I got y = ce^5t
10 = ce^5(0)
c = 10
y=ce^5(1/5ln3)
y=10(3)
y=30
Thanks
What is y if t=(1/5)ln3? ...when dy/dt = 5y, y=10 when t=0.?
2016-08-01 7:28 am
回答 (2)
2016-08-01 10:02 am
✔ 最佳答案
dy/dt = 5y(1/y)dy = 5dt
ln(y) = 5t + C₁
When t = 0, then y = 10 :
ln(10) = 5(0) + C₁
Then, ln(y) = 5t + ln(10)
When t = (1/5)ln(3) :
ln(y) = 5*(1/5)ln(3) + ln(10)
ln(y) = ln(3) + ln(10)
ln(y) = ln(30)
y = 30
2016-08-01 3:55 pm
dy/dt=5y=>
dy/y=5dt=>
ln(y)=5t+C
(0,10)=>C=ln(10)
=>
ln(y)=5t+ln(10)
=>
y=10e^(5t)
When t=ln(3)/5, then
y=10e^[ln(3)]
=>
y=10*3=30
dy/y=5dt=>
ln(y)=5t+C
(0,10)=>C=ln(10)
=>
ln(y)=5t+ln(10)
=>
y=10e^(5t)
When t=ln(3)/5, then
y=10e^[ln(3)]
=>
y=10*3=30
收錄日期: 2021-04-20 16:26:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160731232812AAf7FUY