Power series solution for differential equationby Jen_Jer_888 Tags: power series, power series method, series, series solution 

#1
Jul2811, 09:22 PM

P: 4

1. The problem statement, all variables and given/known data
Solve the fluxional equation (y with a dot on top)/(x with a dot on top) = 2/x + 3  x^2 by first replacing x by (x + 1) and then using power series techniques. 2. Relevant equations dy/dx = 2/x + 3  x^2 3. The attempt at a solution First, I believe the fluxional (y with a dot on top)/(x with a dot on top) was just Newton's language and notation for the derivative dy/dx, so I rewrote the equation as [b]2 above. Then I replaced x by (x + 1) like it says, getting: dy/d(x+1) = 2/(x+1) + 3  (x+1)^2 From there, I attempted to set it equal to the sigma series for the derivative of a power series, so: 2/(x+1) + 3  (x+1)^2 = Sigma n*a_sub_n*(x+1)^(n1) from n=1 to infinity = a_sub_1 + 2*a_sub_2*(x+1)+ 3*a_sub_3*(x+1)^2 + 4*a_sub_4*(x+1)^3 + .... I don't know where to go from there. I'm not even sure how the substitution helps. Since there is no y or higher order derivative, I see no basis to compare series coefficients. Any help would be appreciated! 



#2
Jul2911, 04:12 AM

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Hi Jen_Jer_888! Welcome to PF!
(try using the X^{2} icon just above the Reply box ) First, write z instead of (x+1), it's much easier! Second, have you noticed that you can easily factor x^{3}  3x  2 ? 



#3
Jul2911, 09:25 AM

P: 4

Thank you for the response! Here is my updated attempt. Can you verify it or explain if I go wrong?
Let z = x+1. dy/dx = 2/x + 3  x^{2} => x(dy/dx) = 2+ 3x  x^{3} => x(dy/dx) = x^{3}  3x  2. This factors giving x(dy/dx) = (x+1)^{2}(x2) So dy/dx = [(x+1)^{2}(x2)]/(x). So dy/dx = [z^{2}(z3)]/((z1)) Next I did polynomial long division for (z^{3}  3z^{2})/(z + 1) and I got this: z^{2} + 2z + 2 Remainder 2 or z^{2} + 2z + 2 + 2/(z+1) Then I did long division for the remainder. Flipping the bottom around, I did 2/(1z) and got this: 2  2z  2z^{2}  2z^{3}  2z^{4} + ... Plugging into the original quotient, I get 3z^{2}  2z^{3}  2z^{4}  2z^{5} + ... Next, I lined up coefficients with those of the first derivative of the power series [itex]\sum[/itex] a_{n}z^{n}, since we are assuming a power series solution. That gives a_{1} + 2a_{2}z + 3a_{3}z^{2} + 4a_{4}z^{3} + ... = 3z^{2}  2z^{3}  2z^{4}  2z^{5} + ... Thus, a_{1} = 0, a_{2} = 0, a_{3} = 1, a_{4} = 2/4, a_{5} = 2/5, a_{6} = 2/6, a_{7} = 2/7 and so on Plugging those into the power series for y gives: y = z^{3}  (2/4)z^{4}  (2/5)z^{5}  (2/6)z^{6}  (2/7)z^{7} + ... Is this an acceptable answer to the original question? Have I gone wrong anywhere? I really appreciate your help! 



#4
Jul2911, 09:30 AM

P: 4

Power series solution for differential equation
After I get y = z^{3}  (2/4)z^{4}  (2/5)z^{5}  (2/6)z^{6}  (2/7)z^{7} + ... do I need to plug (x+1) back in and get y in terms of x?
Can I replace x + 1 with x now and get y = x^{3}  (2/4)x^{4}  (2/5)x^{5}  (2/6)x^{6}  (2/7)x^{7} + ... If not, expanding each of those with x + 1 still in there would be a huge pain! Or would there be an easy way to do that? 



#5
Jul2911, 09:53 AM

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Hi Jen_Jer_888!
I'm really not sure what they're expecting you to do … and ln(x) = ln(1z) = (z + z^{2}/2 + z^{3}/3 + … ) = ((1+x) + (1+x)^{2}/2 + (1+x)^{3}/3 + … ). Adding that to the polynomial doesn't seem to give the same result as you have, but I can't see the actual mistake. 


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