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fixed bernoulli and renamed heaviside link

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Sasserisop 2023-10-19 19:53:20 -06:00
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content/Heaviside Unit step function (lec 18).md → content/(Heaviside) Unit step function (lec 18).md
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content/Bernoulli equations (lec 3).md
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@ -17,8 +17,8 @@ $y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$
substituting in we get: substituting in we get:
$y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$ $y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$
and we get a linear equation again: (Handy formula if you wanna solve specific Bernoulli equations quick.) and we get a linear equation again: (Handy formula if you wanna solve Bernoulli equations quick. Just remember that once you find $u(x)$, substitute it back for $y(x)^{1-n}=u(x)$ to get your solution for y.)
$$\frac{1}{1-n}\frac{ du }{ dx }+P(x)=Q(x)\quad \Box$$ $$\frac{1}{1-n}\frac{ du }{ dx }+P(x)u=Q(x)\quad \Box$$
>Remember when I said that when n=1 the equation becomes a separable equation?: >Remember when I said that when n=1 the equation becomes a separable equation?:
>$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$ >$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$
>let $n=1$ >let $n=1$