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Euler Differential Equation 
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#1
May2112, 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. 


#2
May2212, 06:37 AM

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Hi James!
(try using the X^{2} button just above the Reply box ) Yes, that's the method! (i haven't checked your result) 


#3
May2212, 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? 


#4
May2312, 12:26 AM

P: 81

Euler Differential Equation



#5
May2312, 03:25 AM

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Hi James!
(just got up ) then fine , until … hmm … usually exam questions like this factor out nicely let's go back and check … how did you get that? 


#6
May2312, 08:20 AM

P: 13

Cool,
So the assumption of my quadratic formula should be: 2m^{2} + 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 


#7
May2312, 09:08 AM

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2m^{2} + m  15 = 0 


#8
May2312, 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 2m^{2}+ 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 


#9
May2312, 01:13 PM

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the plot was, put x = e^{t} and solve for y against t ! fine until then! 


#10
May2412, 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 


#11
May2412, 01:23 AM

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y = Ae^{5t/2} + Be^{3t}
(now convert to x, then solve for the initial conditions) 


#12
May2412, 04:56 PM

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[itex]2m^2+ m 15= (2m 5)(m+ 3)= 0[/itex] with roots m= 5/2 and m= 3. 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^{m1}[/itex], [itex]y''= m(m1)x^{m2}[/itex] and your differential equation becomes [tex]2x^2(m(m1)x^{m2})+ 3x(mx^{m1}) 15x^m= (2m(m1)+ 3m 15)x^m= 0[/itex] In order that that be 0 for all x, we must have [itex]2m(m1)+ 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
May2512, 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! 


#14
May2512, 09:21 AM

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