mathstresser
Junior Member
- Joined
- Jan 28, 2006
- Messages
- 134
I finished this problem... but will you look at it and tell me if it's right, please?
Solve the linear equation (1+t)x' + x = cos(t) subject to x(0)=1.
By manipulating the equation I get:
(1+t)x'= -x +cos(t) and
x'=(-x + cos(t) )/(1+t) which is equal to
x'= -x/(1+t) +cos(t)/(1+t)
then... I manipulate it some more and get
x' + (1/(1+t))(x) = cos(t)/(1+t)
I then try to find the integrating factor
e^∫(1/(1+t)) and that is e^( ln (t+1))
So the integrating factor is t+1... I take that times the equation and get
[(t+1)x]'= cos(t)
So then I integrate both sides and get:
x(t+1)=sin(t) +c
x= sin(t)/(t+1) + c
c=1
x= sin(t)/(t+1) + 1
Is that right?
Solve the linear equation (1+t)x' + x = cos(t) subject to x(0)=1.
By manipulating the equation I get:
(1+t)x'= -x +cos(t) and
x'=(-x + cos(t) )/(1+t) which is equal to
x'= -x/(1+t) +cos(t)/(1+t)
then... I manipulate it some more and get
x' + (1/(1+t))(x) = cos(t)/(1+t)
I then try to find the integrating factor
e^∫(1/(1+t)) and that is e^( ln (t+1))
So the integrating factor is t+1... I take that times the equation and get
[(t+1)x]'= cos(t)
So then I integrate both sides and get:
x(t+1)=sin(t) +c
x= sin(t)/(t+1) + c
c=1
x= sin(t)/(t+1) + 1
Is that right?