SOLUTION: Find the member of orthogonal trajectories which passes through (1, 2) for the family x^2 + 3y^2 = cy?

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): You can put this solution on YOUR website!
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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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