forked from Sasserisop/MATH201
61 lines
3.0 KiB
Markdown
61 lines
3.0 KiB
Markdown
# Bernoulli's equation:
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### $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad\quad n\in\mathbb{R},\quad n\ne0,1$$
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>I'm calling this #de_b_type1. This is in standard form btw.
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It looks almost like a linear equation! In fact if n=0 it is by definition. We will see further that if n=1 you get a separable equation. So we ignore the cases when $n=0,1$ as these can be solved with prior tools.
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Bernoulli's equations are important as you will see it in biology and in engineering.
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If y is + then y(x)=0 is a solution to the equation:
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$\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$
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Let's move the y to the LHS:
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$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$
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notice that y(x)=0 is no longer a solution! It was lost due to dividing by zero. So from here on out we will have to remember to add it back in our final answers.
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let $y^{1-n}=u$
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Differentiating this with respect to x gives us:
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$(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$
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$y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$
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substituting in we get:
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$y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$
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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.)
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$$\frac{1}{1-n}\frac{ du }{ dx }+P(x)u=Q(x)\quad \Box$$
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>Remember when I said that when n=1 the equation becomes a separable equation?:
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>$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$
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>let $n=1$
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>$y^{-1}\frac{ dy }{ dx }+P(x)=Q(x)$
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>$y^{-1}dy=dx(Q(x)-P(x))$ <-This is indeed a separable equation #de_s_type1
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---
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# Examples of Bernoulli's equation:
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#ex #de_b_type1 Find the general solution to:
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$y'+y=(xy)^2$
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Looks like a Bernoulli equation because when we distribute the $^2$ we get $x^2y^2$ on the RHS. This also tells us that n=2
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$y'+y=x^2y^2$
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$y'y^{-2}+y^{-1}=x^2$
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>Note that we lost the y(x)=0 solution here, we will have to add it back in the end.
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let $u=y^{1-n}=y^{-1}$
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Differentiating wrt. x we get: $\frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}$
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$y^{-2}{\frac{dy}{dx}=-\frac{ du }{ dx }}$
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$y^{-2}{\frac{dy}{dx}+y^{-1}=-\frac{ du }{ dx }}+y^{-1}$
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${x^2=-\frac{ du }{ dx }}+y^{-1}$
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$x^2=-\frac{du}{dx}+u$
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$\frac{du}{dx}-u=-x^2$
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Yay we have a linear equation now! We can solve it using the techniques & formulas we learned for them.
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let $P(x)=-1 \quad Q(x)=-x^2 \qquad I(x)=e^{\int -1 \, dx}=e^{-x}$
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$u=-e^{x}\int e^{-x}x^2 \, dx$
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How to integrate this? You can use integration by parts:
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LIATE: log, inv trig, alg, trig, exp
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$\int fg' \, dx=fg-\int f'g \, dx$
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let $f=x^2 \qquad f'=2x \qquad g'=e^{-x} \qquad g=-e^{-x}$
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$u=-e^{x}\left( x^2(-e^{-x})-\int 2x(-e^{-x}) \, dx \right)$
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$u=-e^{x}\left( -x^2e^{-x}+2\int xe^{-x} \, dx \right)$
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let $f=x \qquad f'=1 \qquad g'=e^{-x} \qquad g=-e^{-x}$
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$u=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-\int -e^{-x} \, dx \right) \right)$
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$\frac{1}{y}=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-e^{-x} +C\right) \right)$
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$\frac{1}{y}=x^2+2(x+1+Ce^x)$
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$\frac{1}{y}=x^2+2x+2+Ce^x$
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The general solution to the DE is:
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$$y(x)=\frac{1}{x^2+2x+2+Ce^x} \quad\text{as well as}\quad y(x)=0$$
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---
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#end of lecture 3 |