Check my work: Find the orthogonal trajectories of the given family of curves:

y^2-x^2=c

So I use implicit differentiation to get $$\frac{ dy }{ dx } = \frac{ x }{ y }$$

Then, I take the negative reciprocal $$\frac{ dy }{ dx } = \frac{ -y }{ x }$$

Next, I separate the terms and integrate so $$\int\limits_{}^{} -\frac{ 1 }{ y }dy = \int\limits_{}^{}\frac{ 1 }{ x }dx$$ giving $$-lny = lnx +c$$

Question

Check my work: Find the orthogonal trajectories of the given family of curves: 

y^2-x^2=c

So I use implicit differentiation to get $$\frac{ dy }{ dx } = \frac{ x }{ y }$$

Then, I take the negative reciprocal $$\frac{ dy }{ dx } = \frac{ -y }{ x }$$

Next, I separate the terms and integrate so $$\int\limits_{}^{} -\frac{ 1 }{ y }dy = \int\limits_{}^{}\frac{ 1 }{ x }dx$$ giving $$-lny = lnx +c$$

anonymous · Answer

exponentiating both sides, Ive got: $$\frac{ 1 }{ y } = xe ^{c}$$. Now I just call e^c C, because its still some constant. So now I've got $$\frac{ 1 }{ y } =cx$$ and then just divide by x to get c alone. So i end with $$\frac{ 1 }{ xy } = c$$ How does this look? Am I close?

anonymous · Answer

Nobody knows how to do this?