Patricia662
New member
- Joined
- Aug 30, 2020
- Messages
- 10
y^2 e^(x^2+y^2)=c find orthogonal trajectory.
Please show us what you have tried and exactly where you are stuck.y^2 e^(x^2+y^2)=c find orthogonal trajectory.
Your second line of solution is incorrect.I dont know how to solve the diferential equation. I tried with wolphram alpha but i could see just a few steps. I dont understand why you can divide with (e^(x^2))/y'View attachment 21294View attachment 21294
Now i have dif. eq. ye^(x^2+y^2)(1+2xy^2))/y'=0 which i still dont know how to solveYour second line of solution is incorrect.
d/dx [e^(x^2+y^2)] =[e^(x^2+y^2)] * [2x + 2yy']
correct it and continue.....
Can you please explain how you got this?[MATH]y = \sqrt{x^2 - 1}[/MATH] are two orthogonal trajectories (Quadrant I and II)
you want more orthogonal trajectories?
[MATH]|y| = \sqrt{Ax^2 - 1}[/MATH]
where [MATH]A > 0[/MATH]
this means you can use
[MATH]y = \sqrt{Ax^2 - 1}[/MATH] (orthogonal trajectories in Quadrant I and II)
or
[MATH]y = -\sqrt{Ax^2 - 1}[/MATH] (orthogonal trajectories in Quadrant III and IV)
I do not arrive at that equation.Now i have dif. eq. ye^(x^2+y^2)(1+2xy^2))/y'=0 which i still dont know how to solve
You and Joshuaa(response #10) are NOT getting the same/similar solutions!I was earlier too fast and again made a mistake. I think I havesolved the equation. Can you please check out if it is correct? Thank you very much for your help.View attachment 21295View attachment 21296