How to solve this Differential Equation?
2015-07-12 2:27 am
I cannot figure this out and it is the last problem i need. Could someone PLEASE help on this? only have half hour!!! http://prntscr.com/7rn347
回答 (1)
2015-07-12 3:02 am
✔ 最佳答案
y' + x*y = 160xthis is just simple plug and chug
y = 1/u(x) * int u(x) * Q(x) dx from y' + P(x)*y = Q(x) & u(x) = e^(int P(x) dx)
u(x) = e^[int x dx] = e^((x^2)/2)
y = 1/e^((x^2)/2) * int 160x*e^((x^2)/2) dx
y = 1/e^((x^2)/2) * int 160*(e^u) du
y = 1/e^((x^2)/2) * [160*(e^u) + C]
y = 1/e^((x^2)/2) * [160*e^((x^2)/2) + C]
y = 160 + [C/e^((x^2)/2)]
2015-07-12 4:06 am
This is the equation given
y' + x*y = 160x. Rewrite it as
y` = 160 x -xy= x(160-y) = - x(y-160)
y`/(y-160) = -x
Integrate it
ln (y-160) = -x^2/2 + c c=constant
y-160 = e^(-x^2/2) + k
y=160 +e^(-x^2/2) +k
This is the answer
y' + x*y = 160x. Rewrite it as
y` = 160 x -xy= x(160-y) = - x(y-160)
y`/(y-160) = -x
Integrate it
ln (y-160) = -x^2/2 + c c=constant
y-160 = e^(-x^2/2) + k
y=160 +e^(-x^2/2) +k
This is the answer
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