forked from Sasserisop/MATH201
539 lines
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539 lines
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<meta name="description" content="Bernoulli’s equation:
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$$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$
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(I’m calling this #de_b_type1) 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 also still get a linear equation.
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Bernoulli’s …">
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<meta property="og:description" content="Bernoulli’s equation: $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$ (I’m calling this #de_b_type1) 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 also still get a linear equation. 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: $\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$ Let’s move the y to the LHS: $y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$ notice that y(x)=0 is no longer a solution!" />
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<meta name="twitter:description" content="Bernoulli’s equation: $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$ (I’m calling this #de_b_type1) 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 also still get a linear equation. 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: $\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$ Let’s move the y to the LHS: $y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$ notice that y(x)=0 is no longer a solution!"/>
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<h1 id="bernoullis-equation">Bernoulli’s equation:</h1>
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<h3 id="frac-dy--dx--pxyqxyn-quadquad-ninmathbbr">$$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$</h3>
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<p>(I’m calling this
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<a class="hashtag" onclick="focusTag(this)">de_b_type1)</a> 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 also still get a linear equation.
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Bernoulli’s equations are important as you will see it in biology and in engineering.</p>
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<p>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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$(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$</p>
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<blockquote>
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<p>notice that when n=1 the above turns into a linear equation:
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$0=\frac{ du }{ dx }$
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$y^{1-n}=u=0+C$
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1=C
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Hold on I dont think I did the above correctly. Anyways.
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So we consider that $n\ne 0,1$ for Bernoulli’s equations as we can solve those cases with earlier tools.</p>
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</blockquote>
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<p>$y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$
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substituting in we get:
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$y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$</p>
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<p>and we get a linear equation again: (Handy formula if you wanna solve specific Bernoulli equations quick.)
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$$\frac{1}{1-n}\frac{ du }{ dx }+P(x)=Q(x)\quad \Box$$</p>
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<hr>
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<h1 id="examples-of-bernoullis-equation">Examples of Bernoulli’s equation:</h1>
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<p><a class="hashtag" onclick="focusTag(this)">ex</a>
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<a class="hashtag" onclick="focusTag(this)">de_b_type1</a> 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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$y’y^{-2}+y^{-1}=x^2$</p>
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<blockquote>
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<p>Note that we lost the y(x)=0 solution here, we will have to add it back in the end.</p>
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</blockquote>
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<p>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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LIATE log, inv, 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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$y(x)=\frac{1}{x^2+2x+2+Ce^x}$ as well as $y(x)=0$</p>
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<hr>
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<p><a class="hashtag" onclick="focusTag(this)">end</a> of lecture 3</p>
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<h3>Referenced in</h3>
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<ul>
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<li>No backlinks found</li>
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