turning into the never-ending algebra problem. Thanks.

**topsquark** · February 21st 2011, 05:27 PM

Originally Posted by Lyov

The Equation is:

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

I have the solution, which thanks to Mathematica, I know to be correct, it is:

$\pm \sqrt{\frac{5t}{2+5t^2c}}$

I have to do a solution check, and keep getting terribly lost in the algebra. I am wondering if anyone has any suggestions or tricks on how to plug this mess back in without it turning into the never-ending algebra problem. Thanks.

My only thought is to solve for $\displaystyle 2 + 5ct^2$ in terms of y:
$\displaystyle 2 + 5ct^2 = \frac{5t}{y}$

That helps a little anyway, since you will have a $\displaystyle (2 + 5ct^2)^2$ in your derivative.

-Dan

**Prove It** · February 21st 2011, 08:43 PM

$\displaystyle t^2\,\frac{dy}{dt} + 2t\,y = y^3$

Make the substitution $\displaystyle v = y^{1-3} = y^{-2} \implies y = v^{-\frac{1}{2}} \implies \frac{dy}{dt} = -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dt}$ and the integral becomes

$\displaystyle -\frac{1}{2}t^2v^{-\frac{3}{2}}\,\frac{dv}{dt} + 2t\,v^{-\frac{1}{2}} = v^{-\frac{3}{2}}$

$\displaystyle -\frac{1}{2}t^2\,\frac{dv}{dt} + 2t\,v = 1$

$\displaystyle -\frac{1}{2}\,\frac{dv}{dt} + 2t^{-1}v = t^{-2}$

$\displaystyle \frac{dv}{dt} - 4t^{-1}v = -2t^{-2}$ .

This is now first-order linear, so mulitplying both sides by the integrating factor $\displaystyle e^{\int{-4t^{-1}\,dt}} = e^{-4\ln{t}} = e^{\ln{\left(t^{-4}\right)}} = t^{-4}$ gives

$\displaystyle t^{-4}\,\frac{dv}{dt} - 4t^{-5}v = -2t^{-6}$

$\displaystyle \frac{d}{dt}\left(t^{-4}v\right) = -2t^{-6}$

$\displaystyle t^{-4}v = \int{-2t^{-6}\,dt}$

$\displaystyle t^{-4}v = \frac{2}{5}t^{-5} + C$

$\displaystyle v = \frac{2}{5}t^{-1} + Ct^4$

$\displaystyle y^{-2} = \frac{2}{5}t^{-1} + Ct^4$

$\displaystyle \frac{1}{y^2} = \frac{2}{5t} + Ct^4$

$\displaystyle \frac{1}{y^2} = \frac{2 + 5Ct^5}{5t}$

$\displaystyle y^2 = \frac{5t}{2 + 5Ct^5}$

$\displaystyle y = \pm \sqrt{\frac{5t}{2 + 5Ct}}$ .

I agree with your final answer.

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