I've got a differential equation to solve: $\frac{dy}{dx} x =\sqrt{x^2 +y^2}$. I substitute $w=\frac{y}{x}$ and I obtain after several steps: $\frac{dw}{\sqrt{1+w^2} -w} = \frac{dx}{x}$, and the integral from left hand side is really unpleasent thing to compute. Do I solve it in the right way? Is my result correct and I should continue doing it this way? Please, can somebody check and help me with solving it?
[Math] Differential equation, a square root and substitution
ordinary differential equations
Related Solutions
In fact it belongs to an Emden-Fowler equation.
First, according to http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf or http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=6 , all Emden-Fowler equations can be transformed into Abel equation of the second kind.
Let $\begin{cases}u=\dfrac{x^3}{y_n^\frac{3}{2}}\\v=\dfrac{x}{y_n}\dfrac{dy_n}{dx}\end{cases}$ ,
Then $\dfrac{dv}{du}=\dfrac{\dfrac{dv}{dx}}{\dfrac{du}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{1}{y_n}\dfrac{dy_n}{dx}-\dfrac{x}{y_n^2}\left(\dfrac{dy_n}{dx}\right)^2}{\dfrac{3x^2}{y_n^\frac{3}{2}}-\dfrac{3x^3}{2y_n^\frac{5}{2}}\dfrac{dy_n}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{v}{x}-\dfrac{v^2}{x}}{\dfrac{3u}{x}-\dfrac{3uv}{2x}}=\dfrac{\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2}{3u\left(1-\dfrac{v}{2}\right)}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}=\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2$
$\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}=3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v$
$\dfrac{d^2y_n}{dx^2}=\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)$
$\therefore\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)-\dfrac{nx}{\sqrt{y_n}}=0$
$\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)=\dfrac{nx}{\sqrt{y_n}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=\dfrac{nx^3}{y_n^\frac{3}{2}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=nu$
$3u\left(\dfrac{v}{2}-1\right)\dfrac{dv}{du}=v^2-v-nu$
Let $w=\dfrac{v}{2}-1$ ,
Then $v=2w+2$
$\dfrac{dv}{du}=2\dfrac{dw}{du}$
$\therefore6uw\dfrac{dw}{du}=(2w+2)^2-(2w+2)-nu$
$6uw\dfrac{dw}{du}=4w^2+6w+2-nu$
$w\dfrac{dw}{du}=\dfrac{2w^2}{3u}+\dfrac{w}{u}+\dfrac{2-nu}{6u}$
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{2}{3uz^2}+\dfrac{1}{uz}+\dfrac{2-nu}{6u}$
$\dfrac{dz}{du}=\dfrac{(nu-2)z^3}{6u}-\dfrac{z^2}{u}-\dfrac{2z}{3u}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2.
$$xy+y+y’^2=0$$ Let : $y’=p$ $$y=-xy’-y’^2=-xp-p^2$$ $$ \frac{dy}{dp} =-p\frac{dx}{dp}-x-2p$$ $$p=y’=\frac{dy}{dx}=\frac{dy}{dp} \frac{dp}{dx} =-p-(x+2p) \frac{dp}{dx}$$ $$2p=-(x+2p) \frac{dp}{dx}$$ $$2p\frac{dx}{dp}+x=-2p$$ The solution of this first order linear ODE is : $$x=-\frac{2}{3}p+\frac{C}{2\sqrt{p}}$$ $$y=-xp -p^2= \frac{2}{3}p^2-\frac{C}{2}\sqrt{p} –p^2 =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p}$$ Finally, the solution of $xy+y+y’^2=0$ expressed on parametric form with parameter $p$ is : $$\begin{cases} x=-\frac{2}{3}p+ \frac{C}{2}\frac{1}{\sqrt{p} } \\ y =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p} \\ \end{cases}$$ If one want to find the explicit function $y(x)$ the parameter $p$ has to be eliminated from the system $\left(x(p),y(p)\right)$. This is possible, but will lead to complicated equations.
Best Answer
So your equation is $$ y'x = \sqrt{x^2+y^2} $$ If you do substitution $w = \frac yx$ as you stated you'll get $$ y = wx \\ y' = w'x+w \\ w'x+w = \text{sign}(x)\sqrt{1+w^2} \\ w' = \frac {\text{sign}(x)\sqrt{1+w^2}-w}x \\ \frac {dw}{{\text{sign}(x)\sqrt{1+w^2}-w}} = \frac {dx}x \\ dw \left({\text{sign}(x)\sqrt{1+w^2}+w} \right) = \frac {dx}x \\ \frac {w^2}2 + \int \left (\text{sign}(x)\sqrt{1+w^2} \right) dw = \ln |x| $$ Now do another substitution $$ w = \sinh t \\ dw = \cosh t\ dt \\ \int \left (\text{sign}(x)\sqrt{1+w^2} \right) dw = \text{sign}(x) \int \cosh^2 tdt = \text{sign}(x) \int \frac {1+\cosh 2t}2dt = \\ = \text{sign}(x) \left( \frac t2 + \frac 14 \sinh 2t\right) = \frac{\text{sign}(x)}2\left ( \text{arcsinh }w + w\sqrt{1+w^2}\right) $$ So final equation is $$ w^2 + \text{sign}(x)\left (\text{arcsinh }w + w\sqrt{1+w^2}\right) = \ln x^2 + C $$ I don't think you can go any further than that, so all is left is substitute $y$ and $x$ to the solution. $$ \frac {y^2}{x^2} + \text{arcsinh }(\frac y{|x|}) + y \sqrt{x^2+y^2} - \ln x^2 = C $$