Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

emma29

  • 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.

  • This Question is Closed
  1. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    think you're making it too hard...

  2. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  3. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    make sense?

  4. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hello?

  5. emma29
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry i'm just trying it out

  6. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    k :)

  7. emma29
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm you're mixing methods:) you have:

  9. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  10. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  12. emma29
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i'll try this out..thanks

  13. Callisto
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 :) \[\alpha = e^{\int\frac{1}{1+t}dt}=e^{\ln|1+t|} = 1+t\]\[y=\frac{1}{\alpha}\int \alpha \times \frac{1}{1+t}cos(t) dt = \frac{1}{1+t}\int (1+t) \times \frac{1}{1+t}cos(t) dt \]\[= \frac{1}{1+t}\int cost dt =...\] Why can't I make my life easier :'(

  14. emma29
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  15. emma29
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks to both for your help!!

  16. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.