Can Euler's Differential Equation Be Solved Using Initial Values?

In summary: Bx^{5/2}+...[/tex] Now, put x= 1 and y= 0. That gives A(1)+...+ B(1)= 0 so B= -A. Also, dy/dx= -3Ax^{-4}+ ...+ 5/2Bx^{3/2}+... sody/dx(1)= -3A(1)+ ...+ 5/2B(1)= 1. That immediately gives B= 0 so A= 0. That is "unfortunate". If you use the substitution u= ln x, then dy/dx= dy/du du/dx= x(dy
  • #1
JamesEllison
13
0
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:
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  • #2
Hi James! :smile:

(try using the X2 button just above the Reply box :wink:)
JamesEllison said:
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! :smile:

(i haven't checked your result)
 
  • #3
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
JamesEllison said:
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!
 
  • #5
Hi James! :smile:

(just got up :zzz:)

JamesEllison said:
… 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) :redface:

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

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

those factors should be on top :wink:

hmm … usually exam questions like this factor out nicely :confused:

let's go back and check …
JamesEllison said:
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0

2m^2 +3m - 1 = 0

ahhh! :biggrin:

how did you get that? :rolleyes:
 
  • #6
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
JamesEllison said:
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0
JamesEllison said:
2m2 + 3m - 15 = 0

is your screen too small? :confused:

2m2 + m - 15 = 0 :smile:
 
  • #8
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
 
  • #9
JamesEllison said:
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 :yuck: …

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

fine until then! :smile:
 
  • #11
y = Ae5t/2 + Be-3t :smile:

(now convert to x, then solve for the initial conditions)
 
  • #12
JamesEllison said:
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]
 
  • #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!
 
Last edited:
  • #14
JamesEllison said:
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.
 

1. What is Euler's differential equation?

Euler's differential equation is a first-order ordinary differential equation that is used to model growth and decay processes in various fields of science and engineering.

2. Who is Leonhard Euler and why is this equation named after him?

Leonhard Euler was a Swiss mathematician who made significant contributions to various areas of mathematics, including differential equations. This equation is named after him because he was the first to introduce and study it in detail.

3. What are the applications of Euler's differential equation?

Euler's differential equation has many applications in physics, chemistry, biology, economics, and other fields. It is commonly used to model population growth, radioactive decay, chemical reactions, and other natural phenomena.

4. What is the general form of Euler's differential equation?

The general form of Euler's differential equation is dy/dx = f(x)y, where f(x) is a function of x and y is the dependent variable. This equation can also be written as y' = f(x)y.

5. How is Euler's differential equation solved?

Euler's differential equation can be solved using various methods, such as separation of variables, substitution, and integration. The solution depends on the specific form of the equation and the initial conditions given.

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