# solve the problem using Bernoulli....

• Feb 10th 2011, 05:42 PM
slapmaxwell1
solve the problem using Bernoulli....
ok i dont see how the book came up with the substitution of u= y^-1

so here is the problem..

(t^2)dy/dt + y^2 = ty

my question is this, if u = y^(1-n) wouldnt n = 0?

i think the book may have made a mistake on this problem....can someone please check the problem and see if the substitution is correct? thanks in advance..
• Feb 10th 2011, 05:52 PM
Krizalid
to make it clearer rewrite the equation as $\displaystyle \dfrac{t^2}{y^2}y'+1=\dfrac ty$ and put $\displaystyle u=\dfrac1y.$
• Feb 10th 2011, 05:56 PM
slapmaxwell1
well i did rewrite it...

i got dy/dt + (y^2/t^2) = y/t but the way i have it written it seems like n would be 1?
• Feb 10th 2011, 06:13 PM
slapmaxwell1
ok i got it..thanks for your help...
• Feb 10th 2011, 07:49 PM
Prove It
To use Bernoulli...

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

$\displaystyle \displaystyle t^2\,\frac{dy}{dt} - t\,y = -y^2$.

Make the substitution $\displaystyle \displaystyle v = y^{1-2} = y^{-1} \implies y = v^{-1}$.

Then $\displaystyle \displaystyle \frac{dy}{dt} = \frac{d}{dt}(v^{-1}) = \frac{d}{dv}(v^{-1})\,\frac{dv}{dt} = -v^{-2}\,\frac{dv}{dt}$.

Substituting into the DE gives

$\displaystyle \displaystyle t^2\left(-v^{-2}\,\frac{dv}{dt}\right) - t\,v^{-1} = -v^{-2}$

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

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

This is now first order linear, so multiplying both sides of the DE by the Integrating Factor $\displaystyle \displaystyle e^{\int{t^{-1}\,dt}} = e^{\ln{t}} = t$ gives

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

$\displaystyle \displaystyle \frac{d}{dt}(t\,v) = t^{-1}$

$\displaystyle \displaystyle t\,v = \int{t^{-1}\,dt}$

$\displaystyle \displaystyle t\,v = \ln{|t|} + C$

$\displaystyle \displaystyle v = \frac{\ln{|t|} + C}{t}$

$\displaystyle \displaystyle y^{-1} = \frac{\ln{|t|} + C}{t}$

$\displaystyle \displaystyle y = \frac{t}{\ln{|t|} + C}$.