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Chapter 8: Problem 2

$$2 x(x-1) y^{\prime \prime}+3(x-1) y^{\prime}-y=0$$

Short Answer

Expert verified
The general solution to the differential equation \(2 x(x-1) y^{\prime \prime}+3(x-1) y^{\prime}-y=0\) is of the form \(y=x^r \summation{n=0}{\infty} a_nx^n\). Exact values of \(a_n\) will depend on initial conditions.

Step by step solution

01

Rewriting the Differential Equation

Begin by writing the given differential equation in standard form: \(2 x(x-1) y^{\prime \prime}+3(x-1) y^{\prime}-y=0\) as \((2x-2) y^{\prime \prime}+(3x-3) y^{\prime}-y=0\)
02

Assume a Power Series Solution

Assume a solution in the power series form, \(y=x^r \summation{n=0}{\infty} a_nx^n\), where \(a_n\) are the coefficients to be determined and r is the point about which the solution is to be developed.
03

Substitute Power Series into the Differential Equation

Differentiate the assumed power series twice and substitute \(y\), \(y^{\prime}\), and \(y^{\prime \prime}\) back into the original differential equation. Multiply out and collect like terms.
04

Set Coefficients to 0

For the sum to be zero, each term of the series must be zero. So set the coefficients of like-power terms to 0 to find values of \(a_n\) for n > 1.
05

Write out the General Solution

Write out the general solution in power series form using the \(a_n\) values found.

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Most popular questions from this chapter

$$3 x y^{\prime \prime}+2(1-x) y^{\prime}-4 y=0$$

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

Use the substitution \(y=x^{r}\) to find \(a\) general solution to the given equation for \(x>0\). \(\frac{d^{2} y}{d x^{2}}=\frac{5}{x} \frac{d y}{d x}-\frac{13}{x^{2}} y\)

\(F(1,1 ; 2 ; x)=-x^{-1} \ln (1-x)\)

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.
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