Simon's copy of Solve a separable first order ODE,

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Simon Thomas 6 years, 1 month ago

Created this as a copy of Solve a separable first order ODE, .

Name	Status	Author	Last Modified
Solve a separable first order ODE,	Ready to use	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Separable variables (1)	draft	Stephen Bowlzer	21/06/2016 16:22
Separable variables 1 violeta	draft	Violeta CIT	23/11/2018 08:51
Aoife's copy of Separable variables 1 violeta	draft	Aoife O'Brien	07/03/2019 11:05
Simon's copy of Solve a separable first order ODE,	draft	Simon Thomas	20/03/2019 09:58
Maria's copy of Separable variables (1)	draft	Maria Aneiros	27/05/2019 09:04
Maria's copy of Solve a separable first order ODE,	draft	Maria Aneiros	27/05/2019 09:04

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Question statement

Find the solution of the differential equation

$(\var{a1}+x)y'=\var{b1}+y,$

satisfying $y(\var{c1})=\var{d1}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	number	8
c1	number	-7
b1	number	7
d1	number	9

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Value

xxxxxxxxxx
 
random(1..9)*sign(random(-1,1))

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

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Generated value: `number`

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Statement
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Variable testing condition
"Unnamed part" → "Unnamed gap" - Correct answer

Parts

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1. Gap Mathematical expression
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$y=$ Unnamed gap (Do not enter decimals in your answer.)

Advice

The differential equation is separable, meaning it can be rearranged to give all the terms in $x$ on one side, and all the terms in $y$ on the other.

So we can write

$\int{\!\frac{1}{\var{b1}+y}\,\mathrm{d}y} = \int{\!\frac{1}{\var{a1}+x}\,\mathrm{d}x},$

then

$\ln\lvert\var{b1}+y\rvert=\ln\lvert\var{a1}+x\rvert+c,$

Exponentiating each side gives:

$e^{\ln\lvert\var{b1}+y\rvert}=e^{\ln\lvert\var{a1}+x\rvert+c},$

$y=\simplify{A({a1}+x)-{b1}},$

which is the general solution of the equation.

Now,

$\var{d1}=y(\var{c1})=\simplify[std]{A({a1}+{c1})-{b1}},$

$A=\simplify[std]{({d1}+{b1})/({a1}+{c1})}=\simplify{{d1+b1}/{a1+c1}},$

and then the full solution is

$y=\simplify[std]{{d1+b1}/{a1+c1}({a1}+x)-{b1}}=\simplify{{(d1*a1-b1*c1)}/{a1+c1}+{(d1+b1)}*x/{a1+c1}}.$

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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 Solve a separable first order ODE,

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Simon Thomas 6 years, 1 month ago

Error in variable testing condition

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Generated value: `number`

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Simon's copy of Solve a separable first order ODE,

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 1 month ago

Error in variable testing condition

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Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`