Question

given (1+t)y' +y= cos(t) , y(0)=1 , I tried solving it using P(x) and q(x). So far I have d/dt(yt) = integral (cost/(1+t))*t dt...I'm stuck with the right side.

Answer 1

think you're making it too hard...

Answer 2

\[(t+1)y' + y = ( (t+1)*y)'\]

Answer 3

make sense?

Answer 4

hello?

Answer 5

sorry i'm just trying it out

Answer 6

k :)

Answer 7

am i suppose to end up with something like this?\[y dy=cost/(t+1) dt\] and then integrate both sides?

Answer 8

hmm you're mixing methods:)

you have:

Answer 9

\[( (t+1)*y)' = \cos(t)\]

Answer 10

\[\int\limits_{ }^{ }( (t+1)*y)' = \int\limits_{ }^{ }\cos(t)\]

Answer 11

\[(t+1)*y= \sin(t) +C\]
whoops

Answer 12

ok i'll try this out..thanks

Answer 13

I'm too stupid to solve in this way...
\[(1+t)y' +y= cos(t)\]\[y' +\frac{1}{1+t}y= \frac{1}{1+t}cos(t)\]
Nice :)
\[\alpha = e^{\int\frac{1}{1+t}dt}=e^{\ln|1+t|} = 1+t\]\[y=\frac{1}{\alpha}\int \alpha \times \frac{1}{1+t}cos(t) dt = \frac{1}{1+t}\int (1+t) \times \frac{1}{1+t}cos(t) dt \]\[= \frac{1}{1+t}\int cost dt =...\]
Why can't I make my life easier :'(

Answer 14

oh wow i made a mistake when taking the integral of \[\int\limits1/(1+t) \] no wonder it wouldn't come out nicely..thanks

Answer 15

thanks to both for your help!!