# Euler Differential Equation

1. May 21, 2012

### 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.

Last edited: May 21, 2012
2. May 22, 2012

### tiny-tim

Hi James!

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

Yes, that's the method!

3. May 22, 2012

### 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?

4. May 23, 2012

### yus310

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

5. May 23, 2012

### tiny-tim

Hi James!

(just got up :zzz:)

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?

6. May 23, 2012

### 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:
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

7. May 23, 2012

### tiny-tim

2m2 + m - 15 = 0

8. May 23, 2012

### 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$\frac{d^2y}{dt^2}$ - 2$\frac{dy}{dt}$ +3$\frac{dy}{dt}$ - 15y = 0

2$\frac{d^2y}{dt^2}$ +$\frac{dy}{dt}$ - 15y = 0

2m2+ m - 15 = 0

m = $\frac{-1±√121}{4}$
m = 5/2
m = -3

y = Aemx + Bemx
y = Ae$\frac{5x}{2}$ + Be-3x

IC1 : y(1) = 0
y = Ae$\frac{5}{2}$ + Be-3 = 0

IC2 : $\frac{dy}{dx}$ = 1

$\frac{dy}{dx}$ = $\frac{5}{2}$ Ae$\frac{5}{2}$ - 3Be-3 = 1

$\frac{d^2y}{dx^2}$ = ?

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

9. May 23, 2012

### tiny-tim

uh-oh, you've lost the plot :yuck: …

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

fine until then!

10. May 24, 2012

### JamesEllison

11. May 24, 2012

### tiny-tim

y = Ae5t/2 + Be-3t

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

12. May 24, 2012

### HallsofIvy

Staff Emeritus
Yes, that's good.

But where did you get this? The characteristic equation for 2y''+ y'- 15= 0 is
$2m^2+ m- 15= (2m- 5)(m+ 3)= 0$ 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 $x^m$. Then $y'= mx^{m-1}$, $y''= m(m-1)x^{m-2}$ and your differential equation becomes
$$2x^2(m(m-1)x^{m-2})+ 3x(mx^{m-1})- 15x^m= (2m(m-1)+ 3m- 15)x^m= 0[/itex] In order that that be 0 for all x, we must have $2m(m-1)+ 3m- 15= 2m^2+ m- 15= 0$, exactly the same characteristic equation as before. Since that has roots -3 and 5/2, the general solution is [tex]y= Ax^{-3}+ Bx^{5/2}$$

13. May 25, 2012

### JamesEllison

Ah. Terriffic.

y = ax-3 + bx$\frac{5}{2}$

Apply IC1: y(1) = 0

y(1) = a(1)-3 + b(1)$\frac{5}{2}$ = 0

y(1) = a + b

=> a = -b

Then
$\frac{dy}{dx}$ = -3ax-3 + $\frac{5}{2}$bx$\frac{5}{2}$

IC2:
y'(1) = 1
1 = -3a + $\frac{5}{2}$b

Since a = -b

3b + $\frac{5}{2}$b = 1

$\frac{6}{2}$b + $\frac{5}{2}$b = 1

11b = 2

b = $\frac{2}{11}$

a = $\frac{-2}{11}$

y(x) = $\frac{-2}{11}$x-3 + $\frac{2}{11}$x$\frac{5}{2}$

Thanks very much for all your help Tim and HallsofIvy, ps your method seems much faster!

Last edited: May 25, 2012
14. May 25, 2012

### HallsofIvy

Staff Emeritus
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.