Determine the principal part of the following 2nd order PDE and change the principal part to canonical form (without mixed derivatives). Uxx+8Uxy+4U₁ = 0. 9. Transform the following equations to a canonical form (i.e. without cross-derivatives): (i) 4U 12Uxy+9Uyy = 0. (ii) 2Uxx - 4Uxy - 6Uyy + Ur = 0. (iii) Urr+2Ury + 17Uyy = 0. To which type do each of the equations above belongs to? 9. (2) use x = x, we Fel y' = −x + Y, 34₁ = 3/4 So PDE becomes Scanned with CS CamScanner 2/41xx + 2-(-6)-4 + Ux. Uy' = 0 2° & > Ux'x' - 8Uy'y' + Ux' + Uy' = 0

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter8: Applications Of Trigonometry
Section: Chapter Questions
Problem 27RE
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Could you please help me solve the question ( that has no solution attached )using a method similar to the one my teacher used in class? I'd like to consolidate this solution with my class notes. For reference,I have linked a similar exercise question and the solution my teacher gave for it.

Determine the principal part of the following 2nd order PDE and change the
principal part to canonical form (without mixed derivatives).
Uxx+8Uxy+4U₁ = 0.
Transcribed Image Text:Determine the principal part of the following 2nd order PDE and change the principal part to canonical form (without mixed derivatives). Uxx+8Uxy+4U₁ = 0.
9. Transform the following equations to a canonical form (i.e. without cross-derivatives):
(i) 4U 12Uxy+9Uyy = 0.
(ii) 2Uxx - 4Uxy - 6Uyy + Ur = 0.
(iii) Urr+2Ury + 17Uyy = 0.
To which type do each of the equations above belongs to?
9.
(2)
use
x = x,
we
Fel
y' = −x + Y,
34₁ = 3/4
So PDE becomes
Scanned with
CS CamScanner
2/41xx +
2-(-6)-4
+ Ux. Uy' = 0
2°
&
> Ux'x' - 8Uy'y' + Ux' + Uy'
= 0
Transcribed Image Text:9. Transform the following equations to a canonical form (i.e. without cross-derivatives): (i) 4U 12Uxy+9Uyy = 0. (ii) 2Uxx - 4Uxy - 6Uyy + Ur = 0. (iii) Urr+2Ury + 17Uyy = 0. To which type do each of the equations above belongs to? 9. (2) use x = x, we Fel y' = −x + Y, 34₁ = 3/4 So PDE becomes Scanned with CS CamScanner 2/41xx + 2-(-6)-4 + Ux. Uy' = 0 2° & > Ux'x' - 8Uy'y' + Ux' + Uy' = 0
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