Find the Orthogonal Trajectories For The Family of Curves

In summary, the solution to the homework statement is y=-x/5(y-3). The solution to the homework equations is F(x,y,C)=0 and G(x,y,K)=0. The family of orthogonal trajectories is found by replacing y' in the differential equation for the given family of curves, y=C*x^5-3, by -(1/y').
  • #1
EmmanuelD
10
0
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:

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!
 
Last edited:
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  • #2
Your first line is wrong. If y=C*x^5+3 then y-3=C*x^5. What is the correct solution for C?
 
  • #3
I apologize; I just edited the post.

Given:
y=C*x^5 - 3

Solution for C:

C=y/x^5 + 3

Then, as shown above, I sub'd in the value for C in the y':

y'=5x^4(y/x^5 + 3)

y'=5y/x + 15x^4

Now, I can't factor out a "y" to make this D.E. separable.
 
  • #4
EmmanuelD said:
I apologize; I just edited the post.

Given:
y=C*x^5 - 3

Solution for C:

C=y/x^5 + 3

Then, as shown above, I sub'd in the value for C in the y':

y'=5x^4(y/x^5 + 3)

y'=5y/x + 15x^4

Now, I can't factor out a "y" to make this D.E. separable.

That's STILL wrong. y=C*x^5-3 -> y+3=C*x^5 -> C=(y+3)/x^5. That's very different from your solution.
 
  • #5
Or, taking a different route (inspired by your initial reply);

C=(y-3)/x^5

Sub'd into y'

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

y'=5/x(y-3)

1/(y-3) dy = 5/x dx

ln(y-3) = 5*ln(x) + C

y-3=x^5 + C

...but this still doesn't look right to me; unless I'm missing something.
 
  • #6
EmmanuelD said:
Or, taking a different route (inspired by your initial reply);

C=(y-3)/x^5

Sub'd into y'

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

y'=5/x(y-3)

1/(y-3) dy = 5/x dx

ln(y-3) = 5*ln(x) + C

y-3=x^5 + C

...but this still doesn't look right to me; unless I'm missing something.

Oops! I forgot what I was doing here. Back to business:

y'=-1/y'

Don't lose hope on me, I'm just burned out.

I'm going to try this on paper.

Thanks for the help thus far! :)
 
  • #7
EmmanuelD said:
Or, taking a different route (inspired by your initial reply);

C=(y-3)/x^5

Sub'd into y'

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

y'=5/x(y-3)

1/(y-3) dy = 5/x dx

ln(y-3) = 5*ln(x) + C

y-3=x^5 + C

...but this still doesn't look right to me; unless I'm missing something.

Exponentiating ln(y-3)=5*ln(x)+C gives you y-3=C*x^5. Don't forget your rules of exponentiation. You might now notice you are right back at your starting point. Hmmm. Didn't you forget to change the y' to -1/y'??
 
  • #8
So, I got:

x^2+5y^2-30y=C

If you solved the problem, is this correct?

Thanks!
 
  • #9
Dick said:
Exponentiating ln(y-3)=5*ln(x)+C gives you y-3=C*x^5. Don't forget your rules of exponentiation. You might now notice you are right back at your starting point. Hmmm. Didn't you forget to change the y' to -1/y'??

Hehe, yes, I did in fact forget :)

I reposted with my answer.

y'=-x/5(y-3)

(y-3) dy = -x/5 dx

y2/2 - 3y = -x^2/10 +C

10*(x^2/10 + y^2/2 - 3y) = C

Ans: x^2 + 5y^2 - 30y = C (?)
 
  • #10
EmmanuelD said:
So, I got:

x^2+5y^2-30y=C

If you solved the problem, is this correct?

Thanks!

I think it's close. Is your given y=C*x^5-3 or y=C*x^5+3? I think the sign on the 3 may have gotten flipped around. Check it.
 
  • #11
Dick said:
I think it's close. Is your given y=C*x^5-3 or y=C*x^5+3? I think the sign on the 3 may have gotten flipped around. Check it.

Grr! Yes, thank you!

I notice I make many silly mistakes when I rush through.

So, it's ...+30y = C :]

Thanks a BUNCH! I know it might be simplistic but our professor stressed how easy it was and just did one example (without the constant).

So, not so bad after all.

Again, thank you SO much!
 

1. What is the concept of orthogonal trajectories?

The concept of orthogonal trajectories involves finding a set of curves that intersect another set of curves at right angles. These curves are known as orthogonal trajectories and can be found for a given family of curves by solving a differential equation.

2. How do you find the orthogonal trajectories for a given family of curves?

To find the orthogonal trajectories for a given family of curves, you need to first express the family of curves in the form of a differential equation. Then, you can use the method of solving orthogonal trajectories, which involves finding the general solution of the differential equation and using it to find the specific solution for the orthogonal trajectories.

3. Can you explain the importance of orthogonal trajectories in mathematics?

Orthogonal trajectories are important in mathematics because they help in understanding the relationship between two sets of curves. They also have applications in various fields such as physics, engineering, and geometry.

4. Are there any specific techniques for finding orthogonal trajectories?

Yes, there are specific techniques for solving for orthogonal trajectories, such as the method of undetermined coefficients and the method of substitution. These techniques involve manipulating the given differential equation to find the general solution and then using it to find the specific solution for the orthogonal trajectories.

5. Can orthogonal trajectories be used to solve real-world problems?

Yes, orthogonal trajectories have various applications in real-world problems, such as finding the paths of electric and magnetic fields, understanding fluid flow, and determining the trajectories of objects under the influence of gravity. They also have applications in optimization and curve fitting problems.

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