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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 $u=\frac{y}{x} \quad \frac{dy}{dx}=u+x{\frac{du}{dx}}$ $\frac{dy}{dx}=\frac{1+u}{1-u}=u+x{\frac{du}{dx}}$ $\frac{dx({f(u)-u})}{x}=du$ $\frac{dx}{x}=\frac{du}{{f(u)-u}}$</p> <blockquote> <p>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!)</p> </blockquote> <p>$\int\frac{dx}{x}=\int\frac{du}{{f(u)-u}}$ $\ln\mid x\mid=\int \frac{du}{\frac{{1+u}}{1-u}-u}$ $\ln\mid x\mid=\int \frac{du}{\frac{{1+u-u+u^2}}{1-u}}$ $\ln\mid x\mid=\int \frac{1-u}{{1+u^2}}du$</p> <blockquote> <p>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.</p> </blockquote> <p>$\ln\mid x\mid=\int \frac{1}{{1+u^2}},du-\int \frac{u}{1+u^2} , du=\arctan\left( \frac{y}{x} \right)+C-I_{0}$ for $I_{0}$ let $v=1+u^2 \quad dv=2udu$ $I_{0}=\int \frac{u}{v} , \frac{dv}{2u}=\frac{1}{2}\int \frac{dv}{v}=\frac{1}{2}\ln(1+u^2)$</p> <blockquote> <p>^Note no abs value needed in the $\ln()$ as $1+u^2$ is always +</p> </blockquote> <p>$\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln(1+u^2)$ $\ln\mid x\mid=\arctan\left( \frac{y}{x} \right)+C-\frac{1}{2}\ln\left( 1+\frac{y^2}{x^2} \right)$ $\mid x\mid=e^{\arctan(\frac{y}{x})+C-\ln(\sqrt{ 1+y^2/x^2 })}$ $x=\frac{e^{\arctan(y/x)}A}{\sqrt{ 1+\frac{y^2}{x^2} }}$ $x\sqrt{ 1+\frac{y^2}{x^2}} ={e^{\arctan(y/x)}A}$ So the final general solution to the problem is:</p> <h4 id="sqrt-x2y2-earctanleft-fracyx-righta">$$\sqrt{ x^2+y^2 }=e^{\arctan\left( \frac{y}{x} \right)}A$$</h4> <hr> <h3 id="heading-1"></h3> <p><a class="hashtag" onclick="focusTag(this)">ex</a> <a class="hashtag" onclick="focusTag(this)">de_h_type2</a>: $$(2x-2y-1)dx+(x-y+1)dy=0$$ 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 <a class="hashtag" onclick="focusTag(this)">de_h_type2</a> let $u=x-y$ $\frac{du}{dx}=1-\frac{dy}{dx}$ $1-\frac{du}{dx}=\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$</p> <blockquote> <p>Obviously we don’t work with x and y as I was entailing above, substitute $u=x-y$ in you silly goose.</p> </blockquote> <p>$1-\frac{du}{dx}=-2\frac{{u-\frac{1}{2}}}{u+1}$ $\frac{du}{dx}=2\frac{{u-\frac{1}{2}}}{u+1}+1$ $\frac{du}{dx}=\frac{2u-1}{u+1}+1$ $\frac{du}{dx}=\frac{{2u-1+u+1}}{u+1}$ $\frac{du}{dx}=\frac{3u}{u+1}$ $\frac{(u+1)du}{3u}=dx$ $\int \frac{(u+1)du}{3u}=\int dx$</p> <blockquote> <p>$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=\ln\mid x\mid+C$ Ah, I made a mistake. $\int dx \ne \ln\mid x\mid+C$</p> </blockquote> <p>$\int \frac{du}{3}+\frac{1}{3}\int \frac{du}{u}=x+C$</p> <blockquote> <p>Okay, now that we have integrated, we can start talking in terms of x and y again</p> </blockquote> <p>$\frac{x-y}{3}+\frac{1}{3}\ln\mid x-y\mid = x+C$ $x-y+\ln\mid x-y\mid=3x+C$ $\ln\mid x-y\mid=C+y+2x$ < this is where he moved the C to the left $\mid x-y\mid=e^Ce^ye^{2x}$ $x-y=Ae^ye^{2x}$ $A(x-y)=e^{y+2x}$</p> <blockquote> <p>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 $\frac{dy}{dx}=-2$</p> </blockquote> <p>Proof: $\lim_{ n \to 0 }e^{y+2x}=n$ $\lim_{ n \to 0 }\ln(n)=y+2x$ $\lim_{ n \to 0 }\frac{d}{dx}\ln(n)=0=\frac{dy}{dx}+2$ $\frac{dy}{dx}=-2$</p> <blockquote> </blockquote> <p>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.</p> <p>We can rearrange to our liking, but we have found the general solution to the DE:</p> <h4 id="x-yae2xy">$$x-y=Ae^{2x+y}$$</h4> <hr> <h3>Referenced in</h3> <ul> <li>No backlinks found</li> </ul> </div> </main> <script type="text/javascript"> </script> </body> </html>