q1_First order differential equations 1 _violeta_separating variables

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Violeta CIT 6 years, 4 months ago

Created this as a copy of MA1002 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

There are 23 other versions that do you not have access to.

Question statement

Separate the variables:

Find the solutions of the following ordinary differential equations satisfying the condition $y=1$ at $x=2$ .

You may find it instructive to sketch your various solutions (but this is not required for this CBA).

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	integer	3
a3	integer	3
a2	integer	6
a4	integer	3

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Generated value: `integer`

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Part a) Gap-fill
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Part b) Gap-fill
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$\displaystyle{\frac{dy}{dx}=\var{a3}\frac{x}{y}}$

The solution can be written in the form $y^2=f(x)$ . Enter $f(x)$ in the box below

$y^2=\;\;$ 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 exams:

Week1DifferentialEquations by matt watkins in matt's workspace.
MATH6059 Tutorial 4 by Catherine Palmer in MATH6059.

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q1_First order differential equations 1 _violeta_separating variables

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Violeta CIT 6 years, 4 months ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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q1_First order differential equations 1 _violeta_separating variables

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Violeta CIT 6 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`