Is this a partial derivatives? Or just simple algebra?

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The discussion clarifies that the transition from the left side to the right side of the equation involves the Product Rule from Calculus I, rather than partial derivatives or simple algebra. The user expresses confusion about the step in their differential equations book, initially considering it might require knowledge of partial derivatives. However, it is confirmed that understanding the Product Rule is sufficient for this problem. The user feels relieved after realizing the solution does not require extensive searching through their book. This highlights the importance of foundational calculus concepts in solving differential equations.
CinderBlockFist
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In my differential equations book there is this step, that i don't know how it goes from one side to the next.


(t^2)y' + 2ty = ((t^2)y)'


cause, on the left side I factor out a t, and i get t(ty'+2y) ...so do i have to learn partial derivatives in order to get from the left side to the right ?
 
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It's neither partial derivatives nor simple algebra. It's the Product Rule from Calculus I.
 
Oh wow I feel stupid, I was searching my whole book. Thanks again.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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