MATH201/content/Homogenous equations (lec 2).md

#start of lecture 2

Homogenous equations:

\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}

Another way you can write a homogenous equation:

\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.

Examples of homogenous equations:

#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 

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!)

\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.

\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 +

\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:

\sqrt{ x^2+y^2 }=e^{\arctan\left( \frac{y}{x} \right)}A$$

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#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.

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

\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

\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 perhaps 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

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\quad \Box 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:

$x-y=Ae^{2x+y}$

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