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 2 years 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.
 2 years 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.

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

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

emma29
 2 years ago
Best ResponseYou've already chosen the best response.0sorry i'm just trying it out

emma29
 2 years ago
Best ResponseYou've already chosen the best response.0am i suppose to end up with something like this?\[y dy=cost/(t+1) dt\] and then integrate both sides?

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

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

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

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

emma29
 2 years ago
Best ResponseYou've already chosen the best response.0ok i'll try this out..thanks

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1I'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 :'(

emma29
 2 years ago
Best ResponseYou've already chosen the best response.0oh wow i made a mistake when taking the integral of \[\int\limits1/(1+t) \] no wonder it wouldn't come out nicely..thanks

emma29
 2 years ago
Best ResponseYou've already chosen the best response.0thanks to both for your help!!
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