let me try another Bernoulli DE (example 3)

Question

idku · Answer

$\large y'+e^{-x}y=y^7$

idku · Answer

$\large y'y^{-7}+e^{-x}y^{-6}=1$


$\large\dfrac{-1}{6}v'+e^{-x}v=1$

$\large v'-6e^{-x}v=-6$

$\large v'e^{e^{6x}}-e^{e^{6x}}6e^{-x}v=-6e^{e^{6x}}$

idku · Answer

$\large ve^{e^{6x}}=\int -6e^{e^{6x}}dx$

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 · Answer

Can someone give me a better example?

idku · Answer

$\large y'-(1/x)y=y^4$

empty · Answer

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 · Answer

yes those are 1/6's

idku · Answer

I didn't know about that taylor series.
Something to work on in my little math world.

idku · Answer

making a sandbox -:(

idku · Answer

I will do that taylor series at some other time, I want to digest the Bernoulli technique first.

empty · Answer

Oh you probably already know this power series! 

$$e^y=\sum_{n=0}^\infty \frac{y^n}{n!}$$
$$y=e^{6x}$$

idku · Answer

oh just a sub, I am so ...don';t want to violate the site's policy.

empty · Answer

Haha hey at least it turned out being easier than you thought rather than it being harder than you thought! xD

idku · Answer

I should have figured that (after having taken calc 2)

idku · Answer

thanks.

idku · Answer

I will proceed to my next example I started to do then for now.....

idku · Answer

$\large y'y^{-4}-(1/x)y^{-3}=1$

$\large v'/(-3)-(1/x)v=1$

$\large v'+(3/x)v=3$

$\large v'e^{3\ln x}+e^{3\ln x}(3/x)v=e^{3\ln x}$

$\large ve^{3\ln x}=e^{3\ln x}$

$\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 · Answer

I left out the integral in line 5

idku · Answer

but I fixed it in line 6

empty · Answer

Are you verifying your solutions by plugging them back into the original differential equation?