# Power series solution for differential equation

1. Jul 28, 2011

### Jen_Jer_888

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 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)^(n-1) 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. Jul 29, 2011

### tiny-tim

Welcome to PF!

Hi Jen_Jer_888! Welcome to PF!

(try using the X2 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 x3 - 3x - 2 ?

3. Jul 29, 2011

### Jen_Jer_888

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 - x2 => x(dy/dx) = 2+ 3x - x3 => -x(dy/dx) = x3 - 3x - 2.
This factors giving -x(dy/dx) = (x+1)2(x-2)
So dy/dx = [(x+1)2(x-2)]/(-x).
So dy/dx = [z2(z-3)]/(-(z-1))
Next I did polynomial long division for (z3 - 3z2)/(-z + 1) and I got this: -z2 + 2z + 2 Remainder -2 or -z2 + 2z + 2 + -2/(-z+1)

Then I did long division for the remainder. Flipping the bottom around, I did -2/(1-z) and got this: -2 - 2z - 2z2 - 2z3 - 2z4 + ...

Plugging into the original quotient, I get -3z2 - 2z3 - 2z4 - 2z5 + ...

Next, I lined up coefficients with those of the first derivative of the power series $\sum$ anzn, since we are assuming a power series solution. That gives a1 + 2a2z + 3a3z2 + 4a4z3 + ... = -3z2 - 2z3 - 2z4 - 2z5 + ...

Thus, a1 = 0, a2 = 0, a3 = -1, a4 = -2/4, a5 = -2/5, a6 = -2/6, a7 = -2/7 and so on

Plugging those into the power series for y gives: y = -z3 - (2/4)z4 - (2/5)z5 - (2/6)z6 - (2/7)z7 + ...

Is this an acceptable answer to the original question? Have I gone wrong anywhere?

4. Jul 29, 2011

### Jen_Jer_888

After I get y = -z3 - (2/4)z4 - (2/5)z5 - (2/6)z6 - (2/7)z7 + ... 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 = -x3 - (2/4)x4 - (2/5)x5 - (2/6)x6 - (2/7)x7 + ...

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?

Last edited: Jul 29, 2011
5. Jul 29, 2011

### tiny-tim

Hi Jen_Jer_888!

I'm really not sure what they're expecting you to do
… the obvious way of solving this is simply to integrate the RHS, which gives you a cubic polynomial in x minus 2ln(x),

and ln(x) = ln(1-z) = -(z + z2/2 + z3/3 + … )

= -((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.

(And I've no idea whether they intend you to expand all those brackets … but personally I wouldn't.)​