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Question 1183747: Find the member of orthogonal trajectories which passes through (1, 2) for the family x^2 + 3y^2 = cy?
Answer by robertb(5830)   (Show Source):
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 ===>  .
Now by implicit differentiating the given equation, one gets
2x + 6yy' = cy', which implies that (2x + 6yy')/y' = c.
Then by equating both expressions for c, one gets
 .
===> 
===> 
<===>  .
Now for orthogonal trajectories, have to solve for
 .
<===>  , which is a Bernoulli D.E. with n = -1.
To solve this, use the substitution  . Hence yy' = v'/2, so that
===> 
This is a 1st order linear D.E. with integrating factor 
===> 
<===> 
===> 
===>  <===>  ===> 
Solving for k for the member of the family of orthogonal trajectories passing through (1,2), we have
 , or k = 3. Therefore, 
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