Simple ODE

**flipperpk** · December 6th 2009, 02:51 PM

Solve using the power series method:

$y' + xy = 0, y(0) = 1$

Can anyone show the steps to do this?

**Prove It** · December 6th 2009, 03:38 PM

Originally Posted by flipperpk

Solve using the power series method:

$y' + xy = 0, y(0) = 1$

Can anyone show the steps to do this?

Why use the power series method? It's seperable and can also be solved using an integrating factor...

Seperating:

$\frac{dy}{dx} + xy = 0$

$\frac{dy}{dx} = -xy$

$\frac{1}{y}\,\frac{dy}{dx} = -x$

$\int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{-x\,dx}$

$\int{\frac{1}{y}\,dy} = -\frac{1}{2}x^2 + C_1$

$\ln{|y|} + C_2 = -\frac{1}{2}x^2 + C_1$

$\ln{|y|} = -\frac{1}{2}x^2 + C$ , where $C = C_1 - C_2$

$|y| = e^{-\frac{1}{2}x^2 + C}$

$|y| = e^Ce^{-\frac{1}{2}x^2}$

$y = \pm e^C e^{-\frac{1}{2}x^2}$

$y = Ae^{-\frac{1}{2}x^2}$ , where $A = \pm e^C$ .

Now use your initial condition to find A.

**flipperpk** · December 6th 2009, 03:58 PM

Originally Posted by Prove It

Why use the power series method? It's seperable and can also be solved using an integrating factor...

Seperating:

$\frac{dy}{dx} + xy = 0$

$\frac{dy}{dx} = -xy$

$\frac{1}{y}\,\frac{dy}{dx} = -x$

$\int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{-x\,dx}$

$\int{\frac{1}{y}\,dy} = -\frac{1}{2}x^2 + C_1$

$\ln{|y|} + C_2 = -\frac{1}{2}x^2 + C_1$

$\ln{|y|} = -\frac{1}{2}x^2 + C$ , where $C = C_1 - C_2$

$|y| = e^{-\frac{1}{2}x^2 + C}$

$|y| = e^Ce^{-\frac{1}{2}x^2}$

$y = \pm e^C e^{-\frac{1}{2}x^2}$

$y = Ae^{-\frac{1}{2}x^2}$ , where $A = \pm e^C$ .

Now use your initial condition to find A.

I appreciate the help, but my professor wants us to use the power series method.

**Prove It** · December 6th 2009, 04:00 PM

Well, you could get the final answer and then express it as a power series...

**TheEmptySet** · December 6th 2009, 04:40 PM

Originally Posted by flipperpk

Solve using the power series method:

$y' + xy = 0, y(0) = 1$

Can anyone show the steps to do this?

Let

$y=\sum_{n=0}^{\infty}a_nx^n$ Then

$y'=\sum_{n=1}^{\infty}na_nx^{n-1}$

Now substituing these into the equation gives

$\sum_{n=1}^{\infty}na_nx^{n-1}+x\sum_{n=0}^{\infty}a_nx^n=0$

$\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^{n+1}=0$

Now we take the first term out of the 1st series to get

$a_1+\sum_{n=2}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^{n+1}=0$

Now reindex the first series with $n \to n-2$ to get

$a_1+\sum_{n=0}^{\infty}(n+2)a_{n+2}x^{n+1}+\sum_{n =0}^{\infty}a_nx^{n+1}=0$

Now combing the sums gives

$a_1+\sum_{n=0}^{\infty}[(n+2)a_{n+2}+a_n]x^{n+1}=0$

$(n+2)a_{n+2}+a_n=0 \implies a_{n+2}=-\frac{a_n}{n+2}$

Also note that the above equation implies that $a_1=0$

Using your intial condition you will find that $a_0=1$

With the relation about you can find all of the coeffeints of the power series.

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