orthogonal trajectory

[Patricia662]

y^2 e^(x^2+y^2)=c find orthogonal trajectory.

 

[Deleted member 4993]

y^2 e^(x^2+y^2)=c find orthogonal trajectory.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[Patricia662]

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'

 

[Deleted member 4993]

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

Your second line of solution is incorrect.

d/dx [e^(x^2+y^2)] =[e^(x^2+y^2)] * [2x + 2yy']

correct it and continue.....

 

[joshuaa]

[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)

 

[Patricia662]

Your second line of solution is incorrect.

d/dx [e^(x^2+y^2)] =[e^(x^2+y^2)] * [2x + 2yy']

correct it and continue.....

Now i have dif. eq. ye^(x^2+y^2)(1+2xy^2))/y'=0 which i still dont know how to solve

 

[Patricia662]

[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)

Can you please explain how you got this?

 

[Deleted member 4993]

Now i have dif. eq. ye^(x^2+y^2)(1+2xy^2))/y'=0 which i still dont know how to solve

I do not arrive at that equation.

Please share your work.

 

[Patricia662]

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.

 

[joshuaa]

first simplify

[MATH]y^2e^{x^2 + y^2} = c[/MATH]
[MATH]\ln(y^2 e^{x^2 + y^2}) = \ln c[/MATH]
[MATH]\ln y^2 + \ln e^{x^2 + y^2} = \ln c[/MATH]
[MATH]\ln y^2 + x^2 + y^2 = \ln c[/MATH]
now, take implicit differentiation

[MATH]\frac{1}{y^2} \cdot 2y \cdot \frac{dy}{dx} + 2x + 2y \cdot \frac{dy}{dx} = 0[/MATH]
[MATH]\frac{2}{y} \cdot \frac{dy}{dx} + 2y \cdot \frac{dy}{dx} = -2x[/MATH]
[MATH]\frac{1}{y} \cdot \frac{dy}{dx} + y \cdot \frac{dy}{dx} = -x[/MATH]
[MATH](\frac{1}{y} + y) \cdot \frac{dy}{dx} = -x[/MATH]
[MATH]\frac{dy}{dx} = \frac{-x}{(\frac{1}{y} + y)}[/MATH]
this means the orthogonal slope is

[MATH]\frac{dy}{dx} = \frac{(\frac{1}{y} + y)}{x}[/MATH]
this is a differential equation. if you want to get [MATH]y[/MATH], rearrange and integrate both sides

[MATH]\frac{dy}{(\frac{1}{y} + y)} = \frac{dx}{x}[/MATH]
[MATH]\int \frac{dy}{(\frac{1}{y} + y)} = \int \frac{dx}{x}[/MATH]
[MATH]\int \frac{1}{(\frac{1}{y} + y)} \ dy = \int \frac{1}{x} \ dx[/MATH]
[MATH]\frac{1}{2}\ln|y^2 + 1| = \ln|x| + K[/MATH]
[MATH]\ln|y^2 + 1| = 2\ln|x| + 2K[/MATH]
[MATH]\ln|y^2 + 1| = \ln|x^2| + 2K[/MATH]
[MATH]y^2 + 1 = x^2e^{2K}[/MATH]
[MATH]e^{2K}[/MATH] is just a constant, you can call it [MATH]A = e^{2K}[/MATH]
then

[MATH]y^2 + 1 = Ax^2[/MATH]
[MATH]y^2 = Ax^2 - 1[/MATH]
[MATH]|y| = \sqrt{Ax^2 - 1}[/MATH]

 

[Patricia662]

Thank you very much. I solved it but still manage to make a small mistake at the end. Now I have just realized that. Thank you for helping me.

 

[Patricia662]

Yes. Thank you for helping me.

 

[Deleted member 4993]

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

You and Joshuaa(response #10) are NOT getting the same/similar solutions!

You got

y² = Ax -1

and Joshuaa got:

y² = Ax² -1

Where is the "problem"?

 

[Patricia662]

I forgot 1/2 on the right side in the integral.