Question

help solving this DE with bernoulli Equation: t^2(dy/dt) +y^2 = ty

Answer 1

I got the part where I have to make it in the form to get y^n where i get dy/dt - (1/t)y = (1/t^2)y^2 but i have a hard time understanding the rest... if u = y^(1-n) then n=2 therefore u= y^(-1) and If i derive that I get du/dx = -y^-2 (dy/dx) but then I'm stuck.... what happens with the y to plug it in, I;m still stuck with it

Answer 2

\[\begin{align*}
t^2\frac{dy}{dt}+y^2&=ty\\\\
\frac{dy}{dt}-\frac{y}{t}&=-\frac{y^2}{t^2}
\end{align*}\]
Setting \(u=y^{-1}\), yo have \(\dfrac{du}{dt}=-y^{-2}\dfrac{dy}{dt}\), or \(-\dfrac{1}{u^2}\dfrac{du}{dt}=\dfrac{dy}{dt}\).
\[\begin{align*}
-\frac{1}{u^2}\frac{du}{dt}-\frac{1}{ut}&=-\frac{1}{u^2t^2}\\\\
\frac{du}{dt}+\frac{1}{t}u&=\frac{1}{t^2}
\end{align*}\]

Answer 3

Note that this is also a homogeneous equation, so you can check your answer another way. Substituting \(y=zt\), so that \(\dfrac{dy}{dt}=t\dfrac{dz}{dt}+z\), and you have
\[\begin{align*}
\frac{dy}{dt}-\frac{y}{t}&=-\frac{y^2}{t^2}\\\\
t\frac{dz}{dt}+z-z&=-z^2\\\\
\frac{dz}{dt}&=-\frac{z^2}{t}\\\\
\frac{dz}{z^2}&=-\frac{dt}{t}\end{align*}\]