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Nonexact Differential Equation (Possible to solve by integrating factor?)

by TranscendArcu
Tags: differential, equation, factor, integrating, nonexact, solve
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TranscendArcu
#1
Apr1-12, 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?
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hikaru1221
#2
Apr1-12, 06:28 PM
P: 799
You can try with this: [itex]\frac{dy}{y^2} + \frac{dt}{t^2} = 0[/itex]
LCKurtz
#3
Apr1-12, 10:48 PM
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Quote Quote by TranscendArcu View Post
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.
Quote Quote by hikaru1221 View Post
You can try with this: [itex]\frac{dy}{y^2} + \frac{dt}{t^2} = 0[/itex]
I think he know that.
@TranscendArcu: It is not a given that a given first order DE can be solved by an integrating factor in any practical fashion, even if you can solve it by separation of variables.


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