The theory of homogeneous linear equations is not limited to only second order equations. It works forany order. No matter how many roots the characteristic polynomial has, the cases work the same wayas for a 2nd order equation:Any non-repeated root r will correspond to a term e" in the general solutionA repeated root r with multiplicity n will correspond to n terms in the general solution: e",te"..., tetA pair of complex conjugate roots a+bi and a-bi will correspond to the terms e“ cos(bt)and et sin(bt) in the general solution3. Consider the differential equation y' +2y=0.a) The equation above is first order linear homogeneous and therefore separable. Solve theequation 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 thecharacteristic polynomial is linear, you should only have one root.

a) write y' =dydx The equation becomes dydx+2y=0 separate the variables and write it again by…

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

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

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" in the general solution
A repeated root r with multiplicity n will correspond to n terms in the general solution: e",
te"
..., tet
A pair of complex conjugate roots a+bi and a-bi will correspond to the terms e“ cos(bt)
and et 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

write $y' = \frac{d y}{d x}$

The equation becomes

$\frac{d y}{d x} + 2 y = 0$

separate the variables and write it again by rearranging it

$\frac{d y}{- 2 y} = d x$

integrate both sides

$\int \frac{d y}{- 2 y} = \int d x$

$- \frac{1}{2} \ln y = x + C$

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