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

by TranscendArcu
Tags: differential, equation, factor, integrating, nonexact, solve
 P: 288 1. The problem statement, all variables and given/known data Solve the differential equation: $t^2 y' + y^2 = 0$ 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: $t^2 y' + y^2 = 0 = t^2 \frac{dy}{dt} + y^2$, implies, $t^2 dy + y^2 dt = 0 = y^2 dt + t^2 dy$. Unfortunately, letting $M(x,y) = y^2$ and $N(x,y) = t^2$ and then taking derivatives shows $M_y = 2y ≠ N_t = 2t$, 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?
 P: 799 You can try with this: $\frac{dy}{y^2} + \frac{dt}{t^2} = 0$
HW Helper
Thanks
PF Gold
P: 7,632
 Quote by TranscendArcu 1. The problem statement, all variables and given/known data Solve the differential equation: $t^2 y' + y^2 = 0$ 3. The attempt at a solution Now, it's definitely possible to solve this via separable of variables.
 Quote by hikaru1221 You can try with this: $\frac{dy}{y^2} + \frac{dt}{t^2} = 0$
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.

 Related Discussions Calculus & Beyond Homework 3 Calculus & Beyond Homework 2 Calculus & Beyond Homework 4 Calculus & Beyond Homework 2 Calculus & Beyond Homework 4