MHB 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:

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

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

$$y=x^r$$

and so:

$$y'=rx^{r-1}$$

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

Now, substituting into the ODE gives us:

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

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

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

$$x^r(r+1)(r-1)=0$$

Thus, the general solution is:

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

Differentiating, we find:

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

Using the initial conditions, we get the linear system:

$$y'(1)=c_1-c_2=2$$

$$y(1)=c_1+c_2=0$$

From which we may determine:

$$c_1=1,\,c_2=-1$$

Thus, the solution satisfying the given IVP is:

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

 

[RAQANDRE]

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

Hello raqandre,

We are given the IVP:

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

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

$$y=x^r$$

and so:

$$y'=rx^{r-1}$$

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

Now, substituting into the ODE gives us:

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

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

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

$$x^r(r+1)(r-1)=0$$

Thus, the general solution is:

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

Differentiating, we find:

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

Using the initial conditions, we get the linear system:

$$y'(1)=c_1-c_2=2$$

$$y(1)=c_1+c_2=0$$

From which we may determine:

$$c_1=1,\,c_2=-1$$

Thus, the solution satisfying the given IVP is:

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

Thank you.

 

[MarkFL]

Glad to help and welcome to MHB! (Cool)