Are there any explicit solutions of $yy' = 5x$ that pass through the origin? First part: Use the fact that $5x^2 − y^2 = c$ is a one-parameter family of solutions of the differential equation: $y y'= 5x$ to find an implicit solution of the initial-value problem: 
$$y \dfrac {dy}{dx} = 5x \\ y(2) = −6$$
I got the answer as $y^2=5x^2+16$. Then it asks if there are any explicit solutions of $yy' = 5x$ that pass through the origin? I'm not sure how to go about this part.
 A: $y=x\sqrt{5}$ is such a solution.
A: If $(0,0)$ satisfies the curve, then $c=0$. Thus the only such curve would be $y^2=5x^2$. 
This is in implicit form. It contains the graphs of various functions. Explicitly you get for example $y=- \sqrt{5}x$ or $y= \sqrt{5}x$.
A: If the curve is to satisfy $y(2)=-6$, it cannot pass through the origin.
But, if we remove that constraint, we do have solutions. The differential equation on solving yields, as you noted, 
$$5x^2-y^2=c$$
where $c$ is constant.
For this to pass through the origin, $(0,0)$ must satisfy it, i.e.,
$$5(0)^2-(0)^2=c\Rightarrow c=0$$
That gives two possible curves: $y=\sqrt 5 x$ and $y+\sqrt 5 x=0$
P.S. A neat trick to see if any curve passes through the origin is to make sure that it doesn't have any constant term.
A: The solution to the initial value problem of your problem does not pass through the origin. To find solution that pass through the origin, you may need to change the initial condition. If you want to keep $x=2$ as your initial time, then the initial value problem condition becomes $y(2)=0$. Integration gives
$$\frac{1}{2}(y^2(x)-y^2(2))=\int^x_2y(s)y'(s)\,ds = 5\int^x_2 s\,ds= \frac{5}{2}(x^2-4)$$
So, if $y(2)=0$, one gets 
$$
y^2(x) = 5x^2-20
$$
This gives two different solutions to the initial value problem, namely
\begin{aligned}
 y_+(x) &=\sqrt{5x^2-20}\\
y_-(x) &= -\sqrt{5x^2-20}
\end{aligned}
In general
$$yy'=5y,\qquad y(x_0)=y_0$$
where $x_0>0$, $y_0>0$, has unique solution $y(x)=\sqrt{5x^2-5x^2_0+ y^2_0}$ defined in the interval $I$ such that $x_0\in I$ and $I\subset\{x: x^2\geq x^2_0-\frac{y^2_0}{5}\}$. If the graph of $y$ passes through the origins, then $x_0=\frac{y_0}{\sqrt{5}}$.
A similar argument works when the initial condition is $y(x_0)=y_0<0$. Then the solution is $y(x)=-\sqrt{5x^2-5x^2_0+ y^2_0}$  defined in an interval $J$ with $x_0\in J$ and $J\subset\{x: x^2\geq x^2_0-\frac{y^2_0}{5}\}$. If $y$ passes through the origin, then $x_0=-\frac{y_0}{\sqrt{5}}$.  
