Question

Find the orthogonal trajectories of xye^(x^2)=C.

Answer 1

0

Answer 2

\[x(2xye ^{x^2}+\frac{ dy }{ dx }e ^{x^2})+ye ^{x^2}=0\]

Answer 3

solve it!

Answer 4

\[2x^2ye ^{x^2}+\frac{ dy }{ dx }xe ^{x^2}+ye ^{x^2}=0\]

Answer 5

\[\frac{ dy }{ dx }xe ^{x^2}=-2x^2ye ^{x^2}-ye ^{x^2}\]

Answer 6

UH

Answer 7

\[\frac{ dy }{ dx }=-2xy-\frac{ y }{ x }=\frac{ -2x^2y-y }{ x }\]

Answer 8

\[\frac{ dy }{ dx }=\frac{ -y(2x^2+1) }{ x }\]

Answer 9

negative reciprocal of dy/dx is...\[\frac{ dy }{ dx }=\frac{ x }{ y(2x^2+1) }\]

Answer 10

\[y dy=\frac{ x }{ 2x^2+1 }dx\]

Answer 11

u=2x^2+1
du=4x dx
(1/4)du=x dx
\[\frac{ 1 }{ 4 }\ln \left| 2x^2+1 \right|+C\]

Answer 12

\[\frac{ y^2 }{ 2 }=\frac{ 1 }{ 4 }\ln \left| 2x^2+1 \right|+C\]

Answer 13

\[y^2=\frac{ 1 }{ 2 }\ln \left| 2x^2+1 \right|+C\]

Answer 14

But the answer in the book is \[y^2=-\frac{ 1 }{ 2 }\ln(1+2x^2)+k\]

Answer 15

And I don't know which answer is right and why the book has the negative sign in front of 1/2.

Answer 16

@SithsAndGiggles

Answer 17

@satellite73

Answer 18

It's probably a typo. Your work is correct.

Answer 19

Thank you for verifying my work.