Anyone help for Differential Equations?
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From: Irvine
I have an idea of how to solve this problem, but there's one thing I simply don't know how to do.
Can anyone tell me the initial steps to drawing a directional field for this equation? (y' = y (prime) or dy/dt)
y' + 3y = t + e^-2t
This is what I'm trying...someone tell me what I'm doing wrong? I'm plugging in numbers to find out the slope/y-axis/x-axis, but not sure if the units are correct. t = x-axis.
t = 0 y(0) = 0
y'(0) + y(0) = 1
y'(0) = 1 <---(this is what I'm not sure about)
Can anyone tell me the initial steps to drawing a directional field for this equation? (y' = y (prime) or dy/dt)
y' + 3y = t + e^-2t
This is what I'm trying...someone tell me what I'm doing wrong? I'm plugging in numbers to find out the slope/y-axis/x-axis, but not sure if the units are correct. t = x-axis.
t = 0 y(0) = 0
y'(0) + y(0) = 1
y'(0) = 1 <---(this is what I'm not sure about)
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Joined: Jul 2001
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From: Yorba Linda, CA
Yes, that's correct.
It might be clearer to write:
y' = t + e^(-2t) - 3y.
Then for each point (t, y) you get a value for dy/dt.
As for solving the equation, it's first-order, linear, non-homogeneous, so you'd first solve the complementary equation:
y' + 3y = 0
to get the complementary solution, then find a particular solution to the original equation: you'll probably want to use undetermined coefficients for the e^(-2t) part and variation of parameters for the t part.
Hope that helps.
It might be clearer to write:
y' = t + e^(-2t) - 3y.
Then for each point (t, y) you get a value for dy/dt.
As for solving the equation, it's first-order, linear, non-homogeneous, so you'd first solve the complementary equation:
y' + 3y = 0
to get the complementary solution, then find a particular solution to the original equation: you'll probably want to use undetermined coefficients for the e^(-2t) part and variation of parameters for the t part.
Hope that helps.
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It's blatantly obvious. It's so plain, it makes Kansas look like the Rockies. 
Just kidding you. You obviously know your stuff, probably better than I do. While I've done a lot of calculus (I think I'm three courses short of a math major - I'm doing mech engineering), I don't actually use this kind of stuff very much at all. Fluids and thermodynamics tend to use the same ol' equations over and over again. Truth be known, I'm probably pretty rusty with early calculus...
Just kidding you. You obviously know your stuff, probably better than I do. While I've done a lot of calculus (I think I'm three courses short of a math major - I'm doing mech engineering), I don't actually use this kind of stuff very much at all. Fluids and thermodynamics tend to use the same ol' equations over and over again. Truth be known, I'm probably pretty rusty with early calculus...
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From: Yorba Linda, CA
I'm reminded of a quote I saw in a hallway in the mathematics department at UCLA (unfortunately, I don't recall the author):
In actual experience,
in particular, before breakfast,
multiplication tends to be noncommutative.
In actual experience,
in particular, before breakfast,
multiplication tends to be noncommutative.
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From: Calgary
Originally posted by magician
Yes, that's correct.
It might be clearer to write:
y' = t + e^(-2t) - 3y.
Then for each point (t, y) you get a value for dy/dt.
As for solving the equation, it's first-order, linear, non-homogeneous, so you'd first solve the complementary equation:
y' + 3y = 0
to get the complementary solution, then find a particular solution to the original equation: you'll probably want to use undetermined coefficients for the e^(-2t) part and variation of parameters for the t part.
Hope that helps.
Yes, that's correct.
It might be clearer to write:
y' = t + e^(-2t) - 3y.
Then for each point (t, y) you get a value for dy/dt.
As for solving the equation, it's first-order, linear, non-homogeneous, so you'd first solve the complementary equation:
y' + 3y = 0
to get the complementary solution, then find a particular solution to the original equation: you'll probably want to use undetermined coefficients for the e^(-2t) part and variation of parameters for the t part.
Hope that helps.
^^^^^^^^^^ The only thing I have to say to that is:
O_o What the HELL did you just say? o_O
Hehehe math is NOT my strong point
