MA1002 First order differential equations 1

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Timur Zaripov

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Timur Zaripov 5 years, 3 months ago

Gave some feedback: Ready to use

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Kieran Mulchrone 7 years, 11 months 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

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	2
a3	integer	4
a2	integer	4
a4	integer	3

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

$y=\;\;$ Unnamed gap

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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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Timur Zaripov 5 years, 3 months ago

Kieran Mulchrone 7 years, 11 months ago

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MA1002 First order differential equations 1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Timur Zaripov 5 years, 3 months ago

Kieran Mulchrone 7 years, 11 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`