Find the orthogonal trajectories of the family of curves. (Use C for any needed constant.) Question on answer, last step.

Question

zenmo · Answer

$$y^2=7kx^3$$

zenmo · Answer

How should I begin this problem?

wavybaby · Answer

Where's C?

zenmo · Answer

C will probably end up in the end-product, but I think K is the constant?

wavybaby · Answer

So you trying to solve the problem by finding k?

zenmo · Answer

attachment

wavybaby · Answer

I would tell you what i got but, there's no answer choices and i'm not a 100% sure. Don't wanna get you wrong. ;)

zenmo · Answer

$$\frac{ y^2 }{ 7x^3 }=k$$Using quotient rule: $$\frac{ 14x^3yy'-21x^2y^2 }{ 49x^6 }=0$$$$\frac{ 2xyy'-3y^2 }{ 7x^4 }=0$$$$xyy'-3y^2=0, 2xy'-3y=0, 2xy'=3y, y' = \frac{ 3y }{ 2x }=> \frac{ -dx }{ dy }= \frac{ 3y }{ 2x }$$$$-2xdx = 3ydy$$$$\int\limits_{}^{}-2xdx = \int\limits_{}^{}3ydy$$

I am at the last step, what shall I do now?

zenmo · Answer

Okay, if I do it the normal way, the answer is$$-x^2+C = \frac{ 3 }{ 2 }y^2, C = x^2+\frac{ 3 }{ 2 }y^2$$
Another slightly different approach, the book's way, $$-x^2+\frac{ C }{ 2 }=\frac{ 3y^ 2}{ 2 }, -2x^2 + C = 3y^2, 2x^2+3y^2=C$$ this is also an answer. I am wondering why it used C/2 instead of just "C"?

photon336 · Answer

$$\frac{ dy }{ dx } => \frac{ 3y^{2} }{ 2xy }$$  how did you get -dx/dy?