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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..
to make it clearer rewrite the equation as $\displaystyle \dfrac{t^2}{y^2}y'+1=\dfrac ty$ and put $\displaystyle u=\dfrac1y.$
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?
ok i got it..thanks for your help...
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
$\displaystyle \displaystyle t\,\frac{dv}{dt} + v = t^{-1}$
$\displaystyle \displaystyle \frac{d}{dt}(t\,v) = t^{-1}$
$\displaystyle \displaystyle t\,v = \int{t^{-1}\,dt}$
$\displaystyle \displaystyle t\,v = \ln{|t|} + C$
$\displaystyle \displaystyle v = \frac{\ln{|t|} + C}{t}$
$\displaystyle \displaystyle y^{-1} = \frac{\ln{|t|} + C}{t}$
$\displaystyle \displaystyle y = \frac{t}{\ln{|t|} + C}$.
