Finding Orthogonal Trajectories of the family curves $y=cx^2+4$.

I want to find Orthogonal Trajectories of the family curves $$y=cx^2+4$$, here is my approach,

$$y=cx^2+4 \Rightarrow\quad \frac{dy}{dx}=2cx$$

Then, in order to find Orthogonal Trajectories we should replace $$y'$$ with $$-\dfrac1{y'}$$,

$$\frac{-1}{y'}=2cx\Rightarrow\quad y'=-\frac1{2cx}\Rightarrow\quad y=\frac{-1}{2c}\int\frac1x dx$$ Hence Iget $$y=\dfrac{-1}{2c}\ln|x|+c'$$. But the final solution in the book is $$y^2-8y+\frac{x^2}2-c_1=0$$.

I can't figure out why my answer is different (maybe wrong). and by comparing, I see the equations are not equivalent.

• In $y'=-\frac1{2cx}$, you need to replace $c$ with $\frac {y - 4} {x^2}$ and then integrate to find orthogonal trajectories. Aug 21, 2021 at 16:02

As mentioned in the comments, you have to get rid of the $$c$$ before you integrate. From $$y=cx^2+4$$, one can see that $$c=\frac{y-4}{x^2}$$ and thus you have
$$\frac{dy}{dx}=2cx=\frac{2x(y-4)}{x^2}=\frac{2y-8}{x}$$
$$\frac{dy}{dx}=-\frac{x}{2y-8}$$