Solve xdy+(y+sqrt{x^2+y^2})dx=0

$$x d y + ( y + \sqrt { x ^ { 2 } + y ^ { 2 } } ) d x = 0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&\left(y\leq 0\text{ and }|x|=-\sqrt{3}y\right)\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply y+\sqrt{x^{2}+y^{2}} by d.

$$xdy+\left(yd+\sqrt{x^{2}+y^{2}}d\right)x=0$$

Use the distributive property to multiply yd+\sqrt{x^{2}+y^{2}}d by x.

$$xdy+ydx+\sqrt{x^{2}+y^{2}}dx=0$$

Combine xdy and ydx to get 2xdy.

$$2xdy+\sqrt{x^{2}+y^{2}}dx=0$$

Combine all terms containing d.

$$\left(2xy+\sqrt{x^{2}+y^{2}}x\right)d=0$$

The equation is in standard form.

$$\left(x\sqrt{x^{2}+y^{2}}+2xy\right)d=0$$

Divide 0 by 2xy+\sqrt{x^{2}+y^{2}}x.

$$d=0$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x=\sqrt{3}y\text{; }x=-\sqrt{3}y\text{, }&y\leq 0\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$