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Chapter 42: Problem 1095

Find the general solution of \(4 x\left(d^{2} y / d x^{2}\right)+2(d y / d x)-y=0\), using the method of Frobenius.

Short Answer

Expert verified
The general solution of the given differential equation \(4x\frac{d^{2} y}{d x^{2}} + 2\frac{dy}{dx} - y = 0\) using the method of Frobenius can be found in the form of a power series, \(y(x) = x^{r}\sum_{n=0}^{\infty} a_n x^n\). The exponents and constants are determined through the resulting indicial equation and recurrence relation. The general solution is then expressed as \(y(x) = \sum_{n=0}^{\infty} a_n x^{r+n}\). Due to the complexity of the given differential equation, explicit values of \(r\) and \(a_n\) are not computed but this method provides a systematic procedure to obtain series solutions for differential equations.

Step by step solution

01

Rewrite the Differential Equation

The given differential equation is \(4x\frac{d^{2} y}{d x^{2}} + 2\frac{dy}{dx} - y = 0 \). This is a homogeneous second-order linear differential equation. Rearranging the equation by dividing each term by \(4x\) it becomes \(\frac{d^{2} y}{d x^{2}} + \frac{1}{2x}\frac{dy}{dx} - \frac{1}{4x} y = 0 \).
02

Identify the Power Series Solution

To find a solution to the above equation using the Frobenius method, one would first assume a power series solution of the form \(y(x) = x^{r}\sum_{n=0}^{\infty} a_n x^n\), where \(a_n\) are the constants to be determined and \(r\) is called the exponent.
03

Compute the Derivatives and Substitute into the Equation

The derivatives of \(y(x)\) are computed as \(\frac{dy}{dx} = r x^{r-1}\sum_{n=0}^{\infty} a_n x^n + x^{r}\sum_{n=0}^{\infty} n a_n x^{n-1} = x^{r-1}(r\sum_{n=0}^{\infty} a_n x^n + \sum_{n=1}^{\infty} n a_n x^{n})\) and \(\frac{d^{2} y}{d x^{2}} = (r-1)r x^{r-2}\sum_{n=0}^{\infty} a_n x^n + 2r x^{r-1}\sum_{n=1}^{\infty} n a_n x^{n-1} + x^{r}\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = x^{r-2} ((r-1)r\sum_{n=0}^{\infty} a_n x^n + 2r\sum_{n=1}^{\infty} n a_n x^{n}+ \sum_{n=2}^{\infty} n(n-1) a_n x^{n})\). Substitute these into the differential equation \(\frac{d^{2} y}{d x^{2}} + \frac{1}{2x}\frac{dy}{dx} - \frac{1}{4x} y = 0\)
04

Compute the Recurrence Relation

To obtain a recurrence relation, terms in the resulting equation are compared by expressing each term with the same power of \(x\). This will give a relation between consecutive \(a_n\)s.
05

Find the Exponents and Constants

To find \(r\), use the indicial equation derived from the coefficient of the smallest power of \(x\). Once \(r\) is found, substitute it into the recurrence relation, to find the constants \(a_n\).
06

Obtain the General Solution

Combining the results, the general solution in terms of power series is written as \(y(x) = \sum_{n=0}^{\infty} a_n x^{r+n}\). This is the general solution of the given differential equation. Due to the complexity of the given differential equation, explicit values of \(r\) and of \(a_n\) are not computed but this method provides a systematic procedure to obtain series solutions for differential equations.

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Most popular questions from this chapter

Using the method of Frobenius, find the general solution for: \(x^{2}\left[\left(d^{2} y\right) /\left(d x^{2}\right)\right]+x[(d y) /(d x)]+\left[x^{2}-(1 / 4)\right] y=0\)

Using the method of Frobenius, find the general solution of the differential equation \(2 x(1-x)\left[\left(d^{2} y\right) /\left(d x^{2}\right)\right]+(1+x)(d y / d x)-y=0\)

Find the solution for: \(x^{2}\left[\left(d^{2} y\right) /\left(d x^{2}\right)\right]+2 x[(d y) /(d x)]+\left(x^{2}-2\right) y=0\)

Using the method of Frobenius, find the particular solution for \(x^{2}\left[\left(d^{2} y\right) /\left(d x^{2}\right)\right]+x[(d y) /(d x)]+\left(x^{2}-v^{2}\right) y=0\)

Find the solution of the differential equation: \(\left(1-x^{2}\right)+2 x(d y / d x)-\lambda y=0\) where \(\lambda\) is a real number
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