idku
  • idku
Another Bernoulli DE (eq 2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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idku
  • idku
\(\large \color{black}{y'+\ln(x)y=y^{2}}\) \(\large \color{black}{y'y^{-2}+\ln(x)y^{-1}=1}\) Substitution \(\large \color{black}{v=y^{-1}}\) \(\large \color{black}{-v'=y'y^{-2}}\) DE becomes, \(\large \color{black}{-v'+v\ln(x)=1}\)
idku
  • idku
Then, the integrating factor is: \(\large \color{black}{H(x)=e^{x\ln(x)-x}}\) oh, boy does that become impossible for me to do?
idku
  • idku
\(\large \color{black}{-v'e^{x\ln(x)-x}+ve^{x\ln(x)-x}\ln(x)=e^{x\ln(x)-x}}\) \(\large \color{black}{ve^{x\ln(x)-x}=\int e^{x\ln(x)-x}dx}\)

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idku
  • idku
\(\large \color{black}{e^x=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}}\) is it possible to use this to get the integral of the right side?
anonymous
  • anonymous
there is no elementary closed-form expression for the integral on the right, but you could use power series to write one in some given interval
anonymous
  • anonymous
but using just that power series alone, no, that will not help
idku
  • idku
well, I guess I am ought to try another example that I can chew:)
idku
  • idku
\(\large \color{black}{y'+xy=y^{4}}\)
idku
  • idku
ok, lets start I guess..... \(\large \color{black}{y'y^{-4}+xy^{-3}=1}\) Substitution: \(\large \color{black}{v=y^{-3}}\) \(\large \color{black}{\dfrac{-1}{3}v'=y'y^{-4}}\) \(\large \color{black}{\dfrac{-1}{3}v'+xv=1}\)
idku
  • idku
\(\large \color{black}{v'-3xv=-3}\)
idku
  • idku
\(\large \color{black}{v'e^{(-3/2)x^2}-3xe^{(-3/2)x^2}v=-3e^{(-3/2)x^2}}\)
idku
  • idku
\(\large \color{black}{ve^{(-3/2)x^2}=\int -3e^{(-3/2)x^2}dx}\)
idku
  • idku
\(\large \color{black}{\displaystyle e^x=\sum_{i=0}^{\infty}\frac{x^i}{i!}}\) \(\large \color{black}{\displaystyle e^{(-3/2)x^2}=\sum_{i=0}^{\infty}\frac{\left((-3/2)x^2\right)^i}{i!}}\) am i correct so far?
idku
  • idku
\(\large \color{black}{\displaystyle e^{(-3/2)x^2}=\sum_{i=0}^{\infty}\frac{(-3/2)^ix^{2i}}{i!}}\) do I treat (-3/2)^i as a constant or I can't do that?
idku
  • idku
\(\large \color{black}{\displaystyle \int e^{(-3/2)x^2}dx=\sum_{i=0}^{\infty}\frac{(-3/2)^ix^{2i+1}}{i!(2i+1)}+C}\)
idku
  • idku
doesn't matter, I guess I get the method, I am just too stupid to come up with an adequate example for me.
idku
  • idku
\(\large \color{black}{\displaystyle y'+y=y^n}\)
idku
  • idku
\(\large \color{black}{\displaystyle y'y^{-n}+y^{-n+1}=1}\) \(\large \color{black}{\displaystyle \frac{1}{1-n}v'=y'y^{-n},~~~~~v=y^{-n+1}=1}\) \(\large \color{black}{\displaystyle \frac{1}{1-n}v'+v=1}\) \(\large \color{black}{\displaystyle v'+(1-n)v=1-n}\) \(\large \color{black}{\displaystyle v'e^{x(1-n)}+e^{x(1-n)}(1-n)v=e^{x(1-n)}(1-n)}\) \(\large \color{black}{\displaystyle ve^{x(1-n)}=\int e^{x(1-n)}(1-n)dx}\)
idku
  • idku
just nvm
idku
  • idku
i suck

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