Another Bernoulli DE (eq 2)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Another Bernoulli DE (eq 2)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\(\large \color{black}{y'+\ln(x)y=y^{2}}\) \(\large \color{black}{y'y^{-2}+\ln(x)y^{-1}=1}\) Substitution \(\large \color{black}{v=y^{-1}}\) \(\large \color{black}{-v'=y'y^{-2}}\) DE becomes, \(\large \color{black}{-v'+v\ln(x)=1}\)
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?
\(\large \color{black}{-v'e^{x\ln(x)-x}+ve^{x\ln(x)-x}\ln(x)=e^{x\ln(x)-x}}\) \(\large \color{black}{ve^{x\ln(x)-x}=\int e^{x\ln(x)-x}dx}\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question