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idku
 one year ago
Another Bernoulli DE (eq 2)
idku
 one year ago
Another Bernoulli DE (eq 2)

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idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{y'+\ln(x)y=y^{2}}\) \(\large \color{black}{y'y^{2}+\ln(x)y^{1}=1}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0Then, 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
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{v'e^{x\ln(x)x}+ve^{x\ln(x)x}\ln(x)=e^{x\ln(x)x}}\) \(\large \color{black}{ve^{x\ln(x)x}=\int e^{x\ln(x)x}dx}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{e^x=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}}\) is it possible to use this to get the integral of the right side?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is no elementary closedform expression for the integral on the right, but you could use power series to write one in some given interval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but using just that power series alone, no, that will not help

idku
 one year ago
Best ResponseYou've already chosen the best response.0well, I guess I am ought to try another example that I can chew:)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{y'+xy=y^{4}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0ok, 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
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{v'3xv=3}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{v'e^{(3/2)x^2}3xe^{(3/2)x^2}v=3e^{(3/2)x^2}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{ve^{(3/2)x^2}=\int 3e^{(3/2)x^2}dx}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\displaystyle e^x=\sum_{i=0}^{\infty}\frac{x^i}{i!}}\) \(\large \color{black}{\displaystyle e^{(3/2)x^2}=\sum_{i=0}^{\infty}\frac{\left((3/2)x^2\right)^i}{i!}}\) am i correct so far?

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\displaystyle e^{(3/2)x^2}=\sum_{i=0}^{\infty}\frac{(3/2)^ix^{2i}}{i!}}\) do I treat (3/2)^i as a constant or I can't do that?

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\displaystyle \int e^{(3/2)x^2}dx=\sum_{i=0}^{\infty}\frac{(3/2)^ix^{2i+1}}{i!(2i+1)}+C}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0doesn't matter, I guess I get the method, I am just too stupid to come up with an adequate example for me.

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\displaystyle y'+y=y^n}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\displaystyle y'y^{n}+y^{n+1}=1}\) \(\large \color{black}{\displaystyle \frac{1}{1n}v'=y'y^{n},~~~~~v=y^{n+1}=1}\) \(\large \color{black}{\displaystyle \frac{1}{1n}v'+v=1}\) \(\large \color{black}{\displaystyle v'+(1n)v=1n}\) \(\large \color{black}{\displaystyle v'e^{x(1n)}+e^{x(1n)}(1n)v=e^{x(1n)}(1n)}\) \(\large \color{black}{\displaystyle ve^{x(1n)}=\int e^{x(1n)}(1n)dx}\)
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