ksaimouli
  • ksaimouli
solve
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!
ksaimouli
  • ksaimouli
\[\frac{ dy }{ dx }=x^2e^{4x}\]
ksaimouli
  • ksaimouli
solve the differential equation
ksaimouli
  • ksaimouli
i tried using by parts

Looking for something else?

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

More answers

anonymous
  • anonymous
i think that just means find the anti derivative of \(x^2e^{4x}\)and my guess is use parts twice
ksaimouli
  • ksaimouli
ya i used
ksaimouli
  • ksaimouli
but not solved let me show
ksaimouli
  • ksaimouli
i will use 7 technique
ksaimouli
  • ksaimouli
|dw:1360895129811:dw|
ksaimouli
  • ksaimouli
|dw:1360895181802:dw|
ksaimouli
  • ksaimouli
|dw:1360895191642:dw|
anonymous
  • anonymous
i think you have it backwards you want to drop the power on the \(x^2\) term
ksaimouli
  • ksaimouli
and now i should choose x^3 and e^4x
anonymous
  • anonymous
no the gimmick is to reduce the power, not raise it. raising the power makes it worse
anonymous
  • anonymous
\(u=x^2, dv = e^{4x}dx, du = 2x, v =\frac{1}{4}e^{4x}\)
ksaimouli
  • ksaimouli
okay i can do it this way so something wrong in my technique right
anonymous
  • anonymous
yes. as you can see your method just gave higher and higher powers of \(x\) you want to make them lower and lower
ksaimouli
  • ksaimouli
so instead of integrating id i take derivative
ksaimouli
  • ksaimouli
but if i follow the LIPET rule it says u= log right
anonymous
  • anonymous
i forget exactly how that box method works, but it is \[\int u dv =uv-\int v du\]
anonymous
  • anonymous
make \(u=x^2, du = 2xdx, dv = e^{4x}dx, v =\frac{1}{4}e^{4x}\) get \[\int x^2e^{4x}=\frac{1}{4}e^{4x}x^2-\frac{1}{2}\int e^{4x}xdx\] then do it again
ksaimouli
  • ksaimouli
okay got u
ksaimouli
  • ksaimouli
thx
anonymous
  • anonymous
yw

Looking for something else?

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