The initial value problem (IVP) given by the equation x²y" + xy' - y = 0, with initial conditions y(1) = 0 and y'(1) = 2, can be solved using the method of guessing a solution of the form y = x^r. Substituting this form into the differential equation leads to the characteristic equation r² - 1 = 0, yielding the general solution y(x) = c₁x + c₂x⁻¹. Applying the initial conditions results in the specific solution y(x) = x - x⁻¹.
PREREQUISITESStudents and professionals in mathematics, particularly those focusing on differential equations, as well as educators teaching ODEs and IVPs.
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:
$$y(x)=x-x^{-1}=x-\frac{1}{x}$$