idku
  • idku
let me try another Bernoulli DE (example 3)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
idku
  • idku
\(\large y'+e^{-x}y=y^7\)
idku
  • idku
\(\large y'y^{-7}+e^{-x}y^{-6}=1\) \(\large\dfrac{-1}{6}v'+e^{-x}v=1\) \(\large v'-6e^{-x}v=-6\) \(\large v'e^{e^{6x}}-e^{e^{6x}}6e^{-x}v=-6e^{e^{6x}}\)
idku
  • idku
\(\large ve^{e^{6x}}=\int -6e^{e^{6x}}dx\) well, I guess I am not yet able to integrate this, but \(\large y^{-6}=\dfrac{\int -6e^{e^{6x}}dx}{e^{e^{6x}}}\) \(\large y^{}=\dfrac{\left(e^{e^{6x}}\right)^6}{\left(\int -6e^{e^{6x}}dx\right)^6}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

idku
  • idku
Can someone give me a better example?
idku
  • idku
\(\large y'-(1/x)y=y^4\)
Empty
  • Empty
Two things! Slight mistake on exponents here: \(\large y^{}=\dfrac{\left(e^{e^{6x}}\right)^{1/6}}{\left(\int -6e^{e^{6x}}dx\right)^{1/6}}\) Also, if you really needed this integral, you could use the taylor series: \[\ e^{e^{6x}}=\sum_{n=0}^\infty \frac{e^{6nx}}{n!}\] Then integrate term by term, until you have a reasonable enough approximation for whatever application you want to use it for, cause like let's face it in the real world 10 digits of accuracy is not bad at all haha
idku
  • idku
yes those are 1/6's
idku
  • idku
I didn't know about that taylor series. Something to work on in my little math world.
idku
  • idku
making a sandbox -:(
idku
  • idku
I will do that taylor series at some other time, I want to digest the Bernoulli technique first.
Empty
  • Empty
Oh you probably already know this power series! \[e^y=\sum_{n=0}^\infty \frac{y^n}{n!}\] \[y=e^{6x}\]
idku
  • idku
oh just a sub, I am so ...don';t want to violate the site's policy.
Empty
  • Empty
Haha hey at least it turned out being easier than you thought rather than it being harder than you thought! xD
idku
  • idku
I should have figured that (after having taken calc 2)
idku
  • idku
thanks.
idku
  • idku
I will proceed to my next example I started to do then for now.....
idku
  • idku
\(\large y'y^{-4}-(1/x)y^{-3}=1\) \(\large v'/(-3)-(1/x)v=1\) \(\large v'+(3/x)v=3\) \(\large v'e^{3\ln x}+e^{3\ln x}(3/x)v=e^{3\ln x}\) \(\large ve^{3\ln x}=e^{3\ln x}\) \(\large ve^{3\ln x}=\int x^3 dx \) \(\large vx^3=(1/4)x^4+C \) \(\large v=\dfrac{\frac{1}{4}x^4+C}{x^3}\) \(\large y^{-3}=\dfrac{\frac{1}{4}x^4+C}{x^3}\) \(\large y=\dfrac{(\frac{1}{4}x^4+C)^{3}}{x^9}\)
idku
  • idku
I left out the integral in line 5
idku
  • idku
but I fixed it in line 6
Empty
  • Empty
Are you verifying your solutions by plugging them back into the original differential equation?

Looking for something else?

Not the answer you are looking for? Search for more explanations.