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Chapter 8: Problem 32

$$x^{2} y^{\prime \prime}-x(1+x) y^{\prime}+y=0$$

Short Answer

Expert verified
The solution process involves recognizing the given differential equation as a Cauchy-Euler equation, conducting a change of variable to simplify it, solve the standard differential equation, and finally substituting back the original variable.

Step by step solution

01

Verify it's a Cauchy-Euler Differential Equation

The general form of a Cauchy-Euler equation is \(ax^2y'' + bxy' + cy = 0\). If we compare this general form with the given equation \(x^2y'' - x(1 + x)y' + y = 0\), it's clear that it falls into this category.
02

Change of Variable

Conduct a change of variable to convert the Cauchy-Euler equation to a standard differential equation. Substitute \(x = e^t\). Therefore, \(dx = e^t dt\) and the differential \(d/dx\) becomes \(d/dt\). Making these substitutions changes our equation into a new form.
03

Solve the Standard Differential Equation

The equation obtained in Step 2 is a standard linear 2nd order differential equation. Solve this equation using standard methods such as the characteristic equation or integrating factor method. These could involve finding the roots of the homogeneous function and apply it to the non-perfectly homogeneous function (if any), Treating the equation as homogeneous if it is, and then using the general solution for second order linear ODE.
04

Substitute Back the Original Variable

Once the solution is found for variable \(t\), remember to substitute back \(t = \ln{x}\) to get the solution in terms of original variable \(x\) and complete the solution process.

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Most popular questions from this chapter

van der Pol Equation. In the study of the vacuum tube, the following equation is encountered:$$y^{\prime \prime}+(0.1)\left(y^{2}-1\right) y^{\prime}+y=0$$ Find the Taylor polynomial of degree 4 approximating the solution with the initial values \(y(0)=1\), \(y^{\prime}(0)=0\).

Show that the Legendre polynomials of even degree are even functions of \(x,\) while those of odd degree are odd functions.

$$x^{2} y^{n}+4 x y^{\prime}+2 y=0, \text { at } x=0$$

The solution to the initial value problem $$\begin{array}{l}{x y^{\prime \prime}(x)+2 y^{\prime}(x)+x y(x)=0} \\\ {y(0)=1, \quad y^{\prime}(0)=0}\end{array}$$ has derivatives of all orders at \(x=0\) (although this is far from obvious). Use L' Hopital's rule to compute the Taylor polynomial of degree 2 approximating this solution.

$$x w^{\prime \prime}-w^{\prime}-x w=0$$
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