Problem states: Find the orthogonal trajectories of the family of curves $ y=-e^{x^2}\, erf \,\,x\,+Ce^{x^2}.\,\\ $ Solve for $ x $ in terms of $ y $.
My attempt to solve this, by rearranging terms first:
$$ \begin{align} y=-e^{x^2}\,\left(\frac{2}{\sqrt{\pi}}\int_0^x{e^{-t^2}dt}\right)+Ce^{x^2}\tag 1 \\y\,e^{-x^2}\,+\,\left(\frac{2}{\sqrt{\pi}}\int_0^x{e^{-t^2}dt}\right)=C\tag 2 \\y'-2xy=\frac{4x\,e^{-x^2}}{\sqrt{\pi}}\tag 3 \\I=-2\int x\, dx \, =\, -x^2\, \implies e^{-x^2}\tag 4 \\y\,e^{-x^2}=\int e^{-x^2}\,\frac{4x\,e^{-x^2}}{\sqrt{\pi}} + C\tag 5 \\y\,e^{-x^2}=-\frac{1}{\sqrt{\pi}}e^{-2x^2}+C\tag 6 \end{align} $$
Should i then differentiate $(6)$ again with respect of $x \,$, find negative reciprocal and finally get $x$ in terms of $y$?
There is a mistake in equation (3), it should've been $$ \begin{align} e^{-x^2}\Big(y' - 2xy + \frac{2}{\sqrt{\pi}}\Big) = 0 \tag 1 \\ \text{as} \hspace{1em} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^x f(t)\mathrm{d}t = f(x) \end{align} $$
Therefore the differential equation for the orthogonal trajectory is: $$ \begin{align} \boxed{y' = \frac{a}{b - xy}}, \hspace{2em} \text{where} \hspace{2em} a = \frac{1}{2}, \ b = \frac{1}{\sqrt{\pi}} \tag 2 \\ \text{which is obtained from (1), as from (1) we get:}\\ y' = 2xy - \frac{2}{\sqrt{\pi}} \end{align} $$
Solving equation (2) will provide the equation for the orthogonal curves.