# Determine the family of a orthogonal trajectories of the family y = x + c*(e^(-x)) wanted a detailed solution till last step

`y = x + ce^(-x)`

Let's first find the derivative of y wrt x.

`(dy)/(dx) = 1 -ce^(-x)`

Consider two families F1 and F2. We can say that they are orthogonal to eachother, if they intesect orthoganlly (the angle in betwen is 90) whenever they intersect.

For F2 to become orthogonal trajectory family of F2,

`(dF2)/(dx) = 1/(-(dF1)/(dx))`

If the family of orthoganal trajectories are z,

`(dz)/(dx) = 1/(-(1 -ce^(-x)))`

`(dz)/(dx) = 1/(ce^(-x)-1)`

we can rearrange the equation by muliplying bothe numerator and denominator by `e^x` .

`(dz)/(dx) = e^x/(c-e^x)`

By integrating wrt to x,

`z = inte^x/(c-e^x)dx`

`z = -int(-e^x)/(c-e^x)dx`

If you look at carefully, you can see that numerator is the derivative of the denominator. Then the integral becomes,

`z = -ln(c-e^x)+ c'`

`z = ln(1/(c-e^x))+ c'`

Again rearranging the equation,

`z = ln((e^(-x))/(ce^(-x)-1)))+ c'`

Therefore the family of orthoganal equations are

`z = ln((e^(-x))/(ce^(-x)-1)))+ c'` where c and c' are arbitrary constants.

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