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

[MarkFL]

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.

 

[MarkFL]

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]

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.

 

[MarkFL]

Glad to help and welcome to MHB! (Cool)

 

[CellCoree]

Hello Raqandre,

Thank you for your question regarding the Cauchy-Euler initial value problem. This type of problem is a common occurrence in differential equations and can be solved using a variety of methods.

To solve this particular problem, we will use the method of undetermined coefficients. This method involves assuming a general form for the solution and then solving for the coefficients using the initial conditions provided.

In this case, we will assume that the solution has the form y = ax^n. Plugging this into the differential equation, we get:

x^2 (ax^n)'' + x(ax^n)' - ax^n = 0

Simplifying this equation, we get:

x^2 a(n)(n-1)x^(n-2) + x a(n)x^(n-1) - ax^n = 0

Collecting like terms, we get:

a(n)(n^2 - n)x^n + a(n)x^n - a(n)x^n = 0

Since the equation must hold for all values of x, we can equate the coefficients to get:

a(n)(n^2 - n) + a(n) = 0

Solving for a(n), we get:

a(n) = 0 or n = 1

Therefore, our general solution is y = ax + b, where a and b are constants.

Now, using the initial conditions provided, we can solve for the values of a and b. Plugging in x = 1 and y = 0, we get:

0 = a(1) + b

Plugging in x = 1 and y' = 2, we get:

2 = a

Therefore, our final solution is y = 2x - 2.

I hope this helps to clarify the solution to your problem. Good luck with your studies!

Best,