y‘+y=3x 求微分方程式之解 （使用一階線性微分方程式）?
回答 (2)
2017-11-15 11:45 pm
y'+y=3x
特徵方程式 r+1=0 ---> r=-1
yh(x) = c e^(-x)
set yp(x) = a x + b, yp'(x) = a
yp'(x) + yp(x) = a + a x + b = 3 x ---> a = 3, b = - 3
yp(x) = 3 x - 3
y(x) = yh(x) + yp(x)
= c e^(-x) + 3x - 3 .....Ans
特徵方程式 r+1=0 ---> r=-1
yh(x) = c e^(-x)
set yp(x) = a x + b, yp'(x) = a
yp'(x) + yp(x) = a + a x + b = 3 x ---> a = 3, b = - 3
yp(x) = 3 x - 3
y(x) = yh(x) + yp(x)
= c e^(-x) + 3x - 3 .....Ans
2017-11-15 8:31 pm
d/dx 3x = 3(1) = 3;
d/dy Y'+y = 1+Y';
d/dY' = 1+y
Hope I am right.
d/dy Y'+y = 1+Y';
d/dY' = 1+y
Hope I am right.
收錄日期: 2021-05-01 15:30:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171115122010AAxby1T