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Messages - Kathryn Bucci

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1
Daniel, Kathryn

The word "homogeneous" is overused. Kathryn is right if equation is homogeneous which means $y'=f(y/x)$. Daniel meanwhile thinks about linear homogeneous, which is a completely different story.

Yes that is what I was referring to - sorry, I should have been more clear.

2
I think that if you had a non-exact homogeneous equation and you can't easily find an integrating factor, you could also solve it by making a substitution y=vx.. then it would be reduced to a separable equation.

3
In that example $W=e^{2t}$ (you made a mistake somewhere). The exponential function is never zero so the Wronskian is always nonzero. Also look at Abel's theorem - that equation satisfies the conditions since it's a 2nd order linear homogenous equation - if you look at the identity in the theorem, you'll see that $W$ must either be always zero or never zero.

4
If 𝑦′=𝑦𝛼 with 0 < 𝛼 < 1 e.g. 𝑦′=3𝑦2/3= f(t,y), then ∂f/∂y=2y-1/3 is not continuous at (0,0).
According to theorem 2.4.2 (existence and uniqueness for 1st order nonlinear equations), both f and ∂f/∂y have to be continuous on an interval containing the initial point (0,0) - ∂f/∂y is not continuous there so you can't infer that there is a unique solution.

5
MAT244--Lectures & Home Assignments / Re: Notation & Linear
« on: September 16, 2018, 10:03:41 PM »
$u_x$ = partial derivative of $u$ with respect to $x$
$u_{xx}$ = second partial derivative of $u$ with respect to $x$

22) is not linear because of the two terms where u is multiplied by $u_x$ and $u_y$

I fixed math expressions for both of you. "Edit" to see how it should be done-V.I.

6
MAT244--Lectures & Home Assignments / Re: Directional field
« on: September 15, 2018, 02:36:04 PM »
I don't think it would be possible to draw a perfect direction field and if you understand the behaviour of the function then you should be able to produce an adequate direction field.

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