I'm trying to find a general solution for this equation
$xy'=y+\sqrt{x^2+y^2}$
I got $\tan\frac{y}{x}+\sec\frac{y}{x} = x + c$ as an answer.
I substituted $y=vx$ and it led me to the trig-related answer.
The answer sheet gives $y+\sqrt{x^2+y^2} = Cx^2$.
I honestly how to derive this. Can I get some help here?
On
You can play a bit with the terms and get$$x^2\left(\frac{y}{x}\right)'= xy'-y = \sqrt{x^2+y^2} = \lvert x \rvert \sqrt{1+\left(\frac{y}{x}\right)^2}.$$ Define $y(x)/x =: z(x)$, then you have $$z' = \frac{\lvert x \rvert}{x^2}\sqrt{1+z^2} $$ and separating variables $$\int\frac{dz}{\sqrt{1+z^2}} = \int\frac{\lvert x \rvert}{x^2}\,dx$$ Putting $z = \sinh w$ yields $\sqrt{1+z^2} = \cosh w$ and $dz = \cosh w \,dw$, so \begin{align}\int \frac{dz}{\sqrt{1+z^2}}=\int \frac{\cosh w}{ \cosh w} dw & = w = \sinh^{-1}z \\ & = \log (z+\sqrt{1+z^2}) \end{align} up to a constant $c$. Then \begin{align} \log (z+\sqrt{1+z^2}) & = \text{sgn}\{x\}\left(\log \lvert x\rvert + \log e^c\right) \\ & = \log (C\lvert x \rvert)^{\text{sgn}\{x\}}, \text{ with }C > 0, \end{align} and finally, since $z(x)=y(x)/x$ you compare the arguments of the logarithms getting $$\frac{y}{x}+ \frac{1}{\lvert x \rvert}\sqrt{x^2+y^2} = (C\lvert x\rvert)^{\text{sgn}\{x\}}. $$ This solution does not depend on the sign of $x$ (just try to explicit $y$ when $x > 0$ and $x < 0$ and you will get the same expression). So the solution for $x > 0$ gives the result: $$y +\sqrt{x^2+y^2} = Cx^2.$$
$$y'=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}$$ $$u=\frac{y}{x}$$ $$u+xu'=u+\sqrt{1+u^2}$$ Separable ODE : $$\int\frac{dx}{x}=\ln|x[+c_1=\int\frac{u'}{\sqrt{1+u^2}}=\sinh^{-1}(u)+c_2$$ $$u=\sinh\left(\ln|cx[\right)=\frac12\left(cx-\frac{1}{cx}\right)$$ $$y=\frac12\left(cx^2-\frac{1}{c}\right)$$ $x^2=\frac{2}{c}y+\frac{1}{c^2}\quad;\quad x^2+y^2=\frac{2}{c}y+\frac{1}{c^2}+y^2=\left(y+\frac{1}{c}\right)^2$
$y+\sqrt{x^2+y^2}=y+(y+\frac{1}{c})=2y+\frac{1}{c}=\left(cx^2-\frac{1}{c}\right)+\frac{1}{c}$ $$y+\sqrt{x^2+y^2}=cx^2$$