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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. -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.
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$y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$ -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.) -$$\frac{1}{1-n}\frac{ du }{ dx }+P(x)=Q(x)\quad \Box$$
-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.
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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
- - - - - - - -#start of lecture 2
-$\frac{dy}{dt}=f\left( \frac{y}{t} \right)$ (I’m calling this de_h_type1) -let $u=\frac{y}{t}$ $y=tu \quad \frac{dy}{dt}=u+t\frac{du}{dt}$ -so $\frac{dy}{dt}=f(u)=u+t{\frac{du}{dt}}$ -The homogenous equation has been converted into a separable DE! -$\frac{du}{dt}=\frac{f(u)-u}{t}$ -$\frac{du}{f(u)-u}=\frac{dt}{t}$
-$\frac{dy}{dx}=G(ax+by)\quad \text{where a, b }\in \mathbb{R}$ (I’m calling this de_h_type2) -Then, let $u=ax+by$ -$\frac{du}{dx}=a+b{\frac{dy}{dx}}$ -$\frac{dy}{dx}=\frac{1}{b}{\frac{du}{dx}}-\frac{a}{b}=G(u)$ -Again, the homogenous equation has been converted to a separable DE! -$dx=\frac{du}{b{G(u)+\frac{a}{b}}}$ -Just integrate both sides as usual and you’re chilling.
-ex -de_h_type1:$\frac{dy}{dx}=\frac{{x+y}}{x-y} \quad x>y\quad\text{This condition is added so the denominator}\ne 0$
-but $\frac{{x+y}}{x-y}\ne f(\frac{y}{x})$… Or is it? How can this be written as a homogenous equation? -divide the top and bottom by x: -$\frac{dy}{dx}=\frac{{1+\frac{y}{x}}}{1-\frac{y}{x}}$ -Yay! now it’s a function of $\frac{y}{x}$ -let $u=\frac{y}{x} \quad \frac{dy}{dx}=u+x{\frac{du}{dx}}$ -$\frac{dy}{dx}=\frac{1+u}{1-u}=u+x{\frac{du}{dx}}$ -$\frac{dx({f(u)-u})}{x}=du$ -$\frac{dx}{x}=\frac{du}{{f(u)-u}}$
---That’s odd, why is it not $\frac{du}{f(u)-u}=\frac{x}{dx}$? I got this by moving the top over. -(it’s because you must move all multiplicative factors when using this technique of moving the top. Be careful!)
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$\int\frac{dx}{x}=\int\frac{du}{{f(u)-u}}$ -$\ln\mid x\mid=\int \frac{du}{\frac{{1+u}}{1-u}-u}$ -$\ln\mid x\mid=\int \frac{du}{\frac{{1+u-u+u^2}}{1-u}}$ -$\ln\mid x\mid=\int \frac{1-u}{{1+u^2}}du$
---let $1+u^2=v \quad dv=2udu$ -$=\int \frac{{1-u}}{v} , du$ Gah, doesn’t work. I didn’t notice I could split the integral up first.
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$\ln\mid x\mid=\int \frac{1}{{1+u^2}},du-\int \frac{u}{1+u^2} , du=\arctan\left( \frac{y}{x} \right)+C-I_{0}$ -for $I_{0}$ let $v=1+u^2 \quad dv=2udu$ -$I_{0}=\int \frac{u}{v} , \frac{dv}{2u}=\frac{1}{2}\int \frac{dv}{v}=\frac{1}{2}\ln(1+u^2)$
---^Note no abs value needed in the $\ln()$ as $1+u^2$ is always +
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$\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln(1+u^2)$ -$\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln\left( 1+\frac{y^2}{x^2} \right)$ -$\mid x\mid=e^{\arctan(\frac{y}{x})+C-\ln(\sqrt{ 1+y^2/x^2 })}$ -$x=\frac{e^{\arctan(y/x)}A}{\sqrt{ 1+\frac{y^2}{x^2} }}$ -$x\sqrt{ 1+\frac{y^2}{x^2}} ={e^{\arctan(y/x)}A}$ -So the final general solution to the problem is:
-ex -de_h_type2: $$(2x-2y-1)dx+(x-y+1)dy=0$$ -Can we write it in the form $\frac{dy}{dx}=G(ax+by)$? -$(x-y+1)dy=-(2x-2y-1)dx$ -$\frac{dy}{dx}=\frac{{2y+1-2x}}{x-y+1}$ -factor out a -2? -$\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$ -Yep! looks like a -de_h_type2 -let $u=x-y$ -$\frac{du}{dx}=1-\frac{dy}{dx}$ -$1-\frac{du}{dx}=\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$
---Obviously we don’t work with x and y as I was entailing above, substitute $u=x-y$ in you silly goose.
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$1-\frac{du}{dx}=-2\frac{{u-\frac{1}{2}}}{u+1}$ -$\frac{du}{dx}=2\frac{{u-\frac{1}{2}}}{u+1}+1$ -$\frac{du}{dx}=\frac{2u-1}{u+1}+1$ -$\frac{du}{dx}=\frac{{2u-1+u+1}}{u+1}$ -$\frac{du}{dx}=\frac{3u}{u+1}$ -$\frac{(u+1)du}{3u}=dx$ -$\int \frac{(u+1)du}{3u}=\int dx$
---$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=\ln\mid x\mid+C$ -Ah, I made a mistake. $\int dx \ne \ln\mid x\mid+C$
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$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=x+C$
---Okay, now that we have integrated, we can start talking in terms of x and y again
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$\frac{x-y}{3}+\frac{1}{3}\ln\mid x-y\mid = x+C$ -$x-y+\ln\mid x-y\mid=3x+C$ -$\ln\mid x-y\mid=C+y+2x$ < this is where he moved the C to the left -$\mid x-y\mid=e^Ce^ye^{2x}$ -$x-y=Ae^ye^{2x}$ -$A(x-y)=e^{y+2x}$
---I know that above step looks illegal, but the prof did this (indirectly, he moved C to the LHS in a prior step without regarding it’s sign). I wonder what happens if A was 0 though? Do we get divide by zero errors? Thinking about it more, we are changing $x-y=0$ to $e^{y+2x}=0$ when $A=0$ The first one has a solution (y=x) the second loses that solution because of ln(0) issues (gives a function that’s undefined for all x). when checking y(x)=x in the DE, it is a valid solution. So it is an illegal step! Because we lost a valid solution. I’ll have to check with the prof. -Interestingly, if we act like $e^{y+2x}=0$ is defined, we get $\frac{dy}{dx}=-2$
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Proof: -$\lim_{ n \to 0 }e^{y+2x}=n$ -$\lim_{ n \to 0 }\ln(n)=y+2x$ -$\lim_{ n \to 0 }\frac{d}{dx}\ln(n)=0=\frac{dy}{dx}+2$ -$\frac{dy}{dx}=-2$
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so from $\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$ we get: -$-2=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$ -$x-y+1=x-y-\frac{1}{2}$ -$1=-\frac{1}{2}$ -So what does this all mean? I honestly have no idea. I think it means we assumed that $e^{y+2x}=0$ is defined and because we arrived at a contradiction, our assumption was wrong. That didn’t really get us to show if it was a valid solution or not like I imagined.
-We can rearrange to our liking, but we have found the general solution to the DE:
-I have written these notes for myself, I thought it would be cool to share them. These notes may be inaccurate, incomplete, or incoherent. No warranty is expressed or implied. Reader assumes all risk and liabilities.
- - - - - - - -$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$ -imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!
-so can we make them 0? -let $x=u+k$ -$y=v+l$ -where $k,l$ are constants -$(a_{1}u+b_{1}vK+\underbrace{\cancel{ c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l } }_{ 0 })dv=0$ -$a_{1}k+b_{1}l=-c_1$ -$a_{2}k+b_{2}l=-c_{2}$ -if $\det(a_{1},b_{1},a_{2},b_{2})\ne 0$ turn into homogenous -if $\det(\dots)=0 \Rightarrow$ equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)
-ex de_LC_type1 -$$(-3x+y+6)dx+(x+y+2)dy=0$$ -let $x=u+k$ -$y=v+l$ -$(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$ -we want $6-3k+l$ and $2+k+l$ to equal 0 -so: -$-3k+l=-6$ -$k+l=-2$ -$det(-3,1,1,1)=-4$ //he can call it a dinosaur if he wanted to :D -solving gives us: -$k=1,l=-3$ -so $x=u+1 \quad y=v-3$ -$(-3u+v)du+(u+v)dv=0$ //Beutiful1! it’s homogenous now -$\frac{ dv }{ du }=\frac{{3u-v}}{u+v}=\frac{{3-\frac{v}{u}}}{1+\frac{v}{u}}$ -$\frac{v}{u}=w \quad v=uw \quad \frac{ dv }{ du }=w+u\frac{ dw }{ du }$ -$w+u\frac{ dw }{ du }=\frac{{3-w}}{1+w}$ This is the equation we have to solve -$u\frac{ dw }{ du }=\frac{{3-2w-w^2}}{1+w}$ -$-\frac{{w+1}}{w^2+2w-3}dw=\frac{du}{u}$ -$\int-\frac{{w+1}}{w^2+2w-3}dw=\int\frac{du}{u}$ -let $z=w^2+2w-3$ -$dz=2(w+1)dw$ -$\frac{1}{2}\int \frac{dz}{z}=\ln\mid u\mid^{-1}$ -$\ln\mid z\mid^{1/2}-\ln\mid u\mid^{-1}=C$ -$\ln(\mid z\mid^{1/2}\mid u\mid)=C$ -$\mid z\mid^{1/2}u=e^C$ -$\mid z\mid u^2=e^{2C}$ -$zu^2=A$ -$\left( \left( \frac{v}{u} \right)^2+\frac{2v}{u}-3 \right)u^2=A$ -remember $u=x-1 \quad v=y+3$ -$$\left( \left( \frac{{y+3}}{x-1} \right)^2+\frac{2(y+3)}{x-1}-3 \right)(x-1)^2=A$$ -you can “simplify” it to: $(y+3)^2+2(y+3)(x-1)-3(x-1)^2=A$
-two variable equations -$dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ suppose it equals to zero (as shown in the equation) you get a horizontal plane (a constant) -so $F(x,y)=C$ -the solution to these exact equations is given by $F()$ but how do we get F from the derivatives? -Equation of the form: $$M(x,y)dx=N(x,y)dy=0$$ -is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$ -then differentiating we get: -$\frac{ \partial M }{ \partial y }=\frac{ \partial^{2} F }{ \partial y\partial x }$ -$\frac{ \partial N }{ \partial x }=\frac{ \partial^{2} F }{ \partial x\partial y }$ Order of going in x then y vs y then x doesn’t matter as it lands you on the same point (idk how this is related yet) -Exact equation$\Rightarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ if it’s continuous (?) -also: Exact equation$\Leftarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ -Test for exactness: -exact $\iff \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ (this can be proved, but it wasnt proved in class) -end of lecture 4
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