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Find the Orthogonal Trajectories For The Family of Curves 
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#1
Sep1511, 10:13 PM

P: 10

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:
1. The problem statement, all variables and given/known data Find the orthogonal trajectories for the family of curves: y=C*(x^5)  3 2. Relevant 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. 3. 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!! 


#2
Sep1511, 10:32 PM

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Your first line is wrong. If y=C*x^5+3 then y3=C*x^5. What is the correct solution for C?



#3
Sep1511, 10:40 PM

P: 10

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
Sep1511, 10:43 PM

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Find the Orthogonal Trajectories For The Family of Curves



#5
Sep1511, 10:50 PM

P: 10

Or, taking a different route (inspired by your initial reply);
C=(y3)/x^5 Sub'd into y' y'=5x^4((y3)/x^5)) y'=5/x(y3) 1/(y3) dy = 5/x dx ln(y3) = 5*ln(x) + C y3=x^5 + C ...but this still doesn't look right to me; unless I'm missing something. 


#6
Sep1511, 11:00 PM

P: 10

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
Sep1511, 11:03 PM

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#8
Sep1511, 11:04 PM

P: 10

So, I got:
x^2+5y^230y=C If you solved the problem, is this correct? Thanks! 


#9
Sep1511, 11:13 PM

P: 10

I reposted with my answer. y'=x/5(y3) (y3) 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
Sep1511, 11:19 PM

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#11
Sep1511, 11:26 PM

P: 10

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!!!!!!! 


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