## emma29 3 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.

1. Algebraic!

think you're making it too hard...

2. Algebraic!

$(t+1)y' + y = ( (t+1)*y)'$

3. Algebraic!

make sense?

4. Algebraic!

hello?

5. emma29

sorry i'm just trying it out

6. Algebraic!

k :)

7. emma29

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

8. Algebraic!

hmm you're mixing methods:) you have:

9. Algebraic!

$( (t+1)*y)' = \cos(t)$

10. Algebraic!

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

11. Algebraic!

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

12. emma29

ok i'll try this out..thanks

13. Callisto

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 :'(

14. emma29

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

15. emma29

thanks to both for your help!!