<s>[INST] Solve $yy' + x =\sqrt{x^2 + y^2}$ - Substitution for Diff Eqs. I have this problem
$$yy' + x = \sqrt{x^{2} + y^{2}}$$
I wanted to know if my work and answer are correct.
Let $v = x^{2} + y^{2}. v' = 2x + 2yy'$
$$ \frac{v'}{2} = x + yy'$$
Substituting everything in gives me:
$$\frac{v'}{2} = \sqrt{v}$$
From here it becomes a separable equation:
$$\frac{dv}{\sqrt{v}} = 2dx$$
$$2\sqrt{v} = 2x +C$$
Subbing in $x^2 + y^2$ back gives me: $$2\sqrt{x^2 + y^2} = 2x + C$$
And now if at this point my work is correct, we just solve for $y$ right?
$$\sqrt{x^2 + y^2} = x + \frac{C}{2}$$
Square both sides:
$$x^2 + y^2 = \bigg(x+\frac{C}{2}\bigg)^{2}$$
$$y^2 = \bigg(x+\frac{C}{2}\bigg)^{2} - x^2$$
And finally,
$$y = \pm\sqrt{\bigg(x+\frac{C}{2}\bigg)^{2} - x^2}$$
 [/INST] Your solution is correct.
At $$y^2 = \bigg(x+\frac{C}{2}\bigg)^{2} - x^2$$
You may simplify $$ y^2 = \bigg(x+\frac{C}{2}\bigg)^{2} - x^2=Cx+\frac {C^2}{4}$$
Which is a family of parabolas.$$y^2 =Cx+\frac {C^2}{4}.$$
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