135 lines
5.0 KiB
Markdown
135 lines
5.0 KiB
Markdown
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#start of lecture 2
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## Homogenous equations:
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$$\frac{dy}{dt}=f\left( \frac{y}{t} \right)$$
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>I'm calling this #de_h_type1
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let $u=\frac{y}{t}$ $y=tu \quad \frac{dy}{dt}=u+t\frac{du}{dt}$
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so $\frac{dy}{dt}=f(u)=u+t{\frac{du}{dt}}$
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The homogenous equation has been converted into a separable DE!
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$\frac{du}{dt}=\frac{f(u)-u}{t}$
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$\frac{du}{f(u)-u}=\frac{dt}{t}$
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## Another way you can write a homogenous equation:
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$$\frac{dy}{dx}=G(ax+by)\quad \text{where a, b }\in \mathbb{R}$$
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>I'm calling this #de_h_type2
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Then, let $u=ax+by$
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$\frac{du}{dx}=a+b{\frac{dy}{dx}}$
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$\frac{dy}{dx}=\frac{1}{b}{\frac{du}{dx}}-\frac{a}{b}=G(u)$
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Again, the homogenous equation has been converted to a separable DE!
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$dx=\frac{du}{b{G(u)+\frac{a}{b}}}$
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Just integrate both sides as usual and you're chilling.
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## Examples of homogenous equations:
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#ex #de_h_type1
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$$\frac{dy}{dx}=\frac{{x+y}}{x-y} \quad x>y\quad\text{This condition is added so the denominator}\ne 0$$
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but $\frac{{x+y}}{x-y}\ne f(\frac{y}{x})$... Or is it? How can this be written as a homogenous equation?
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divide the top and bottom by x:
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$\frac{dy}{dx}=\frac{{1+\frac{y}{x}}}{1-\frac{y}{x}}$
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Yay! now it's a function of $\frac{y}{x}$
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let $u=\frac{y}{x} \quad \frac{dy}{dx}=u+x{\frac{du}{dx}}$
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$\frac{dy}{dx}=\frac{1+u}{1-u}=u+x{\frac{du}{dx}}$
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$\frac{dx({f(u)-u})}{x}=du$
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$\frac{dx}{x}=\frac{du}{{f(u)-u}}$
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>That's odd, why is it not $\frac{du}{f(u)-u}=\frac{x}{dx}$? I got this by moving the top over.
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(it's because you must move all multiplicative factors when using this technique of moving the top. Be careful!)
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$\int\frac{dx}{x}=\int\frac{du}{{f(u)-u}}$
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$\ln\mid x\mid=\int \frac{du}{\frac{{1+u}}{1-u}-u}$
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$\ln\mid x\mid=\int \frac{du}{\frac{{1+u-u+u^2}}{1-u}}$
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$\ln\mid x\mid=\int \frac{1-u}{{1+u^2}}du$
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>let $1+u^2=v \quad dv=2udu$
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$=\int \frac{{1-u}}{v} \, du$ Gah, doesn't work. I didn't notice I could split the integral up first.
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$\ln\mid x\mid=\int \frac{1}{{1+u^2}}\,du-\int \frac{u}{1+u^2} \, du=\arctan\left( \frac{y}{x} \right)+C-I_{0}$
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for $I_{0}$ let $v=1+u^2 \quad dv=2udu$
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$I_{0}=\int \frac{u}{v} \, \frac{dv}{2u}=\frac{1}{2}\int \frac{dv}{v}=\frac{1}{2}\ln(1+u^2)$
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> ^Note no abs value needed in the $\ln()$ as $1+u^2$ is always +
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$\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln(1+u^2)$
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$\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln\left( 1+\frac{y^2}{x^2} \right)$
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$\mid x\mid=e^{\arctan(\frac{y}{x})+C-\ln(\sqrt{ 1+y^2/x^2 })}$
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$x=\frac{e^{\arctan(y/x)}A}{\sqrt{ 1+\frac{y^2}{x^2} }}$
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$x\sqrt{ 1+\frac{y^2}{x^2}} ={e^{\arctan(y/x)}A}$
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So the final general solution to the problem is:
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#### $$\sqrt{ x^2+y^2 }=e^{\arctan\left( \frac{y}{x} \right)}A$$
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---
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#ex #de_h_type2 $$(2x-2y-1)dx+(x-y+1)dy=0$$
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Can we write it in the form $\frac{dy}{dx}=G(ax+by)$?
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$(x-y+1)dy=-(2x-2y-1)dx$
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$\frac{dy}{dx}=\frac{{2y+1-2x}}{x-y+1}$
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factor out a -2?
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$\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$
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Yep! looks like a #de_h_type2
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let $u=x-y$
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$\frac{du}{dx}=1-\frac{dy}{dx}$
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$1-\frac{du}{dx}=\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$
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>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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$1-\frac{du}{dx}=-2\frac{{u-\frac{1}{2}}}{u+1}$
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$\frac{du}{dx}=2\frac{{u-\frac{1}{2}}}{u+1}+1$
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$\frac{du}{dx}=\frac{2u-1}{u+1}+1$
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$\frac{du}{dx}=\frac{{2u-1+u+1}}{u+1}$
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$\frac{du}{dx}=\frac{3u}{u+1}$
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$\frac{(u+1)du}{3u}=dx$
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$\int \frac{(u+1)du}{3u}=\int dx$
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>$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=\ln\mid x\mid+C$
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>Ah, I made a mistake. $\int dx \ne \ln\mid x\mid+C$
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$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=x+C$
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> Okay, now that we have integrated, we can start talking in terms of x and y again
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$\frac{x-y}{3}+\frac{1}{3}\ln\mid x-y\mid = x+C$
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$x-y+\ln\mid x-y\mid=3x+C$
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$\ln\mid x-y\mid=C+y+2x$ < this is where he moved the C to the left
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$\mid x-y\mid=e^Ce^ye^{2x}$
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$x-y=Ae^ye^{2x}$
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$A(x-y)=e^{y+2x}$
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> 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 perhaps it is an illegal step! Because we lost a valid solution. I'll have to check with the prof.
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> Interestingly, if we act like $e^{y+2x}=0$ is defined, we get $\frac{dy}{dx}=-2$
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>
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>Proof:
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>$\lim_{ n \to 0 }e^{y+2x}=n$
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>$\lim_{ n \to 0 }\ln(n)=y+2x$
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>$\lim_{ n \to 0 }\frac{d}{dx}\ln(n)=0=\frac{dy}{dx}+2$
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>$\frac{dy}{dx}=-2\quad \Box$
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>plugging into the equation $(2x-2y-1)dx+(x-y+1)dy=0$ yields:
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>$1=\frac{2(x-y+1)}{2x-2y-1}$
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>$2x-2y-1=2(x-y+1)$
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>$-1=2$
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>So what does this all mean? I think it means that even if we imagine that $\frac{dy}{dx}$ exists, the equation is not satisfied and $y=x$ is definitely not a solution even when we try to cheat a little.
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We can rearrange to our liking, but we have found the general solution to the DE:
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#### $$x-y=Ae^{2x+y}$$
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---
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