Solve the Differential equation $x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$ Solve the Differential equation $$x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$$
i reduced the equation as
$$x^3\frac{dy}{dx}=y^3\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right)$$ $\implies$
$$\frac{x^3}{y^3}\frac{dy}{dx}=\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right) \tag{1}$$ Next put $$\frac{x}{y}=v$$ we get
$$ x=vy$$ then
$$ \frac{dx}{dy}=v+y\frac{dv}{dy}$$ Then $(1)$ becomes
$$\frac{v^3}{v+y\frac{dv}{dy}}=1+\sqrt{1+v^2}$$ Reciprocating we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{1}{\sqrt{1+v^2}+1}$$  Rationalizing RHS we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{\sqrt{1+v^2}-1}{v^2}$$ Rearranging we get
$$\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\frac{dy}{y}$$ 
EDIT: i am posting here the clue given by paul using Substitution $v=\tan z$:
$$\int\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\int\frac{\sec^2 z\: dz}{\tan z(\sec z-2)}=\int\frac{dz}{\sin z(1-2\cos z)}$$ So
$$\int\frac{dz}{\sin z(1-2\cos z)}=\int \frac{\sin z\: dz}{\sin^2 z(1-2\cos z)}$$
Put $\cos z=t$ and use partial fractions
 A: Well, let $\text{y}\left(x\right)=x\cdot\text{r}\left(x\right)$, which gives $\text{y}'\left(x\right)=x\cdot\text{r}'\left(x\right)+\text{r}\left(x\right)$:
$$x^3\cdot\text{y}'\left(x\right)=\text{y}\left(x\right)^3+\text{y}\left(x\right)^2\cdot\sqrt{x^2+\text{y}\left(x\right)^2}\space\Longleftrightarrow\space$$
$$x^3\cdot\left(x\cdot\text{r}'\left(x\right)+\text{r}\left(x\right)\right)=\left(x\cdot\text{r}\left(x\right)\right)^3+\left(x\cdot\text{r}\left(x\right)\right)^2\cdot\sqrt{x^2+\left(x\cdot\text{r}\left(x\right)\right)^2}\tag1$$
Solve for $\text{r}'\left(x\right)$:
$$\text{r}'\left(x\right)=\frac{\text{r}\left(x\right)\cdot\left(\text{r}\left(x\right)^2-1+\text{r}\left(x\right)\cdot\sqrt{1+\text{r}\left(x\right)^2}\right)}{x}\tag2$$
Devide both sides by the numerator of the RHS and integrate both sides:
$$\int\frac{\text{r}'\left(x\right)}{\text{r}\left(x\right)\cdot\left(\text{r}\left(x\right)^2-1+\text{r}\left(x\right)\cdot\sqrt{1+\text{r}\left(x\right)^2}\right)}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\tag3$$
For the integrals:


*

*Substitute $\text{u}=\text{r}\left(x\right)$:
$$\int\frac{\text{r}'\left(x\right)}{\text{r}\left(x\right)\cdot\left(\text{r}\left(x\right)^2-1+\text{r}\left(x\right)\cdot\sqrt{1+\text{r}\left(x\right)^2}\right)}\space\text{d}x=$$
$$\int\frac{1}{\text{u}\cdot\left(\text{u}^2-1+\text{u}\cdot\sqrt{1+\text{u}^2}\right)}\space\text{d}\text{u}=$$
$$\frac{\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2\right|}{3}-\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-1\right|+\frac{2\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-3\right|}{3}+\text{C}_1\tag4$$

*$$\int\frac{1}{x}\space\text{d}x=\ln\left|x\right|+\text{C}_2\tag5$$


So, we get:
$$\frac{\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2\right|}{3}-\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-1\right|+\frac{2\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-3\right|}{3}=\ln\left|x\right|+\text{C}\tag6$$.
Now, use:
$$\text{u}=\frac{\text{y}\left(x\right)}{x}\tag7$$
A: The solution of the ODE is presented below on parametric form. The explicit form can be obtained but involves the roots of a cubic polynomial equation which would lead to a big formula.

