Answered: Solve this linear differential equation…

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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Exercise 3. ANALYSIS

Solve this linear differential equation of the 1st order with the given initial condition:

y '+ (1/x)y = 3

y (1) = 2

Note: We will look for a particular solution in the form yp = ax + b

Expert Solution

Step 1

The given differential equation is $y' + (\frac{1}{x}) y = 3$ and the initial condition is $y (1) = 2$ .

The differential equation $y' + (\frac{1}{x}) y = 3$ is of the form $y' + p (x) y = q (x)$ where $p (x) = \frac{1}{x}$ and $q (x) = 3$ .

The solution of the differential equation of the form $y' + p (x) y = q (x)$ is $y = \frac{1}{e^{\int p d x}} (\int q e^{\int p d x} d x)$ .

Step 2

The integrating factor is,

$\begin{array}{rcl} e^{\int p d x} & = & e^{\int \frac{1}{x} d x} \\ = & e^{\ln x} \\ = & x \end{array}$

Then the solution of the equation $y' + (\frac{1}{x}) y = 3$ is,

$\begin{array}{rcl} y & = & \frac{1}{e^{\int p d x}} (\int q e^{\int p d x} d x) \\ = & \frac{1}{x} (\int 3 x d x) \\ = & \frac{1}{x} (\frac{3 x^{2}}{2} + C) \\ = & \frac{3 x}{2} + \frac{C}{x} \end{array}$

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