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Chapter 8: Problem 34

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

Short Answer

Expert verified
The general solution to the differential equation is \(y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}\) where \(r_1\) and \(r_2\) are roots of characteristic equation given by the roots of the quadratic equation \(r^2 + (1 + \frac{2}{x})r - \frac{1}{x} = 0\). The exact form will depend on the roots of this quadratic equation.

Step by step solution

01

Rewrite in standard form

A standard form of the linear second order differential equation is \(y'' + p(x)y' + q(x)y = 0\). To bring the given equation into standard form, divide every term by \(x\), giving us \(y'' + (1 + \frac{2}{x})y' - \frac{1}{x}y = 0\) .
02

Solve the homogeneous equation

In this case, we only have a homogeneous differential equation as there is no forcing function. The characteristic polynomial for this differential equation is given by \(r^2 + (1 + \frac{2}{x})r - \frac{1}{x} = 0\).Solve this quadratic equation, using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, \(a=1\), \(b = 1 + \frac{2}{x}\), and \(c = - \frac{1}{x}\). Apply the formula for \(r\) and simplify the roots as much as possible. Then the homogeneous solution will be \(y_h = c_1 e^{r_1 x} + c_2 e^{r_2 x}\) where \(r_1\) and \(r_2\) are roots of characteristic equation and \(c_1\) and \(c_2\) are arbitrary constants.
03

Construct the general solution

As no particular solution is necessary, the general solution is the same as the homogeneous solution. It is given by \(y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}\).

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Most popular questions from this chapter

Use the method of Frobenius and a reduction of order procedure (see page 197\()\) to find at least the first three nonzero terms in the series expansion about the irregular singular point \(x=0\) for a general solution to the differential equation \(x^{2} y^{\prime \prime}+y^{\prime}-2 y=0\)

$$x^{2} z^{\prime \prime}+\left(x^{2}+x\right) z^{\prime}-z=0$$

\(\left(1+x^{2}\right) y^{\prime \prime}-x y^{\prime}+y=e^{-x}\)

$$3 x y^{\prime \prime}+(2-x) y^{\prime}-y=0$$

\(f(x)=\frac{1+x}{1-x}, \quad x_{0}=0\)
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