# Thread: Why not find the solution?

1. ## Why not find the solution?

Find the orthogonal trajectories of the family of curves $y = C_1e^{-x}$

Well, by definition, the steps are these:

First solve for $C$. and $C = \displaystyle\frac{y}{e^ {-x}}$

differentiate the equation:
$
y * (y\prime) + Ce^{-x} = 0$
, you get rid of the constant C and finally going to be that, by definition,

$y_2' = - \displaystyle\frac{1}{y'}$

But I have problems to find the answer of the book is

$\displaystyle\frac{y^2}{2} = x + C$

Thank you very much,

greetings

2. Remember that if two lines are perpendicular then the product of their gradients is –1.

The calculation goes like this. If $y = Ce^{-x}$ then $y'=-Ce^{-x} = -y$. If the curve has gradient $-y$ then the orthogonal curve will have gradient $-1/(-y) = 1/y$. So we need to solve the differential equation $y'=1/y$.

Write it as $yy'=1$ and integrate to get $\tfrac12y^2 = x + C$.

3. Hello! Yes, and I could clarify,
greetings