idku
  • idku
my first Bernoulli DE ... (just looked at it on lamar tutorial and will see how will I do this one)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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idku
  • idku
\(\large \color{black}{y'=4y+e^xy^3}\)
idku
  • idku
(this example is completely made up, not on the tutorial)
amistre64
  • amistre64
how do you know it satisifies a Berny then?

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amistre64
  • amistre64
im thinking (couild be misremembering) that a Berny is a reduction process
idku
  • idku
\(\large \color{black}{y'=4y+e^xy^{3}}\) \(\large \color{black}{y'-4y=e^xy^{3}}\) \(\large \color{black}{y'y^{-3}-4y^{-2}=e^x}\) 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
  • idku
how would I then integrate both sides, if I have that on the RIGHT side?
idku
  • idku
oh, I missed a minus on the right side (when I multiplied by -2)
idku
  • idku
\(\large \color{black}{dv/dx(ve^{4v^2})=-e^{4v^2}2e^x}\)
idku
  • idku
the integrating facto is wrong? is it just e^x because integrating factor is with wespect to x, not v?
amistre64
  • amistre64
v e^(8v) = v' e^(8v) + 8v e^(8v)
amistre64
  • amistre64
hmm, dont think thats quite right either
idku
  • idku
\(\large \color{black}{v'+8v=-2e^x}\) \(\large \color{black}{v'e^{8x}+e^{8x}8v=-2e^{8x}e^x}\)
idku
  • idku
like that?
amistre64
  • amistre64
that seems better yes
idku
  • idku
\(\large \color{black}{ve^{8x}=\int -2e^{9x}dx}\) \(\large \color{black}{ve^{8x}= (-2/9)e^{9x}+c}\) \(\large \color{black}{v= (-2/9)e^{x}+c/e^{8x}}\)
idku
  • idku
Like this?
amistre64
  • amistre64
so far so good
idku
  • idku
\(\large \color{black}{v= (-2/9)e^{x}+C/e^{8x}}\) \(\large \color{black}{y^{-2}= (-2/9)e^{x}+C/e^{8x}}\) 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
  • amistre64
the process seems fair
idku
  • idku
Awesome, I prevailed it! Thank you amistre for your confirmation!
amistre64
  • amistre64
youre welcome, i didnt scrutinize it, but it seemed to be in good form overall.
idku
  • idku
Yes, fine for the first time. (Especially considering the fact that I am not really a "math guy")
idku
  • idku
Thank you once again.

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