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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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Algebraic!Best ResponseYou've already chosen the best response.1
think you're making it too hard...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
\[(t+1)y' + y = ( (t+1)*y)'\]
 one year ago

emma29Best ResponseYou've already chosen the best response.0
sorry i'm just trying it out
 one year ago

emma29Best ResponseYou've already chosen the best response.0
am i suppose to end up with something like this?\[y dy=cost/(t+1) dt\] and then integrate both sides?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
hmm you're mixing methods:) you have:
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
\[( (t+1)*y)' = \cos(t)\]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
\[\int\limits_{ }^{ }( (t+1)*y)' = \int\limits_{ }^{ }\cos(t)\]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
\[(t+1)*y= \sin(t) +C\] whoops
 one year ago

emma29Best ResponseYou've already chosen the best response.0
ok i'll try this out..thanks
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
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 :) \[\alpha = e^{\int\frac{1}{1+t}dt}=e^{\ln1+t} = 1+t\]\[y=\frac{1}{\alpha}\int \alpha \times \frac{1}{1+t}cos(t) dt = \frac{1}{1+t}\int (1+t) \times \frac{1}{1+t}cos(t) dt \]\[= \frac{1}{1+t}\int cost dt =...\] Why can't I make my life easier :'(
 one year ago

emma29Best ResponseYou've already chosen the best response.0
oh wow i made a mistake when taking the integral of \[\int\limits1/(1+t) \] no wonder it wouldn't come out nicely..thanks
 one year ago

emma29Best ResponseYou've already chosen the best response.0
thanks to both for your help!!
 one year ago
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