#start of lecture 2

2 new tricks: homogenous and linear DE

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:$\frac{dy}{dx}=\frac{{x+y}}{x-y} \quad x>y\quad\text{This condition is added so the denominator}\ne 0$

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!)

$\int\frac{dx}{x}=\int\frac{du}{{f(u)-u}}$ $\ln\mid x\mid=\int \frac{du}{\frac{{1+u}}{1-u}-u}$ $\ln\mid x\mid=\int \frac{du}{\frac{{1+u-u+u^2}}{1-u}}$ $\ln\mid x\mid=\int \frac{1-u}{{1+u^2}}du$

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:

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

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 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: $\lim_{ n \to 0 }e^{y+2x}=n$ $\lim_{ n \to 0 }\ln(n)=y+2x$ $\lim_{ n \to 0 }\frac{d}{dx}\ln(n)=0=\frac{dy}{dx}+2$ $\frac{dy}{dx}=-2$

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:

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

Referenced in