A community for students.
Here's the question you clicked on:
 0 viewing
idku
 one year ago
my first Bernoulli DE ... (just looked at it on lamar tutorial and will see how will I do this one)
idku
 one year ago
my first Bernoulli DE ... (just looked at it on lamar tutorial and will see how will I do this one)

This Question is Closed

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{y'=4y+e^xy^3}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1(this example is completely made up, not on the tutorial)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1how do you know it satisifies a Berny then?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im thinking (couild be misremembering) that a Berny is a reduction process

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{y'=4y+e^xy^{3}}\) \(\large \color{black}{y'4y=e^xy^{3}}\) \(\large \color{black}{y'y^{3}4y^{2}=e^x}\) and then I substitute, \(\large \color{black}{v=y^{2}}\) \(\large \color{black}{v'=2y'y^{3}}\) > \(\large \color{black}{(1/2)v'=y'y^{3}}\) my DE becomes, \(\large \color{black}{\dfrac{1}{2}v'4v=e^x}\) For convenience, \(\large \color{black}{v'+8v=2e^x}\) Integrating factor: \(\large \color{black}{e^{4v^2}}\) \(\large \color{black}{v'e^{4v^2}+e^{4v^2}8v=e^{4v^2}2e^x}\) \(\large \color{black}{dv/dx(ve^{4v^2})=e^{4v^2}2e^x}\) wait I am getting stuck here a bit

idku
 one year ago
Best ResponseYou've already chosen the best response.1how would I then integrate both sides, if I have that on the RIGHT side?

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh, I missed a minus on the right side (when I multiplied by 2)

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{dv/dx(ve^{4v^2})=e^{4v^2}2e^x}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1the integrating facto is wrong? is it just e^x because integrating factor is with wespect to x, not v?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1v e^(8v) = v' e^(8v) + 8v e^(8v)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1hmm, dont think thats quite right either

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{v'+8v=2e^x}\) \(\large \color{black}{v'e^{8x}+e^{8x}8v=2e^{8x}e^x}\)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1that seems better yes

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{ve^{8x}=\int 2e^{9x}dx}\) \(\large \color{black}{ve^{8x}= (2/9)e^{9x}+c}\) \(\large \color{black}{v= (2/9)e^{x}+c/e^{8x}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{v= (2/9)e^{x}+C/e^{8x}}\) \(\large \color{black}{y^{2}= (2/9)e^{x}+C/e^{8x}}\) and then, \(\large \color{black}{y^2=\dfrac{1}{ (2/9)e^{x}+C/e^{8x}}}\) \(\large \color{black}{y=\pm\sqrt{\dfrac{1}{ (2/9)e^{x}+C/e^{8x}}}}\) and the rest is a matter of algebraic manipulation if I have done it correclty.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the process seems fair

idku
 one year ago
Best ResponseYou've already chosen the best response.1Awesome, I prevailed it! Thank you amistre for your confirmation!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1youre welcome, i didnt scrutinize it, but it seemed to be in good form overall.

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yes, fine for the first time. (Especially considering the fact that I am not really a "math guy")
Ask your own question
Sign UpFind 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
 Engagement 19 Mad Hatter
 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.