Julie's copy of Second order differential equations 5

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Julie Crowley

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Julie Crowley 8 years, 5 months ago

Published this.

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Julie Crowley 11 years ago

Created this as a copy of Second order differential equations 5.

Name	Status	Author	Last Modified
Julie's copy of Second order differential equations 5	draft	Julie Crowley	08/10/2016 20:26
Second order differential equations - sine forcing	Ready to use	Daniel Nucinkis	01/06/2016 09:47
Simon's copy of Julie's copy of Second order differential equations 5	draft	Simon Thomas	20/03/2019 11:58

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Question statement

Find the complete general solution of the equation:

$\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y=x}$
in the form

$\displaystyle{Ae^{ax}+Be^{bx}+y_{PI}(x)}$ where

$A$ and

$B$ are arbitrary constants and

$y_{PI}$ is a particular integral.

Note that we require that $a \gt b$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

q1	integer	1
p	integer	4
q	integer	1

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Value

xxxxxxxxxx
 
random(1..8)

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Description

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Generated value: `integer`

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Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part b) Gap-fill
1. Gap Mathematical expression
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Gap-fill

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$a=\;\;$ Gap 0. $\;\;\;b=\;\;$ Gap 1.. Remember that we require $a \gt b$

Advice

First we find the Complementary Function (CF) i.e. the solution of:

$\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y=0}$

The auxillary equation is $\simplify[std]{lambda^2+{2*p}lambda+{p^2-q^2}=0} \Rightarrow \simplify[std]{(lambda+{p-q})(lambda+{p+q})=0}\Rightarrow \lambda = \var{q-p}\textrm{ or }\lambda = \var{-q-p}$

Hence the CF is $\displaystyle{y_{CF}(x)=\simplify[std]{A*e^({q-p}x)+ B*e^({-q-p}x)}}$ for $A$ , $B$ arbitrary constants.

So $a=\var{q-p}$ and $b=\var{-q-p}$ as we required $a \gt b$ .

To find the Particular Integral (PI) for $\displaystyle{\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y=x}}$ , we observe that

$y_{PI}(x)=cx+d$ is a possible PI for constants $c,\;d$ to be found.

Substituting $y_{PI}(x)=cx+d$ in to the equation gives:

$\displaystyle \simplify[std]{{2*p}c+{p^2-q^2}*(cx+d)} =x \Rightarrow \simplify[std]{{p^2-q^2}cx+{p^2-q^2}d+{2*p}c}= x \Rightarrow c=\simplify[std]{1/{p^2-q^2}},\;\;d = \simplify[std]{{-2p}/{(p^2-q^2)^2}}$

So the particular integral is $\displaystyle{\simplify[std]{x/{p^2-q^2}-{2*p}/{(p^2-q^2)^2}}}$

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Julie's copy of Second order differential equations 5

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Taxonomy: mathcentre

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Taxonomy: Context

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Julie Crowley 8 years, 5 months ago

Julie Crowley 11 years ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

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Julie's copy of Second order differential equations 5

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Julie Crowley 8 years, 5 months ago

Julie Crowley 11 years ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`