1)Find the solution of the differential equation
(ln(y)^5)(dy/dx)=x^5y
which satisfies the initial condition y(1)=e^2.
2)Solve the initial value problem
dy/dt+2y=20sin(t)+20cos(t) with y(0)=1.
3)Solve the initial value problem
(11(t+1)dy/dt)−8y=24t, for t>−1 with y(0)=6.
4)y′+2ty={4t 0≤t≤22 ,y(0)=3
={0 2<t , y(0)=3
y(t)=2+(1/(e^(t^2))) for 0≤t≤22
y(t)=? for 2<t
Differential equation?
2015-01-27 1:16 am
回答 (1)
2015-01-27 1:51 am
✔ 最佳答案
1.)((ln(y))^5) (dy/dx) = x^5*y
(ln(y))^5*(1/y) dy = (x^5) dx
Integrating both sides:
1/6*(ln(y))^6 = (x^6)/6 + C
y = e^((x^6 + C)^(1/6))
Plugging in initial condition:
y(1) = e^2
e^2 = e^((1 + C)^(1/6))
C = 63
y = e^((x^6 + 63)^(1/6))
收錄日期: 2021-04-15 18:05:17
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https://hk.answers.yahoo.com/question/index?qid=20150126171657AAYXemW