Simon's copy of Julie's copy of First order differential equations 1

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Simon Thomas 6 years ago

Created this as a copy of Julie's copy of First order differential equations 1.

Name	Status	Author	Last Modified
Julie's copy of First order differential equations 1	draft	Julie Crowley	08/10/2016 20:24
MA1002 First order differential equations 1	Ready to use	Kieran Mulchrone	16/04/2020 10:38
First order differential equations	draft	Matthew Mears	27/03/2020 15:02
q1_First order differential equations 1 _violeta_separating variables	draft	Violeta CIT	05/12/2018 20:45
Simon's copy of Julie's copy of First order differential equations 1	draft	Simon Thomas	18/03/2019 10:33
Maria's copy of MA1002 First order differential equations 1	draft	Maria Aneiros	27/05/2019 09:04

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Question statement

Separation of variables

Use separation of variables to solve the following ordinary differential equations, satisfying the condition $y=1$ at $x=2$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	integer	2
a3	integer	5
a2	integer	3
a4	integer	5

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..5)

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

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Advice
"Part a)" - prompt
"Part a)" → "Unnamed gap" - Correct answer

Parts

Part a) Gap-fill
1. Gap Mathematical expression
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Part b) Gap-fill
1. Gap Mathematical expression
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Part c) Gap-fill
1. Gap Mathematical expression
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Part d) Gap-fill
1. Gap Mathematical expression
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

$\displaystyle{\frac{dy}{dx}=\frac{y}{\var{a1}x}}$

$y=\;\;$ Unnamed gap

Do not enter decimals in your answer; use only fractions or integers.

Advice

These are all separable first order differential equations.

$\displaystyle{\frac{dy}{dx}=\frac{y}{\var{a1}x} \Rightarrow \int \frac{1}{y}\;dy = \frac{1}{\var{a1}}\int\frac{1}{x}\;dx \Rightarrow \ln(y)=\frac{1}{\var{a1}}\ln(x)+C}$

Exponentiation of both sides then gives $y=Ax^{1/\var{a1}}$ where we have renamed the constant of integration.

To find the particular solution satisfying $y=1$ at $x=2$ , we have $\displaystyle{1=A \times 2^{1/\var{a1}} \Rightarrow A = \frac{1}{2^{1/\var{a1}}}}$

Hence the solution is $\displaystyle{y=\left(\frac{x}{2}\right)^{1/\var{a1}}}$

$\displaystyle{\frac{dy}{dx}=-\var{a2}\frac{y}{x} \Rightarrow \int \frac{1}{y}\;dy = -\var{a2}\int\frac{1}{x}\;dx \Rightarrow \ln(y)=-\var{a2}\ln(x)+C}$

Exponentiation of both sides then gives $y=Ax^{-\var{a2}}$ where we have renamed the constant of integration.

The particular solution satisfying $y=1$ at $x=2$ , gives $A = 2^{\var{a2}}$

Hence the solution is $\displaystyle{y=\left(\frac{2}{x}\right)^{\var{a2}}}$

$\displaystyle{\frac{dy}{dx}=\var{a3}\frac{x}{y} \Rightarrow \int y\;dy = \var{a3}\int x\;dx \Rightarrow \frac{y^2}{2}=\var{a3}\frac{x^2}{2}+C\Rightarrow y^2=\var{a3}x^2+A}$

The particular solution satisfying $y=1$ at $x=2$ , gives $A = \var{1-4*a3}$ .

Hence the solution is $\displaystyle{y^2=\simplify[std]{{a3}x^2+{1-4*a3}}}$ .

$\displaystyle{\frac{dy}{dx}=-\var{a4}\frac{x}{y} \Rightarrow \int y\;dy = -\var{a4}\int x\;dx \Rightarrow y^2=-\var{a4}x^2+A}$

The particular solution satisfying $y=1$ at $x=2$ , gives $A = \var{1+4*a4}$ .

Hence the solution is $\displaystyle{y^2=\simplify[std]{{-a4}x^2+{1+4*a4}}}$ .

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This question is used in the following exam:

Differential Equations (Multiple types) by Simon Thomas in Differential Equations.

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Simon's copy of Julie's copy of First order differential equations 1

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Taxonomy: mathcentre

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Simon Thomas 6 years ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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Simon's copy of Julie's copy of First order differential equations 1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`