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Euler Differential Equation

by JamesEllison
Tags: 2nd order pde, euler differential
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JamesEllison
#1
May21-12, 09:22 PM
P: 13
Problem: Solve the initial Value:
when x=1, y=0
dy/dx = 1

2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0

My attempt:

x = e^t
dx/dt = e ^t

dy/dt = dy/dx * dx/dt
dy/dt = x*dy/dx

d^2y/dt^2
= d/dt(dy/dt)
= d/dt(x*dy/dx)
=d/dx(x*dy/dx)*(dx/dt)
since dx/dt = x
=(x^2*d^2y/dx^2) + (x*dy/dx)
=(x^2*d^2y/dx^2) + (dy/dt)
(x^2*d^2y/dx^2) = (d^2y/dt^2) - (dy/dt)

From this point, do I simply substitute the Values for the 1st/2nd derivatives in ? i.e.

2((d^2y/dt^2) - (dy/dt)) + 3(dy/dt) - 15y = 0
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0

2m^2 +3m - 1 = 0

Then solve for m,

Then create the solution in the form of :

Ae^mt + Be^mt = 0 ??

Is that the right path? Any help is appreciated.

Cheers.
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tiny-tim
#2
May22-12, 06:37 AM
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Hi James!

(try using the X2 button just above the Reply box )
Quote Quote by JamesEllison View Post
2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0

My attempt:

x = e^t

From this point, do I simply substitute the Values for the 1st/2nd derivatives in ?

Ae^mt + Be^mt = 0 ??
(you mean = y, not 0 !)

Yes, that's the method!

(i haven't checked your result)
JamesEllison
#3
May22-12, 07:08 PM
P: 13
Excellent. Sorry for the lack of real looking equations, I am on my phone.
2m^2 + 3m - 1 = 0
m = [-3+/- sqrt( 9 + 8 )]/4
= 1.1231 / 4
=0.2808
m = -7.1231/4
= -1.7808

Should y be assigned a coefficient?

Ae^mx + Be^mx = Cy
Ae^0.2808x + Be^-1.7808x = Cy
Sub in IC 1
x = 1
y = 0

Ae^0.2808 + Be^-1.7808 = 0

And IC 2
dy/dx = 1

Ae^0.2808/0.2808 -
Be^-1.7808/1.7808 = 1

Is that going in the right direction?

yus310
#4
May23-12, 12:26 AM
P: 81
Euler Differential Equation

Quote Quote by JamesEllison View Post
Excellent. Sorry for the lack of real looking equations, I am on my phone.
2m^2 + 3m - 1 = 0
m = [-3+/- sqrt( 9 + 8 )]/4
= 1.1231 / 4
=0.2808
m = -7.1231/4
= -1.7808

Should y be assigned a coefficient?

Ae^mx + Be^mx = Cy
Ae^0.2808x + Be^-1.7808x = Cy
Sub in IC 1
x = 1
y = 0

Ae^0.2808 + Be^-1.7808 = 0

And IC 2
dy/dx = 1

Ae^0.2808/0.2808 -
Be^-1.7808/1.7808 = 1

Is that going in the right direction?
Excellent! You are thinking like mathematican! Keeping going! gOOD JOB!
tiny-tim
#5
May23-12, 03:25 AM
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Hi James!

(just got up )

Quote Quote by JamesEllison View Post
Should y be assigned a coefficient?

Ae^mx + Be^mx = Cy
no, it's unnecessary, it makes no difference (and it'll probably lose you a mark in the exam)

then fine , until
And IC 2
dy/dx = 1

Ae^0.2808/0.2808 -
Be^-1.7808/1.7808 = 1
those factors should be on top

hmm usually exam questions like this factor out nicely

let's go back and check
Quote Quote by JamesEllison View Post
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0

2m^2 +3m - 1 = 0
ahhh!
how did you get that?
JamesEllison
#6
May23-12, 08:20 AM
P: 13
Cool,

So the assumption of my quadratic formula should be:

2m2 + 3m - 15 = 0

m = (-3√129)/4
m = 2.089
m = -3.589

So that general assumption is:
Aemx + Bemx = y
Ae2.089x + Be-3.589x = y

IC 1
y = Ae2.089 + Be-3.589 = 0

IC2
dy/dx = 2.089Ae2.089 - 3.589Be-3.589 = 1

How do I go about finding the A and B co efficients??

PS

Thanks for the responses :D
tiny-tim
#7
May23-12, 09:08 AM
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Quote Quote by JamesEllison View Post
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0
Quote Quote by JamesEllison View Post
2m2 + 3m - 15 = 0
is your screen too small?
2m2 + m - 15 = 0
JamesEllison
#8
May23-12, 01:04 PM
P: 13
Thanks very much. Officially on a computer now. My screen was far to small, and i was writing out by hand first to see what they'd really look like. :)
Thanks for being patient.

2[itex]\frac{d^2y}{dt^2}[/itex] - 2[itex]\frac{dy}{dt}[/itex] +3[itex]\frac{dy}{dt}[/itex] - 15y = 0

2[itex]\frac{d^2y}{dt^2}[/itex] +[itex]\frac{dy}{dt}[/itex] - 15y = 0


2m2+ m - 15 = 0

m = [itex]\frac{-1√121}{4}[/itex]
m = 5/2
m = -3

y = Aemx + Bemx
y = Ae[itex]\frac{5x}{2}[/itex] + Be-3x

IC1 : y(1) = 0
y = Ae[itex]\frac{5}{2}[/itex] + Be-3 = 0

IC2 : [itex]\frac{dy}{dx}[/itex] = 1

[itex]\frac{dy}{dx}[/itex] = [itex]\frac{5}{2}[/itex] Ae[itex]\frac{5}{2}[/itex] - 3Be-3 = 1

[itex]\frac{d^2y}{dx^2}[/itex] = ?

Not entirely sure where to go now with substitution..

Gah, i am getting stuck too often. I am so tired :( heading off to bed, its 4am here. Really appreciate your help tim :D
tiny-tim
#9
May23-12, 01:13 PM
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Quote Quote by JamesEllison View Post
2[itex]\frac{d^2y}{dt^2}[/itex] - 2[itex]\frac{dy}{dt}[/itex] +3[itex]\frac{dy}{dt}[/itex] - 15y = 0

2[itex]\frac{d^2y}{dt^2}[/itex] +[itex]\frac{dy}{dt}[/itex] - 15y = 0


2m2+ m - 15 = 0

m = [itex]\frac{-1√121}{4}[/itex]
m = 5/2
m = -3

y = Aemx + Bemx
y = Ae[itex]\frac{5x}{2}[/itex] + Be-3x
uh-oh, you've lost the plot

the plot was, put x = et and solve for y against t !
fine until then!
JamesEllison
#10
May24-12, 12:24 AM
P: 13
On phone again xD

"Put x =e^t ; solve for Y against t."

y = Ae^mx + Be^mx
y = Ae^me^t + Be^me^t? Just before I go ahead..

http://tutorial.math.lamar.edu/class...equations.aspx
tiny-tim
#11
May24-12, 01:23 AM
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y = Ae5t/2 + Be-3t

(now convert to x, then solve for the initial conditions)
HallsofIvy
#12
May24-12, 04:56 PM
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Quote Quote by JamesEllison View Post
Problem: Solve the initial Value:
when x=1, y=0
dy/dx = 1

2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0

My attempt:

x = e^t
dx/dt = e ^t

dy/dt = dy/dx * dx/dt
dy/dt = x*dy/dx

d^2y/dt^2
= d/dt(dy/dt)
= d/dt(x*dy/dx)
=d/dx(x*dy/dx)*(dx/dt)
since dx/dt = x
=(x^2*d^2y/dx^2) + (x*dy/dx)
=(x^2*d^2y/dx^2) + (dy/dt)
(x^2*d^2y/dx^2) = (d^2y/dt^2) - (dy/dt)

From this point, do I simply substitute the Values for the 1st/2nd derivatives in ? i.e.

2((d^2y/dt^2) - (dy/dt)) + 3(dy/dt) - 15y = 0
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0
Yes, that's good.

2m^2 +3m - 1 = 0
But where did you get this? The characteristic equation for 2y''+ y'- 15= 0 is
[itex]2m^2+ m- 15= (2m- 5)(m+ 3)= 0[/itex] with roots m= 5/2 and m= -3.

Then solve for m,

Then create the solution in the form of :

Ae^mt + Be^mt = 0 ??

Is that the right path? Any help is appreciated.
-
Cheers.
Right path- wrong solution to the equation.

By the way, while I am a strong advocate of changing the variable to convert the equation to one with constant coefficients, a "short cut" is to "try" a solution to the original equation of the form [itex]x^m[/itex]. Then [itex]y'= mx^{m-1}[/itex], [itex]y''= m(m-1)x^{m-2}[/itex] and your differential equation becomes
[tex]2x^2(m(m-1)x^{m-2})+ 3x(mx^{m-1})- 15x^m= (2m(m-1)+ 3m- 15)x^m= 0[/itex]
In order that that be 0 for all x, we must have [itex]2m(m-1)+ 3m- 15= 2m^2+ m- 15= 0[/itex], exactly the same characteristic equation as before. Since that has roots -3 and 5/2, the general solution is
[tex]y= Ax^{-3}+ Bx^{5/2}[/tex]
JamesEllison
#13
May25-12, 07:56 AM
P: 13
Ah. Terriffic.

y = ax-3 + bx[itex]\frac{5}{2}[/itex]

Apply IC1: y(1) = 0

y(1) = a(1)-3 + b(1)[itex]\frac{5}{2}[/itex] = 0

y(1) = a + b

=> a = -b

Then
[itex]\frac{dy}{dx}[/itex] = -3ax-3 + [itex]\frac{5}{2}[/itex]bx[itex]\frac{5}{2}[/itex]

IC2:
y'(1) = 1
1 = -3a + [itex]\frac{5}{2}[/itex]b

Since a = -b

3b + [itex]\frac{5}{2}[/itex]b = 1

[itex]\frac{6}{2}[/itex]b + [itex]\frac{5}{2}[/itex]b = 1

11b = 2

b = [itex]\frac{2}{11}[/itex]

a = [itex]\frac{-2}{11}[/itex]

y(x) = [itex]\frac{-2}{11}[/itex]x-3 + [itex]\frac{2}{11}[/itex]x[itex]\frac{5}{2}[/itex]

Thanks very much for all your help Tim and HallsofIvy, ps your method seems much faster!
HallsofIvy
#14
May25-12, 09:21 AM
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Quote Quote by JamesEllison View Post
Ah. Terriffic.

y = ax-3 + bx[itex]\frac{5}{2}[/itex]

Apply IC1: y(1) = 0

y(1) = a(1)-3 + b(1)[itex]\frac{5}{2}[/itex] = 0

y(1) = a + b
You mean a+ b= 0.
=> a = -b

Then
[itex]\frac{dy}{dx}[/itex] = -3ax-3 + [itex]\frac{5}{2}[/itex]bx[itex]\frac{5}{2}[/itex]
Probably a typo: -3x-4 but fortunately at x= 1, it doesn't matter.

IC2:
y'(1) = 1
1 = -3a + [itex]\frac{5}{2}[/itex]b

Since a = -b

3b + [itex]\frac{5}{2}[/itex]b = 1

[itex]\frac{6}{2}[/itex]b + [itex]\frac{5}{2}[/itex]b = 1

11b = 2

b = [itex]\frac{2}{11}[/itex]

a = [itex]\frac{-2}{11}[/itex]

y(x) = [itex]\frac{-2}{11}[/itex]x-3 + [itex]\frac{2}{11}[/itex]x[itex]\frac{5}{2}[/itex]

Thanks very much for all your help Tim and HallsofIvy, ps your method seems much faster!
As long as your characteristic equation has only distinct real roots it is. But if you have multiple roots or complex roots (or a non-homogeneous equation) it may be simpler to use what you have learned for linear d.e s with constant coefficients.


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