Another Bernoulli DE (eq 2)

Question

idku · Answer

$\large \color{black}{y'+\ln(x)y=y^{2}}$

$\large \color{black}{y'y^{-2}+\ln(x)y^{-1}=1}$

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 · Answer

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 · Answer

$\large \color{black}{-v'e^{x\ln(x)-x}+ve^{x\ln(x)-x}\ln(x)=e^{x\ln(x)-x}}$

$\large \color{black}{ve^{x\ln(x)-x}=\int e^{x\ln(x)-x}dx}$

idku · Answer

$\large \color{black}{e^x=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}}$

is it possible to use this to get the integral of the right side?

anonymous · Answer

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 · Answer

but using just that power series alone, no, that will not help

idku · Answer

well, I guess I am ought to try another example that I can chew:)

idku · Answer

$\large \color{black}{y'+xy=y^{4}}$

idku · Answer

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 · Answer

$\large \color{black}{v'-3xv=-3}$

idku · Answer

$\large \color{black}{v'e^{(-3/2)x^2}-3xe^{(-3/2)x^2}v=-3e^{(-3/2)x^2}}$

idku · Answer

$\large \color{black}{ve^{(-3/2)x^2}=\int -3e^{(-3/2)x^2}dx}$

idku · Answer

$\large \color{black}{\displaystyle e^x=\sum_{i=0}^{\infty}\frac{x^i}{i!}}$

$\large \color{black}{\displaystyle e^{(-3/2)x^2}=\sum_{i=0}^{\infty}\frac{\left((-3/2)x^2\right)^i}{i!}}$

am i correct so far?

idku · Answer

$\large \color{black}{\displaystyle e^{(-3/2)x^2}=\sum_{i=0}^{\infty}\frac{(-3/2)^ix^{2i}}{i!}}$

do I treat (-3/2)^i as a constant or I can't do that?

idku · Answer

$\large \color{black}{\displaystyle \int e^{(-3/2)x^2}dx=\sum_{i=0}^{\infty}\frac{(-3/2)^ix^{2i+1}}{i!(2i+1)}+C}$

idku · Answer

doesn't matter, I guess I get the method, I am just too stupid to come up with an adequate example for me.

idku · Answer

$\large \color{black}{\displaystyle y'+y=y^n}$

idku · Answer

$\large \color{black}{\displaystyle y'y^{-n}+y^{-n+1}=1}$

$\large \color{black}{\displaystyle \frac{1}{1-n}v'=y'y^{-n},~~~~~v=y^{-n+1}=1}$

$\large \color{black}{\displaystyle \frac{1}{1-n}v'+v=1}$

$\large \color{black}{\displaystyle v'+(1-n)v=1-n}$

$\large \color{black}{\displaystyle v'e^{x(1-n)}+e^{x(1-n)}(1-n)v=e^{x(1-n)}(1-n)}$

$\large \color{black}{\displaystyle ve^{x(1-n)}=\int e^{x(1-n)}(1-n)dx}$

idku · Answer

just nvm

idku · Answer

i suck