Euler Differential Equation

JamesEllison

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.

tiny-tim

Hi James!

(try using the X² button just above the Reply box )
(you mean = y, not 0 !)

Yes, that's the method!

(i haven't checked your result)

JamesEllison

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

Excellent! You are thinking like mathematican! Keeping going! gOOD JOB!

tiny-tim

Hi James!

(just got up )

no, it's unnecessary, it makes no difference (and it'll probably lose you a mark in the exam)

then fine , until …
those factors should be on top

hmm … usually exam questions like this factor out nicely

let's go back and check …
ahhh!

how did you get that?

JamesEllison

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:
Ae^mx + Be^mx = y
Ae^2.089x + Be^-3.589x = y

IC 1
y = Ae^2.089 + Be^-3.589 = 0

IC2
dy/dx = 2.089Ae^2.089 - 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

is your screen too small?

2m² + m - 15 = 0

JamesEllison

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 = Ae^mx + Be^mx
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

uh-oh, you've lost the plot …

the plot was, put x = e^t and solve for y against t !

fine until then!

JamesEllison

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

y = Ae^5t/2 + Be^-3t

(now convert to x, then solve for the initial conditions)

HallsofIvy

Yes, that's good.

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.

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

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

You mean a+ b= 0.
Probably a typo: -3x^-4 but fortunately at x= 1, it doesn't matter.

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.