Power series solution for differential equation

[Jen_Jer_888]

Homework Statement

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.

Homework Equations

dy/dx = 2/x + 3 - x^2

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!

 

[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 ?

 

[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 $\sum$ a_nzⁿ, 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?
I really appreciate your help!

 

[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?

 

[tiny-tim]

Hi Jen_Jer_888!

I'm really not sure what they're expecting you to do …

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.

… 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.)​