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

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Ivp
AI Thread Summary
The initial value problem x^2 y" + xy' - y = 0 with conditions y(1) = 0 and y'(1) = 2 can be solved by assuming a solution of the form y = x^r. Substituting this into the differential equation leads to the characteristic equation r^2 - 1 = 0, yielding the general solution y(x) = c_1 x + c_2 x^{-1}. Applying the initial conditions results in a system of equations that determines the constants c_1 and c_2 as 1 and -1, respectively. Therefore, the specific solution to the IVP is y(x) = x - 1/x.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$y(x)=x-x^{-1}=x-\frac{1}{x}$$
 
Re: raqandre's question at Yahoo! Answers regarding a Cauchy-euler IVP

MarkFL said:
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.
 
Glad to help and welcome to MHB! (Cool)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top