RE: Particular Solution to Differential Equation
November 4, 2015 at 3:08 pm
(This post was last modified: November 4, 2015 at 3:09 pm by TheRealJoeFish.)
(November 4, 2015 at 2:54 pm)The_Flying_Skeptic Wrote: 5dy/dt + 10y(t) = 2x(t)
x(t) = 2
so would a particular solution be y(t)=0.4 ?
Oooooh boy it's been a while since my DiffEq days. I think it would be because, if y(t) is the constant .4, its derivative is 0, so the first term would be 5*0, the second term would be 10*.4, and the sum would be 2*2, for 0+4=4.
I think that's how maths work, but someone feel free to correct me otherwise. I'm not sure if you can just do it like that.
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