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.

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! 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: $(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$

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’s equations as we can solve those cases with earlier tools.

$y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$ substituting in we get: $y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$

and we get a linear equation again: (Handy formula if you wanna solve specific Bernoulli equations quick.) $$\frac{1}{1-n}\frac{ du }{ dx }+P(x)=Q(x)\quad \Box$$

Examples of Bernoulli’s equation:

ex de_b_type1 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 $y'+y=x^2y^2$ $y’y^{-2}+y^{-1}=x^2$

Note that we lost the y(x)=0 solution here, we will have to add it back in the end.

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 & 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 $\int fg' , dx=fg-\int f’g , dx$ 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 $y(x)=0$

end of lecture 3

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