Solving an ODE: The Pwer Series and Seperation of Variables

[TFM]

Homework Statement

Solve the following equation by a power series and also by separation of variables. Check that the two agree.

Homework Equations

N/A

The Attempt at a Solution

Power Series:

[tex] (1+x) \frac{dy}{dx} = y [/tex]

[tex] (1+x) \frac{1}{dx} = y \frac{1}{dy} [/tex]

The power series is:

[tex] (1+x) \equiv 1+x+0x^2+0x^x [/tex] ...

Thus

[tex] \frac{1 + x}{dx} = \frac{y}{dy} [/tex]


Separation by Variables:

[tex] (1+x) y' = y [/tex]

[tex] y' = \frac{1}{((1+x)} y [/tex]

[tex] x'=g(t)h(x) [/tex]

[tex] H(x) = G(t) + C [/tex]

[tex] H=\int \frac{dx}{h(x)} ; G=\int g(t)dt [/tex]

[tex] H=\int \frac{dy}{y} \equiv \int \frac{1}{y} dy = ln y [/tex]

[tex] G = \int \frac{1}{1+x} dx = ln(1 + x) [/tex]

[tex] ln y = ln (1+x) + c [/tex]

[tex] y = 1 + x + c [/tex]

These two methods haven't agreed for this question. i think the problem lays in my Power Series.

Anyone got any idesa?

TFM

 

[CompuChip]

Your second solution looks right, up to
ln(y) = ln(1 + x) + c
But if you exponentiate that, you won't get
y = 1 + x + c,
do that step again.

As for your real question, I suppose that they mean: plug in a solution
[tex]y(x) = \sum_{n = 0}^\infty a_n x^n[/tex]
and determine the coefficients [itex]a_n[/itex] from the differential equation.
Note that you can differentiate by terms, so
[tex]\frac{dy}{dx} = \sum_{n = 0}^\infty n a_n x^{n - 1}[/tex]
etc.

 

[TFM]

Indeed it won't:

ln(y) = ln(1 + x) + c

take exponentials of both sides leaves:

[tex] y = 1 + x + e^c [/tex]

Okay so for the power series:

[tex] y(x) = \sum_{n = 0}^\infty a_n x^n [/tex]

so would the coeffieients coming from here:

[tex] (1+x) \frac{1}{dx} = y \frac{1}{dy} [/tex]

be 1 and 1, since you have x^0 has a coefficient of 1, and x has a coefficent of 1 also?

TFM

 

[CompuChip]

No,
e^(ln(1 + x) + c) = e^ln(1+x) * e^c = ...

And what do you mean by: "the coefficients being 1 and 1"? You have infinitely many of them! Start by plugging the sum into
[tex]
(1+x) \frac{dy}{dx} = y [/tex]
- what do you get?

 

[TFM]

No,
e^(ln(1 + x) + c) = e^ln(1+x) * e^c = ...

Since e^c is another constant, wouldn't it not be:

= k + kx

?

And what do you mean by: "the coefficients being 1 and 1"? You have infinitely many of them! Start by plugging the sum into
[tex]
(1+x) \frac{dy}{dx} = y [/tex]
- what do you get?

So I should have:

[tex] (1+x) \frac{dy}{dx} = \sum_{n = 0}^\infty a_n x^n [/tex]

?

TFM

 

[gabbagabbahey]

Since e^c is another constant, wouldn't it not be:

= k + kx

?

Yes, that would be fine.

So I should have:

[tex] (1+x) \frac{dy}{dx} = \sum_{n = 0}^\infty a_n x^n [/tex]

?

TFM

Well, that is a true statement, but not very useful to you in that form. If [tex]y(x)= \sum_{n = 0}^\infty a_n x^n[/tex]...what is [tex]\frac{dy}{dx}[/tex]?

 

[TFM]

Would this be:

[tex] y(x)= \sum_{n = 0}^\infty a_n x^n [/tex]

[tex] \frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n - 1) x^{n-1} [/tex]

Does this look right?

TFM

 

[CompuChip]

Almost; is [itex]2 x^2[/itex] the derivative of [itex]x^3[/itex]?

Now write down your complete differential equation.

 

[TFM]

Okay so:

x^3 differentiates to 3x^2

[tex] \frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n) x^{n-1} [/tex]

Now write down your complete differential equation.

so would this be:

[tex] \frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n) x^{n-1} + a_{n-1}(n+1) x^{n} + a_{n-2}(n+2) x^{n + 1} + ... [/tex]

?

TFM