Need help with parts a, b, and c. Thank you. 1.Solve the following (first order & linear!) differential equations:a) dy+2y(x) = 2e-*b) du -2 y(x) + (1+x²)²c) dy+ay(x) = b sin(wx)d) x²y' - 4xy = x³ sin(x), x>0(a, b, w are constants.)e) y' - 2y = f(x), y(0) = 1, where f(x)f) sin(x)y' + y cos(x) = sin(2x), y(π/2) = 3=1- x,0,if x < 1if x ≥ 1

Answered: 1. da da Solve the following (first…

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

Need help with parts a, b, and c. Thank you.

1.
Solve the following (first order & linear!) differential equations:
a) dy+2y(x) = 2e-*
b) du -2 y(x) + (1+x²)²
c) dy+ay(x) = b sin(wx)
d) x²y' - 4xy = x³ sin(x), x>0
(a, b, w are constants.)
e) y' - 2y = f(x), y(0) = 1, where f(x)
f) sin(x)y' + y cos(x) = sin(2x), y(π/2) = 3
=
1- x,
0,
if x < 1
if x ≥ 1

Expert Solution

Step 1: We give standard form of general solution of first order linear differential equation.

(.) Given,

(a). $fraction numerator d y over denominator d x end fraction plus 2 y left parenthesis x right parenthesis equals 2 e to the power of negative x end exponent$

(b). $fraction numerator d y over denominator d x end fraction space equals space minus fraction numerator 2 x over denominator open parentheses 1 plus x squared close parentheses end fraction y left parenthesis x right parenthesis plus 2 over open parentheses 1 plus x squared close parentheses squared$

(c). $fraction numerator d y over denominator d x end fraction plus a y left parenthesis x right parenthesis equals b sin open parentheses omega x close parentheses space space semicolon space space space left parenthesis a comma b comma omega space a r e space c o n s tan t s right parenthesis space$

(.) Solution of first order linear differential equation of the form $fraction numerator d y over denominator d x end fraction plus P left parenthesis x right parenthesis y left parenthesis x right parenthesis plus equals Q left parenthesis x right parenthesis$ is given by,

$y left parenthesis x right parenthesis open parentheses I. F. close parentheses space equals space space integral space Q left parenthesis x right parenthesis open parentheses I. F. close parentheses d x space plus space c$

where $c$ is an arbitrary constant and $I. F.$ is an integrating factor given by,

$I. F. space equals space e to the power of integral P left parenthesis x right parenthesis d x end exponent$