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raqandre's question at Yahoo! Answers regarding a Cauchy-Euler IVP

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MarkFL

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Feb 24, 2012
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Here is the question:

Solve the initial value problem x^2 y" + xy' - y = 0?

solve the initial value problem x^2 y" + xy' - y = 0 with initial conditions: y(1)= 0 & y'(1) = 2
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Re: raqandre's question at Yahoo! Answers regarding a Cauchy-euler IVP

Hello raqandre,

We are given the IVP:

\(\displaystyle x^2y"+xy'-y=0\) where \(\displaystyle y'(1)=2,\,y(1)=0\)

One way to proceed is the guess a solution of the form:

\(\displaystyle y=x^r\)

and so:

\(\displaystyle y'=rx^{r-1}\)

\(\displaystyle y''=r(r-1)x^{r-2}\)

Now, substituting into the ODE gives us:

\(\displaystyle x^2\left(r(r-1)x^{r-2} \right)+x\left(rx^{r-1} \right)-\left(x^r \right)=0\)

\(\displaystyle x^r\left(r(r-1)+r-1 \right)=0\)

\(\displaystyle x^r\left(r^2-1 \right)=0\)

\(\displaystyle x^r(r+1)(r-1)=0\)

Thus, the general solution is:

\(\displaystyle y(x)=c_1x+c_2x^{-1}\)

Differentiating, we find:

\(\displaystyle y'(x)=c_1-c_2x^{-2}\)

Using the initial conditions, we get the linear system:

\(\displaystyle y'(1)=c_1-c_2=2\)

\(\displaystyle y(1)=c_1+c_2=0\)

From which we may determine:

\(\displaystyle c_1=1,\,c_2=-1\)

Thus, the solution satisfying the given IVP is:

\(\displaystyle y(x)=x-x^{-1}=x-\frac{1}{x}\)
 

RAQANDRE

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Jun 16, 2013
1
Re: raqandre's question at Yahoo! Answers regarding a Cauchy-euler IVP

Hello raqandre,

We are given the IVP:

\(\displaystyle x^2y"+xy'-y=0\) where \(\displaystyle y'(1)=2,\,y(1)=0\)

One way to proceed is the guess a solution of the form:

\(\displaystyle y=x^r\)

and so:

\(\displaystyle y'=rx^{r-1}\)

\(\displaystyle y''=r(r-1)x^{r-2}\)

Now, substituting into the ODE gives us:

\(\displaystyle x^2\left(r(r-1)x^{r-2} \right)+x\left(rx^{r-1} \right)-\left(x^r \right)=0\)

\(\displaystyle x^r\left(r(r-1)+r-1 \right)=0\)

\(\displaystyle x^r\left(r^2-1 \right)=0\)

\(\displaystyle x^r(r+1)(r-1)=0\)

Thus, the general solution is:

\(\displaystyle y(x)=c_1x+c_2x^{-1}\)

Differentiating, we find:

\(\displaystyle y'(x)=c_1-c_2x^{-2}\)

Using the initial conditions, we get the linear system:

\(\displaystyle y'(1)=c_1-c_2=2\)

\(\displaystyle y(1)=c_1+c_2=0\)

From which we may determine:

\(\displaystyle c_1=1,\,c_2=-1\)

Thus, the solution satisfying the given IVP is:

\(\displaystyle y(x)=x-x^{-1}=x-\frac{1}{x}\)
Thank you.
 
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MarkFL

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Feb 24, 2012
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Glad to help and welcome to MHB! (Cool)