- #1

EmmanuelD

- 10

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Hello, forum! I'm a newbie here. I've been visiting this site for a while but just recently joined. Anyways, I was wondering if anyone could help with this problem. I can find the orthogonal trajectories, however, this one is killing me because there is a constant. Allow me to type it below:

Find the orthogonal trajectories for the family of curves:

y=C*(x^5) - 3

F(x,y,C)=0 and G(x,y,K)=0

1. Determine the differential equation for the given family F(x,y,C)=0

2. Replace y' in that equation by -(1/y'); the resulting equation is the differential equation for the family of orthogonal trajectories.

3. Find the general solution of the new differential equation. This is the family of orthogonal trajectories.

Given:

y=C*(x^5) - 3 --> C=(y/(x^5))-3

y'=5*(x^4)*C - 3

Now, substituting C:

y'=5*(x^4)*[(y/(x^5))-3]

y'=(5y/x)-15x^4

Conclusion:

1. Am I even on the right track?

2. Is "separable" differentiation/integration the only method that applies?

Thank you so much for taking the time to review this!

## Homework Statement

Find the orthogonal trajectories for the family of curves:

y=C*(x^5) - 3

## Homework Equations

F(x,y,C)=0 and G(x,y,K)=0

1. Determine the differential equation for the given family F(x,y,C)=0

2. Replace y' in that equation by -(1/y'); the resulting equation is the differential equation for the family of orthogonal trajectories.

3. Find the general solution of the new differential equation. This is the family of orthogonal trajectories.

## The Attempt at a Solution

Given:

y=C*(x^5) - 3 --> C=(y/(x^5))-3

y'=5*(x^4)*C - 3

Now, substituting C:

y'=5*(x^4)*[(y/(x^5))-3]

y'=(5y/x)-15x^4

Conclusion:

1. Am I even on the right track?

2. Is "separable" differentiation/integration the only method that applies?

Thank you so much for taking the time to review this!

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