Finding orthogonal trajectories. Original equation $$y^4=Cx^5$$
Ok my issue is that i am not sure how I would continue after taking the derivative. $$\frac{y^4}{x^5}=C$$
$$\frac{4y^3x^5\frac{dy}{dx}-5y^4x^4}{x^{10}}dx$$ I am not sure how I would continue to seperate the variable to be able to take the integral to the find the orthogonal trjectory.
 A: For the family of curves $y(x, C)$ where
$$
y^4 = C \, x^5
$$
and $C$ is a constant, we want to find the orthogonal curves $z(x, C)$.
$y$ and $z$ should intersect at $x_0$, thus
$$
y(x_0) = z(x_0) \quad (*)
$$
and if $y$ has the tangent 
$$
T(x) = y(x_0) + y'(x_0)\, x
$$ 
at the point $x_0$ then $z$ should have an orthogonal tangent there
$$
U(x) = z(x_0) + z'(x_0)\, x
$$
Using the scalar product we get the condition for orthogonal tangent vectors
$$
0 = (1, T'(x_0)) \cdot (1, U'(x_0)) = 1 + T'(x_0) U'(x_0) \Rightarrow \\
U'(x_0) = - \frac{1}{T'(x_0)}
$$
and this means
$$
z'(x_0) = - \frac{1}{y'(x_0)} \quad (**)
$$
Solving for $y$ and differentiating, we get
$$
y = (C \, x^5)^{1/4} \Rightarrow \\
y' = \frac{5}{4} (C \, x)^{1/4}
$$
and therefore
$$
z' = - \frac{4}{5} (C \, x)^{-1/4}
$$
Integrating gives
$$
z = -\frac{16}{15} C^{-1/4} \, x^{3/4} + D
$$
with an integration constant $D$ such that equation $(*)$ holds:
$$
(C \, x_0^5)^{1/4} = -\frac{16}{15} C^{-1/4} \, x_0^{3/4} + D \Rightarrow \\
D = (C \, x_0^5)^{1/4} + \frac{16}{15} C^{-1/4} \, x_0^{3/4}
$$
which gives
$$
z = -\frac{16}{15} C^{-1/4} \, (x^{3/4} - x_0^{3/4}) + (C \, x_0^5)^{1/4}
$$
Here is the positive part of the curve for $C=1$ (green) and $C=2$ (cyan) and the orthogonal curves at $x_0 \in \{ 0.5, 1, 1.5, 2 \}$.

A: You have $\frac{4y^3x^5\frac{dy}{dx}- 5y^4x^4}{x^{10}}dx$.  This is wrong.  There should be no "dx" at the end and it should be an equation, equal to 0:
$\frac{4y^3x^5\frac{dy}{dx}- 5y^4x^4}{x^{10}}= 0$
Multiplying both sides by the $x^{10}$, $4y^3x^5\frac{dy}{dx}- 5y^4x^4= 0$  so $4y^3x^5\frac{dy}{dx}= 5y^4x^4$ and $\frac{dy}{dx}= \frac{5y^4x^4}{4y^3x^5}= \frac{5y}{4x}$.
The orthogonal complement is then given by $\frac{dy}{dx}= -\frac{4x}{5y}$.
A: Using Quotient Rule to benefit
$$ \dfrac{y^4}{x^5} = \dfrac{4 y^3 y^{'}}{5 x^4} 
 \rightarrow  \dfrac{y}{x} = \dfrac{4 y^{'}}{5} $$
To obtain orthogonal trajectory  differential equation we replace the derivative by its negative reciprocal
$$ \dfrac{y}{x} = \dfrac{-4 }{5y^{'}} $$ Re-arrange
$$ 5 \,y\, dy + 4\ x \,dx =0 $$ Integrate
$$ \dfrac {5 y^2}{2} + 2 x^2 = c^2. $$ 
It is convenient to handle $ \dfrac{P}{Q} =\dfrac{P'}{Q'} $ for quotients. In fact also useful as  $ \dfrac{P}{Q} =-\dfrac{P'}{Q'} $ for products.
