Solution of $(x^2 + y^2)\ dx -2xy\ dy$ = 0
Solve $(x^2 + y^2)dx -2xydy = 0$
The answer is $x^2 - y^2 = Cx$
I've tried the following methods but I'm not getting the answer :
- Variable Separable (n/a)
- Homogenous Differential Equation (Coefficients not of degree 1)
- Linear Differential Equation; Not Linear due to $n \ne 1$
- Exact Differential Equation: Not Equal
- Unexact Differential Equaiton: Doesn't fit
- Bernoulli's: getting $y^2 + x^2/3 = C/x$
What else can I try? Any hint?
$\endgroup$ 18 Answers
$\begingroup$$$x^2 + y^2 = 2xy \dfrac{dy}{dx} \implies x^2 + y^2 = x\dfrac{d(y^2)}{dx} .$$ Substitute $y^2 = z$ and you will get the nice form: $$x \dfrac{dz}{dx}-z = x^2.$$
$\endgroup$ $\begingroup$Geometry also helps. $2xy$ looks like a diagonal term, rotation by $45^\circ$ will simplify the equation. Use
$$x=u+v$$ $$y=u-v$$ (substitution of new variables $u=(x+y)/2$ and $v=(x-y)/2$).
We get $$2(u^2+v^2)(du+dv)-2(u^2-v^2)(du-dv)=0$$ $$v^2du+u^2dv=0$$ This is actually as good as it can possibly get. You can separate the variables!
$$\int \frac{du}{u^2}=-\int\frac{dv}{v^2}$$ $$1/u+1/v=C$$ $$u+v=Cuv$$ $$C'x=x^2-y^2$$
When I forgot the minus in the second term ($-2xy$) I got the same wrong solution you got with Bernoulli, so you should just check your method again and be careful with the signs.
$\endgroup$ 0 $\begingroup$Hint: $${x^2} + {y^2} = 2xy.y' \Rightarrow y' = \frac{{{x^2} + {y^2}}}{{2xy}} = \frac{{1 + {{\left( {\frac{y}{x}} \right)}^2}}}{{2\frac{y}{x}}}.$$ Put $u=\frac{y}{x}$, we have $y=ux$, and so $y'=u'x+u$. The latter equation becomes $$u'x + u = \frac{{1 + {u^2}}} {u} = \frac{1} {u} + \frac{u} {2} \Rightarrow u'x = \frac{1} {u} - \frac{u} {2} = \frac{{2 - {u^2}}} {{2u}}.$$
$\endgroup$ $\begingroup$With the integrand factor $\dfrac1{x^2}$, $$\frac{(x^2+y^2)\,dx-2xy\,dy}{x^2}=dx-d\frac{y^2}x,$$and $$\frac{x^2-y^2}x=C.$$
$\endgroup$ $\begingroup$Switching to polar coordinates $$r^2(\cos\theta\,dr-r\sin\theta\,d\theta)=2r^2\cos\theta\sin\theta(\sin\theta\,dr+r\cos\theta\,d\theta)$$ makes the equation separable: $$\frac{dr}r=\frac{\sin\theta+2\cos^2\theta\sin\theta}{\cos\theta-2\cos\theta\sin^2\theta}d\theta=-\frac{\sin\theta\,(2\cos^2\theta+1)}{\cos\theta\,(2\cos^2\theta-1)}d\theta=\frac{2t^2+1}{t(2t^2-1)}dt.$$ Then $$\ln(r)=\ln\left(\frac t{2t^2-1}\right)+C$$
$$r(2\cos^2\theta-1)=C\cos\theta,$$ $$2r^2\cos^2\theta-r^2=Cr\cos\theta,$$ $$x^2-y^2=Cx.$$
$\endgroup$ 4 $\begingroup$This seem to be the most simple answer: Substituting $u = x^2$ and $v = y^2$ gives $(u+v)du - 2udv = 0$ which has the solution $v = u + Cu^{1/2}$ or $y^2 - x^2 = Cx$.
$\endgroup$ $\begingroup$$$ x' = \frac{2xy}{x^2+y^2} $$ Let $x=vy$ then we get $$ v'y + v = \frac{2v}{v^2+1} $$ Which is a nicer form don't you think?
$\endgroup$ $\begingroup$$(x^2+y^2)dx−2xydy=0$
$\frac{dy}{dx}=\frac{x^2+y^2}{2xy} $..(i)
This is a homogeneous differential equation because it has homogeneous functions of same degree 2.
homogeneous functions are: $(x^2+y^2)$ and $2xy$, both functions have degree 2.
Solution of differential equation: Equation (i) can be written as,
$\frac{dy}{dx}=\frac{1+\frac{y}{x}}{2(\frac{y}{x})}$..(ii)
Substitute, $\frac{y}{x}=v \implies\frac{dy}{dx}=x\frac{dv}{dx}+v$
Now, equation(ii)becomes,
$x\frac{dv}{dx}+v=\frac{1+v^{2}}{2v}\implies x\frac{dv}{dx}=\frac{1+v^{2}}{2v}-v\implies x\frac{dv}{dx}=\frac{1-v^{2}}{2v} $
Integrate both side:
$\int {\frac{2v}{1-v^2}dv}=\int {\frac{1}{x}}dx$...(iii)
Substitute, $1-v^2 =t\implies2v\;dv=-dt$
$\int {\frac{-1}{t}dt}=\int {\frac{1}{x}}dx$
$ -lnt=lnx + lnC\implies t^{-1}=xC$
Put the value of t
$(1-v^{2})^{-1}=xC\implies \frac{1}{xC}=1-v^{2}$
Put the value of v,
$\frac{1}{xC}=1-\frac{y^{2}}{x^{2}}$
$C=\frac{1}{x(1-\frac{y^{2}}{x^{2}})}$
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