Solve a constant coefficient second order ODE

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Simon Thomas 6 years, 2 months ago

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Question statement

You are given the differential equation

$\simplify{{a1}*y''+{b1}*y'+{c1}*y=0}.$

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	integer	8
correctform	string	$A\mathrm{e}^{\lambda_1 x}+B\m
incorrectform	list	List of 2 items
forms	list	List of 3 items
disc	number	121
b1	integer	5
c1	integer	-3
lambda1	number	0.375
lambda2	number	-1.000
ltgteq	string	greater than

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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Which of the following choices defines the form of the general solution of the differential equation?

In each case $A$ and $B$ are arbitrary constants, and $\lambda_1$ , $\lambda_2$ , $\lambda$ , $\alpha$ , and $\beta$ are other constants arising from the solution of the auxiliary equation (their actual values are not important for this part of the question).

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{correctform}

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Advice

To determine the form of the general solution of the equation

$ay''+by'+c=0,$

first set $y=\mathrm{e}^{\lambda x}$ , and substitute to obtain

$a\lambda^2+b\lambda+c=0,$

for which the solutions are

$\lambda_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

If $b^2-4ac>0$ , then the roots are real and distinct, and the solution takes the form

$y(x)=\var{forms[0]}.$

If $b^2-4ac=0$ , then the roots are real and repeated, and the solution takes the form

$y(x)=\var{forms[1]}.$

If $b^2-4ac<0$ , then the roots are complex, and the solution takes the form

$y(x)=\var{forms[2]},$

where $\lambda_1=\alpha+i\beta$ and $\lambda_2=\alpha-i\beta$ .

a)

In this question we have $\simplify{{a1}*y''+{b1}*y'+{c1}*y=0}$ , and then

$b^2-4ac=\var{b1^2}-4\times(\var{a1*c1})=\var{disc},$

which is {ltgteq} zero, so the general solution takes the form

$y(x)=\var{correctform}.$

b)

Making the substitution $y=\mathrm{e}^{\lambda x}$ , then gives

$\simplify{{a1}*lambda^2+{b1}*lambda+{c1}=0},$

which has solutions

$\lambda_1=\frac{\var{-b1}+\sqrt{\var{disc}}}{\var{2*a1}}=\var{lambda1} \text{ to 3 d.p.,}$

and

$\lambda_2=\frac{\var{-b1}-\sqrt{\var{disc}}}{\var{2*a1}}=\var{lambda2} \text{ to 3 d.p.}$

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Solve a constant coefficient second order ODE

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Taxonomy: mathcentre

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Simon Thomas 6 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Error

Solve a constant coefficient second order ODE

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 2 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Choose one from a list

Advanced settings

a)

b)

Generated value: `integer`