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Chapter 5: Problem 10

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \((x-2)^{2} y^{\prime \prime}+5(x-2) y^{\prime}+8 y=0\)

Short Answer

Expert verified
Question: Find the general solution of the given differential equation \((x-2)^2y'' + 5(x-2)y' + 8y = 0\). Answer: The general solution of the given differential equation is \(y(x) = A (x-2)^{-2+2i} + B (x-2)^{-2-2i}\), where A and B are arbitrary constants.

Step by step solution

01

Introduce a substitution

First, let us introduce a substitution to simplify the differential equation. Let \(z = x - 2\). Then, our equation becomes: \((z)^{2} y^{\prime \prime}+5(z) y^{\prime}+8 y=0\)
02

Standard form of Equation

Now, let's rewrite the differential equation in the standard form. Divide the equation by \(z^2\) to get: \(y^{\prime \prime}+\frac{5}{z} y^{\prime}+\frac{8}{z^2} y=0\)
03

Solution Method

We can now see that this is a Cauchy-Euler differential equation. To solve this, we propose a solution of the form \(y(z) = z^r\), and find the corresponding values of \(r\).
04

Plug in the proposed solution

Replace the function \(y(z)\) and its derivatives into the differential equation: \(y'(z) = rz^{r-1}\) \(y''(z) = r(r-1)z^{r-2}\) Plug these into the differential equation: \(r(r-1)z^{r-2}+\frac{5}{z} rz^{r-1}+\frac{8}{z^2} z^r=0\)
05

Simplify and find characteristic equation

Notice that all terms have at least \(z^{r-2}\) as a factor. We can divide by \(z^{r-2}\) and simplify: \(r(r-1)+5r+8=0\) This equation is the characteristic equation. Solve for r: \(r^2 - r + 5r + 8 = 0\) \(r^2 + 4r + 8 = 0\)
06

Solve the characteristic equation

To find the roots of the quadratic equation, we can use the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) \(r = \frac{-4 \pm \sqrt{(-4)^2 - 4(1)(8)}}{2(1)}\) \(r = \frac{-4 \pm \sqrt{16 - 32}}{2}\) Since \(16-32=-16\), we have a complex solution: \(r = \frac{-4 \pm \sqrt{-16}}{2} = -2 \pm 2i\)
07

General solution

As the roots are complex, the general solution is of the form: \(y(z) = A z^{r_1} + B z^{r_2}\) \(y(z) = A z^{-2+2i} + B z^{-2-2i}\)
08

Transform back to x

Now let's revert back to the variable \(x\) using the substitution \(z = x - 2\): \(y(x) = A (x-2)^{-2+2i} + B (x-2)^{-2-2i}\) This is the general solution of the given differential equation that is valid in any interval not including the singular point.

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Most popular questions from this chapter

Show that the given differential equation has a regular singular point at \(x=0 .\) Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. Find the series solution \((x>0)\) corresponding to the larger root. If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also. \(x^{2} y^{\prime \prime}+x y^{\prime}+(x-2) y=0\)

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that, if \(\alpha\) is zero or a positive even integer \(2 n,\) the series solution \(y_{1}\) reduces to a polynomial of degree \(2 n\) containing only even powers of \(x\). Find the polynomials corresponding to \(\alpha=0,2,\) and \(4 .\) Show that, if \(\alpha\) is a positive odd integer \(2 n+1,\) the series solution \(y_{2}\) reduces to a polynomial of degree \(2 n+1\) containing only odd powers of \(x .\) Find the polynomials corresponding to \(\alpha=1,3,\) and \(5 .\)

First Order Equations. The series methods discussed in this section are directly applicable to the first order linear differential equation \(P(x) y^{\prime}+Q(x) y=0\) at a point \(x_{0}\), if the function \(p=Q / P\) has a Taylor series expansion about that point. Such a point is called an ordinary point, and further, the radius of convergence of the series \(y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\) is at least as large as the radius of convergence of the series for \(Q / P .\) In each of Problems 16 through 21 solve the given differential equation by a series in powers of \(x\) and verify that \(a_{0}\) is arbitrary in each case. Problems 20 and 21 involve nonhomogeneous differential equations to which series methods can be easily extended. Where possible, compare the series solution with the solution obtained by using the methods of Chapter 2 . $$ y^{\prime}+x y=1+x $$

In this section we showed that one solution of Bessel's equation of order zero, $$ L[y]=x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0 $$ is \(J_{0}\), where \(J_{0}(x)\) is given by Fa. ( 7) with \(a_{0}=1\). According to Theorem 5.7 .1 a second solution has the form \((x>0)\) $$ y_{2}(x)=J_{0}(x) \ln x+\sum_{n=1}^{\infty} b_{n} x^{n} $$ (a) Show that $$ L\left[y_{2}\right](x)=\sum_{n=2}^{\infty} n(n-1) b_{n} x^{n}+\sum_{n=1}^{\infty} n b_{n} x^{n}+\sum_{n=1}^{\infty} b_{n} x^{n+2}+2 x J_{0}^{\prime}(x) $$ (b) Substituting the series representation for \(J_{0}(x)\) in Eq. (i), show that $$ b_{1} x+2^{2} b_{2} x^{2}+\sum_{n=3}^{\infty}\left(n^{2} b_{n}+b_{n-2}\right) x^{n}=-2 \sum_{n=1}^{\infty} \frac{(-1)^{n} 2 n x^{2 n}}{2^{2 n}(n !)^{2}} $$ (c) Note that only even powers of \(x\) appear on the right side of Eq. (ii). Show that \(b_{1}=b_{3}=b_{5}=\cdots=0, b_{2}=1 / 2^{2}(1 !)^{2},\) and that $$ (2 n)^{2} b_{2 n}+b_{2 n-2}=-2(-1)^{n}(2 n) / 2^{2 n}(n !)^{2}, \quad n=2,3,4, \ldots $$ Deduce that $$ b_{4}=-\frac{1}{2^{2} 4^{2}}\left(1+\frac{1}{2}\right) \quad \text { and } \quad b_{6}=\frac{1}{2^{2} 4^{2} 6^{2}}\left(1+\frac{1}{2}+\frac{1}{3}\right) $$ The general solution of the recurrence relation is \(b_{2 n}=(-1)^{n+1} H_{n} / 2^{2 n}(n !)^{2}\). Substituting for \(b_{n}\) in the expression for \(y_{2}(x)\) we obtain the solution given in \(\mathrm{Eq} .(10) .\)

Use the method of Problem 23 to solve the given equation for \(x>0 .\) \(x^{2} y^{\prime \prime}+7 x y^{\prime}+5 y=x\)
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