anonymous
  • anonymous
Diff Eq problem: \[\frac{ dy }{ dt } + t^2y = 1\] So, first I find my integrating factor. \[e ^{\int\limits_{}^{}t^2dt} = e ^{2t+c}\] Next, I multiply both sides by the integrating factor. \[(\frac{ dy }{ dt } + t^2y)e ^{2t+c} = e ^{2t+c} \] Then I want to integrate both sides so I should have : \[\int\limits_{}^{}d(e ^{2t+c} )y = \int\limits_{}^{} e ^{2t+c} dt\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Is this looking good so far @phi ?
anonymous
  • anonymous
so now I've got \[e ^{2t+c}y = \frac{ 1 }{ 2 }e ^{2t+c} \] hmmmmm, at this point its not looking right, and not headed toward the answer in the book.
anonymous
  • anonymous
I should have \[y = e ^{-1/3t^3} \int\limits_{}^{}e ^{-1/3t^3}dt+c\]

Looking for something else?

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

More answers

inkyvoyd
  • inkyvoyd
line 2 is incorrect
inkyvoyd
  • inkyvoyd
\(e ^{\int\limits_{}^{}t^2dt} = e ^{t^3/3+c}\)
austinL
  • austinL
I don't believe you need the "+C" in the exponential.
inkyvoyd
  • inkyvoyd
Austin is correct probably, seeing as you can multiply both sides by whatever expression you want as long as you don't multiply by zero and divide after
austinL
  • austinL
\(y\prime + t^2y=1\) \(\mu(t)=e^{\int p(t)dt}\) \(p(t)=t^2\) \(\int t^2dt=\dfrac{t^3}{3}\) \(\mu(t)y\prime+t^2\mu(t)y=\mu(t)\)
inkyvoyd
  • inkyvoyd
staph! I wanted the medal
austinL
  • austinL
Then I do believe that you simply integrate both sides and solve for y if feasible?
anonymous
  • anonymous
Thanks a ton guys!
austinL
  • austinL
No problemo!

Looking for something else?

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