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

1. 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, $$\large \color{black}{-v'+v\ln(x)=1}$$

2. 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?

3. idku

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

4. 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?

5. 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

6. anonymous

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

7. idku

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

8. idku

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

9. idku

ok, lets start I guess..... $$\large \color{black}{y'y^{-4}+xy^{-3}=1}$$ Substitution: $$\large \color{black}{v=y^{-3}}$$ $$\large \color{black}{\dfrac{-1}{3}v'=y'y^{-4}}$$ $$\large \color{black}{\dfrac{-1}{3}v'+xv=1}$$

10. idku

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

11. idku

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

12. idku

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

13. 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?

14. 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?

15. idku

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

16. idku

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

17. idku

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

18. idku

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

19. idku

just nvm

20. idku

i suck