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Find the orthogonal trajectories of the family of curves:

y = k*(e^-x)
---------------
so k = y/(e^-x)

differentiating we get:

1 = -k(e^-x)*(dx/dy)
1/(dx/dy) = -k(e^-x)
dy/dx = -k(e^-x)...substituting for k:
dy/dx = -(y/(e^-x))*(e^-x)
dy/dx = -y
Integral(1/=y)dy = Integral dx
-ln(y) = x + C
x = -ln(y) + C

Is this correct? Thanks.

Yes, correct.

Yes, correct.

Thanks for checking!!!

Yes, your solution is correct. The process you followed to find the orthogonal trajectories of the family of curves y = k * e^(-x) is accurate. You differentiated the equation with respect to x, substituted for k using k = y / (e^(-x)), and simplified the expression to dy/dx = -y. Integrating this expression leads to the solution x = -ln(y) + C, where C is the constant of integration. This solution represents the orthogonal trajectories. Well done!