Is my method in solving this equation correct and also are there any mistakess
ANY HELP WOULD BE MUCH APPRECIATED
thanksss
Differential Equations HELPPP?
回答 (1)
2021-04-16 3:27 pm
x²y' - 3xy - y = 0
x²y' - y.(3x + 1) = 0
x²y' = y.(3x + 1)
y'/y = (3x + 1)/x²
y'/y = (3x/x²) + (1/x²)
y'/y = (3/x) + (1/x²)
Ln(y) = 3.Ln(x) + (- 1/x) + k → where k is constant
Ln(y) = 3.Ln(x) - (1/x) + k
Ln(y) = Ln(x³) - (1/x) + k
e^[Ln(y)] = e^[Ln(x³) - (1/x) + k]
y = e^[Ln(x³)] * e^(- 1/x) * e^(k)
y = x³ * e^(- 1/x) * e^(k) → where: e^(k) = K
y = [x³/e^(1/x)] * K
y = [x³/e^(1/x)] * K → where K is constant
x²y' - y.(3x + 1) = 0
x²y' = y.(3x + 1)
y'/y = (3x + 1)/x²
y'/y = (3x/x²) + (1/x²)
y'/y = (3/x) + (1/x²)
Ln(y) = 3.Ln(x) + (- 1/x) + k → where k is constant
Ln(y) = 3.Ln(x) - (1/x) + k
Ln(y) = Ln(x³) - (1/x) + k
e^[Ln(y)] = e^[Ln(x³) - (1/x) + k]
y = e^[Ln(x³)] * e^(- 1/x) * e^(k)
y = x³ * e^(- 1/x) * e^(k) → where: e^(k) = K
y = [x³/e^(1/x)] * K
y = [x³/e^(1/x)] * K → where K is constant
收錄日期: 2021-04-24 08:44:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20210416045105AArC2ET