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Nonexact Differential Equation (Possible to solve by integrating factor?) 
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#1
Apr112, 09:38 AM

P: 288

1. The problem statement, all variables and given/known data
Solve the differential equation: [itex]t^2 y' + y^2 = 0[/itex] 3. The attempt at a solution Now, it's definitely possible to solve this via separable of variables. But I am curious to know if I can solve it with an integrating factor. Having done some reading, I noticed that this equation is nearly in the form of an exact differential. Rewriting: [itex]t^2 y' + y^2 = 0 = t^2 \frac{dy}{dt} + y^2[/itex], implies, [itex]t^2 dy + y^2 dt = 0 = y^2 dt + t^2 dy[/itex]. Unfortunately, letting [itex]M(x,y) = y^2[/itex] and [itex]N(x,y) = t^2[/itex] and then taking derivatives shows [itex]M_y = 2y ≠ N_t = 2t[/itex], so it looks like an exact equation isn't going to emerge from this. In the event that the equation is not exact, an integrating factor is typically sought. The problem is, I don't know how to go about finding such an thing. Can someone help me? 


#2
Apr112, 06:28 PM

P: 799

You can try with this: [itex]\frac{dy}{y^2} + \frac{dt}{t^2} = 0[/itex]



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