y"=D04x2has a solution yı = x- sin æ. Find fundamental pair of solutions using D'Alembert method.

The differential equation y''+1xy'+1-14x2y=0 has a solution y1=x-12sin x. Find the fundamental pair…

Answered: y" + =y + (1 –)y = 0 4x2 has a solution…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

y"
=D0
4x2
has a solution yı = x- sin æ. Find fundamental pair of solutions using D'Alembert method.

Expert Solution

Step 1

The differential equation $y'' + \frac{1}{x} y' + (1 - \frac{1}{4 x^{2}}) y = 0$ has a solution $y_{1} = x^{- \frac{1}{2}} \sin x$ . Find the fundamental pair of solution using D'Alembert method.

Since the differential equation is a second order differential equation, it has two linearly independent solutions. The two functions that generates the solution of the given differential equation is called as fundamental pair of solution. The one of the solution $y_{1} = x^{- \frac{1}{2}} \sin x$ is given. Find the other solution $y_{2}$ .

Since $y_{1}$ is a solution of the differential equation, it satisfies the given equation, hence,

$y_{1}^{''} + \frac{1}{x} y_{1}^{'} + (1 - \frac{1}{4 x^{2}}) y_{1} = 0$ ... (1)

According to D'Alembert method, assume that $y_{2} = v (x) y_{1} (x)$ be the other solution of the given equation. Hence, it satisfies the differential equation.

$y_{2}^{''} + \frac{1}{x} y_{2}^{'} + (1 - \frac{1}{4 x^{2}}) y_{2} = 0$ ... (2)

Find the first derivative of $y_{2} = v (x) y_{1} (x)$ using the product rule $(u v)' = u v' + v u'$ :

$y_{2}^{'} = v (x) y_{1}^{'} + y_{1} v' (x)$

Differentiate again using product rule:

$\begin{array}{rcl} y_{2}^{''} & = & (v (x) y_{1}^{'})' + (y_{1} v' (x))' \\ = & v (x) y_{1}^{''} + y_{1}^{'} v' (x) + y_{1} v'' (x) + v' (x) y_{1}^{'} \\ = & v (x) y_{1}^{''} + 2 v' (x) y_{1}^{'} + y_{1} v'' (x) \end{array}$

Substitute $y_{2} = v (x) y_{1} (x)$ , $y_{2}^{'} = v (x) y_{1}^{'} + y_{1} v' (x)$ and $\begin{array}{rcl} y_{2}^{''} & = & v (x) y_{1}^{''} + 2 v' (x) y_{1}^{'} + y_{1} v'' (x) \end{array}$ in the equation (2):

$\begin{array}{rcl} [v (x) y_{1}^{''} + 2 v' (x) y_{1}^{'} + y_{1} v'' (x)] + \frac{1}{x} [v (x) y_{1}^{'} + y_{1} v' (x)] + (1 - \frac{1}{4 x^{2}}) v (x) y_{1} (x) & = & 0 \\ [v (x) y_{1}^{''} + \frac{1}{x} v (x) y_{1}^{'} + (1 - \frac{1}{4 x^{2}}) v (x) y_{1} (x)] + [2 v' (x) y_{1}^{'} + y_{1} v'' (x)] + \frac{1}{x} y_{1} v' (x) & = & 0 \\ [y_{1}^{''} + \frac{1}{x} y_{1}^{'} + (1 - \frac{1}{4 x^{2}}) y_{1} (x)] v (x) + [2 v' (x) y_{1}^{'} + y_{1} v'' (x)] + \frac{1}{x} y_{1} v' (x) & = & 0 \end{array}$

Using (1), substitute $y_{1}^{''} + \frac{1}{x} y_{1}^{'} + (1 - \frac{1}{4 x^{2}}) y_{1} = 0$ in the equation:

$\begin{array}{rcl} [0] v (x) + [2 v' (x) y_{1}^{'} + y_{1} v'' (x)] + \frac{1}{x} y_{1} v' (x) & = & 0 \\ y_{1} v'' (x) + 2 v' (x) y_{1}^{'} + \frac{1}{x} y_{1} v' (x) & = & 0 \\ y_{1} v'' (x) + (2 y_{1}^{'} + \frac{1}{x} y_{1}) v' (x) & = & 0 \end{array}$