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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.

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

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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

$$\frac{dy}{dx}=2cx=\frac{2x(y-4)}{x^2}=\frac{2y-8}{x}$$

When you take the negative reciprocal of this, as you did, you now get

$$\frac{dy}{dx}=-\frac{x}{2y-8}$$

You can take the rest from here.

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  • $\begingroup$ Thanks for the answer. I think you missed a negative sign at the end. $\endgroup$
    – Etemon
    Aug 21, 2021 at 16:12
  • $\begingroup$ @Soheil Oops; you're right. Sorry about that. $\endgroup$
    – Kman3
    Aug 21, 2021 at 16:13

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