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idku
 one year ago
let me try another Bernoulli DE (example 3)
idku
 one year ago
let me try another Bernoulli DE (example 3)

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idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large y'+e^{x}y=y^7\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large y'y^{7}+e^{x}y^{6}=1\) \(\large\dfrac{1}{6}v'+e^{x}v=1\) \(\large v'6e^{x}v=6\) \(\large v'e^{e^{6x}}e^{e^{6x}}6e^{x}v=6e^{e^{6x}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large ve^{e^{6x}}=\int 6e^{e^{6x}}dx\) well, I guess I am not yet able to integrate this, but \(\large y^{6}=\dfrac{\int 6e^{e^{6x}}dx}{e^{e^{6x}}}\) \(\large y^{}=\dfrac{\left(e^{e^{6x}}\right)^6}{\left(\int 6e^{e^{6x}}dx\right)^6}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1Can someone give me a better example?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Two 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
 one year ago
Best ResponseYou've already chosen the best response.1I didn't know about that taylor series. Something to work on in my little math world.

idku
 one year ago
Best ResponseYou've already chosen the best response.1I will do that taylor series at some other time, I want to digest the Bernoulli technique first.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Oh you probably already know this power series! \[e^y=\sum_{n=0}^\infty \frac{y^n}{n!}\] \[y=e^{6x}\]

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh just a sub, I am so ...don';t want to violate the site's policy.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Haha hey at least it turned out being easier than you thought rather than it being harder than you thought! xD

idku
 one year ago
Best ResponseYou've already chosen the best response.1I should have figured that (after having taken calc 2)

idku
 one year ago
Best ResponseYou've already chosen the best response.1I will proceed to my next example I started to do then for now.....

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large y'y^{4}(1/x)y^{3}=1\) \(\large v'/(3)(1/x)v=1\) \(\large v'+(3/x)v=3\) \(\large v'e^{3\ln x}+e^{3\ln x}(3/x)v=e^{3\ln x}\) \(\large ve^{3\ln x}=e^{3\ln x}\) \(\large ve^{3\ln x}=\int x^3 dx \) \(\large vx^3=(1/4)x^4+C \) \(\large v=\dfrac{\frac{1}{4}x^4+C}{x^3}\) \(\large y^{3}=\dfrac{\frac{1}{4}x^4+C}{x^3}\) \(\large y=\dfrac{(\frac{1}{4}x^4+C)^{3}}{x^9}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1I left out the integral in line 5

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Are you verifying your solutions by plugging them back into the original differential equation?
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