Maria's copy of Simon's copy of Julie's copy of First order differential equations 5

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Maria Aneiros 5 years, 10 months ago

Created this as a copy of Simon's copy of Julie's copy of First order differential equations 5.

Name	Status	Author	Last Modified
Julie's copy of First order differential equations 5	draft	Julie Crowley	08/10/2016 20:25
Simon's copy of Julie's copy of First order differential equations 5	draft	Simon Thomas	20/03/2019 09:09
Maria's copy of Simon's copy of Julie's copy of First order differential equations 5	draft	Maria Aneiros	27/05/2019 09:04
Integrating Factor 1	Ready to use	Nick McCullen	12/10/2020 17:27
Integrating Factor 2	Ready to use	Nick McCullen	06/10/2020 11:21
1st order ODE - Initial Conditions and new value	Ready to use	Nick McCullen	09/11/2020 22:02

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Question statement

Integrating Factor Method:

Find the solution of:

$x\frac{dy}{dx}+\var{a}y=\simplify[std]{{b}x^{n}}$

which satisfies $\displaystyle{y(1)=\simplify[std]{{b*(c+1)}/{a+n}}}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	6
c	integer	4
b	integer	2
n	integer	5

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Generated value: `integer`

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"Unnamed part" → "Unnamed gap" - Correct answer

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1. Gap Mathematical expression
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Solution is:

$y=\;\;$ Unnamed gap

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Advice

$\displaystyle{x\frac{dy}{dx}+\var{a}y=\simplify[std]{{b}x^{n}}}$ can be solved by the integrating factor method.

First we divide both sides by $x$ to obtain $\displaystyle{\frac{dy}{dx}+\frac{\var{a}}{x}y=\simplify[std]{{b}x^{n-1}}}$

This is now in the standard form $\displaystyle{\frac{dy}{dx}+a(x)y=b(x)}$ which we can solve by multiplying through by the integrating factor $e^{\int a(x)dx}$

In our example, $a(x) = \frac{\var{a}}{x}$ so $e^{\int a(x)dx}=e^{\int{\frac{\var{a}}{x}dx}}=e^{\var{a}\ln(x)}=x^\var{a}$

Hence multiplying both sides of $\displaystyle{\frac{dy}{dx}+\frac{\var{a}}{x}y=\simplify[std]{{b}x^{n-1}}}$ by $x^\var{a}$ gives

$\displaystyle{x^\var{a}\frac{dy}{dx}+\var{a}x^{\var{a-1}}y=\simplify[std]{{b}x^{n-1+a}}}$

We note that we can rewrite the left hand side as $\frac{d(x^{\var{a}}y)}{dx}$ (product rule) and hence:

$\frac{d(x^{\var{a}}y)}{dx}=\simplify[std]{{b}x^{a+n-1}}$
We can integrate both sides to get:

$x^{\var{a}}y=\simplify[std]{{b}/{a+n}x^{a+n}+A}$
to determine

$A$ we use the condition

$\displaystyle{y(1)=\simplify[std]{{b*(c+1)}/{a+n}}}$ and we see that:

$\simplify[std]{{b*(c+1)}/{a+n}}=\simplify[std]{{b}/{a+n}+A}$

$A = \simplify[std]{{b*(c+1)}/{a+n}} - \simplify[std]{{b}/{a+n} = {b*c}/{a+n}}$
and so the solution is:

$x^{\var{a}}y=\simplify[std]{{b}/{a+n}x^{a+n}+{b*c}/{a+n}} \Rightarrow y=\simplify[std]{{b}/{a+n}*(x^{n}+{c}*x^{-a})}$

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Maria's copy of Simon's copy of Julie's copy of First order differential equations 5

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Taxonomy: mathcentre

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Maria Aneiros 5 years, 10 months ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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Error

Maria's copy of Simon's copy of Julie's copy of First order differential equations 5

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Maria Aneiros 5 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`