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$$\frac{ dy }{ dx } &#43;P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$
(I&rsquo;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&rsquo;s …">
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If y is &#43; then y(x)=0 is a solution to the equation: $\frac{dy}{dx}&#43;0=0\quad\Rightarrow \quad0=0$ Let&rsquo;s move the y to the LHS: $y^{-n}\frac{ dy }{ dx }&#43;P(x)y^{1-n}=Q(x)$ notice that y(x)=0 is no longer a solution!" />
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<meta name="twitter:description" content="Bernoulli&rsquo;s equation: $$\frac{ dy }{ dx } &#43;P(x)y=Q(x)y^n \quad,\quad n\in\mathbb{R}$$ (I&rsquo;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&rsquo;s equations are important as you will see it in biology and in engineering.
If y is &#43; then y(x)=0 is a solution to the equation: $\frac{dy}{dx}&#43;0=0\quad\Rightarrow \quad0=0$ Let&rsquo;s move the y to the LHS: $y^{-n}\frac{ dy }{ dx }&#43;P(x)y^{1-n}=Q(x)$ notice that y(x)=0 is no longer a solution!"/>
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<h1 id="bernoullis-equation">Bernoulli&rsquo;s equation:</h1>
<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>
<p>(I&rsquo;m calling this
<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.
Bernoulli&rsquo;s equations are important as you will see it in biology and in engineering.</p>
<p>If y is + then y(x)=0 is a solution to the equation:
$\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$
Let&rsquo;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! 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.
let $y^{1-n}=u$
Differentiating this with respect to x gives us:
$(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$</p>
<blockquote>
<p>notice that when n=1 the above turns into a linear equation:
$0=\frac{ du }{ dx }$
$y^{1-n}=u=0+C$
1=C
Hold on I dont think I did the above correctly. Anyways.
So we consider that $n\ne 0,1$ for Bernoulli&rsquo;s equations as we can solve those cases with earlier tools.</p>
</blockquote>
<p>$y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$
substituting in we get:
$y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$</p>
<p>and we get a linear equation again: (Handy formula if you wanna solve specific Bernoulli equations quick.)
$$\frac{1}{1-n}\frac{ du }{ dx }+P(x)=Q(x)\quad \Box$$</p>
<hr>
<h1 id="examples-of-bernoullis-equation">Examples of Bernoulli&rsquo;s equation:</h1>
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
<a class="hashtag" onclick="focusTag(this)">de_b_type1</a> Find the general solution to:
$y'+y=(xy)^2$
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
$y'+y=x^2y^2$
$y&rsquo;y^{-2}+y^{-1}=x^2$</p>
<blockquote>
<p>Note that we lost the y(x)=0 solution here, we will have to add it back in the end.</p>
</blockquote>
<p>let $u=y^{1-n}=y^{-1}$
Differentiating wrt. x we get: $\frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}$
$y^{-2}{\frac{dy}{dx}=-\frac{ du }{ dx }}$
$y^{-2}{\frac{dy}{dx}+y^{-1}=-\frac{ du }{ dx }}+y^{-1}$
${x^2=-\frac{ du }{ dx }}+y^{-1}$
$x^2=-\frac{du}{dx}+u$
$\frac{du}{dx}-u=-x^2$
Yay we have a linear equation now! We can solve it using the techniques &amp; formulas we learned for them.
let $P(x)=-1 \quad Q(x)=-x^2 \qquad I(x)=e^{\int -1 , dx}=e^{-x}$
$u=-e^{x}\int e^{-x}x^2 , dx$
LIATE log, inv, alg, trig, exp
$\int fg' , dx=fg-\int f&rsquo;g , dx$
let $f=x^2 \qquad f'=2x \qquad g'=e^{-x} \qquad g=-e^{-x}$
$u=-e^{x}\left( x^2(-e^{-x})-\int 2x(-e^{-x}) , dx \right)$
$u=-e^{x}\left( -x^2e^{-x}+2\int xe^{-x} , dx \right)$
let $f=x \qquad f'=1 \qquad g'=e^{-x} \qquad g=-e^{-x}$
$u=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-\int -e^{-x} , dx \right) \right)$
$\frac{1}{y}=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-e^{-x} +C\right) \right)$
$\frac{1}{y}=x^2+2(x+1+Ce^x)$
$\frac{1}{y}=x^2+2x+2+Ce^x$
The general solution to the DE is:
$y(x)=\frac{1}{x^2+2x+2+Ce^x}$ as well as $y(x)=0$</p>
<hr>
<p><a class="hashtag" onclick="focusTag(this)">end</a> of lecture 3</p>
<h3>Referenced in</h3>
<ul>
<li>No backlinks found</li>
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