The discussion revolves around solving the initial value problem (IVP) defined by the Cauchy-Euler differential equation \(x^2 y'' + xy' - y = 0\) with the initial conditions \(y(1) = 0\) and \(y'(1) = 2\). The focus is on exploring potential solutions and methods for solving this type of differential equation.
There is no explicit disagreement in the discussion, as both participants present the same method and arrive at the same solution. However, no alternative methods or solutions are explored, leaving the discussion somewhat limited in scope.
The discussion does not address potential limitations or alternative approaches to solving the IVP, nor does it explore the implications of the solution found.
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
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:
$$y(x)=x-x^{-1}=x-\frac{1}{x}$$