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3. Consider the differential equation y' +2y= 0. a) The equation above is first order linear homogeneous and therefore separable. Solve the equation by separation of variables. b) Because the equation is linear and homogeneous, it also has a characteristic polynomial. What is the characteristic polynomial? c) Solve the differential equation using its characteristic polynomial. Note that since the characteristic polynomial is linear, you should only have one root.

3. Consider the differential equation y' +2y= 0. a) The equation above is first order linear homogeneous and therefore separable. Solve the equation by separation of variables. b) Because the equation is linear and homogeneous, it also has a characteristic polynomial. What is the characteristic polynomial? c) Solve the differential equation using its characteristic polynomial. Note that since the characteristic polynomial is linear, you should only have one root.

Advanced Engineering Mathematics
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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The theory of homogeneous linear equations is not limited to only second-order equations. It works for any order. No matter how many roots the characteristic polynomial has, the cases work the same way as for a 2nd order equation: - Any non-repeated root \( r \) will correspond to a term \( e^{rt} \) in the general solution. - A repeated root \( r \) with multiplicity \( n \) will correspond to \( n \) terms in the general solution: \( e^{rt} \), \( te^{rt} \), ..., \( t^{n-1}e^{rt} \). - A pair of complex conjugate roots \( a+bi \) and \( a-bi \) will correspond to the terms \( e^{at} \cos(bt) \) and \( e^{at} \sin(bt) \) in the general solution. 3. Consider the differential equation \( y' + 2y = 0 \). a) The equation above is first-order linear homogeneous and therefore separable. Solve the equation by separation of variables. b) Because the equation is linear and homogeneous, it also has a characteristic polynomial. What is the characteristic polynomial? c) Solve the differential equation using its characteristic polynomial. Note that since the characteristic polynomial is linear, you should only have one root.
Transcribed Image Text:The theory of homogeneous linear equations is not limited to only second-order equations. It works for any order. No matter how many roots the characteristic polynomial has, the cases work the same way as for a 2nd order equation: - Any non-repeated root \( r \) will correspond to a term \( e^{rt} \) in the general solution. - A repeated root \( r \) with multiplicity \( n \) will correspond to \( n \) terms in the general solution: \( e^{rt} \), \( te^{rt} \), ..., \( t^{n-1}e^{rt} \). - A pair of complex conjugate roots \( a+bi \) and \( a-bi \) will correspond to the terms \( e^{at} \cos(bt) \) and \( e^{at} \sin(bt) \) in the general solution. 3. Consider the differential equation \( y' + 2y = 0 \). a) The equation above is first-order linear homogeneous and therefore separable. Solve the equation by separation of variables. b) Because the equation is linear and homogeneous, it also has a characteristic polynomial. What is the characteristic polynomial? c) Solve the differential equation using its characteristic polynomial. Note that since the characteristic polynomial is linear, you should only have one root.
Expert Solution
Step 1

a)

write y' =dydx

The equation becomes

dydx+2y=0

separate the variables and write it again by rearranging it 

dy-2y=dx

integrate both sides 

∫dy-2y=∫dx

-12lny =x +C

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