Differential Equation Question Type

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Andrew Iskauskas

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Andrew Iskauskas 6 years, 1 month ago

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Andrew Iskauskas 6 years, 1 month ago

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Question statement

This question concerns finding solutions to a second-order constant-coefficient ODE.

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			Name	Type	Generated Value

root1	integer	5
root2	integer	3
sol	string	Ae^(5t)+Be^(3t)

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Name

Data type

Value

xxxxxxxxxx
 
random(-5..5)

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Description

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Generated value: `integer`

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Parts

Differential Equation

Differential Equation

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Find the general solution to the differential equation for $x(t)$ :

$\simplify{x''(t)+{-root1-root2}x'(t)+{root1*root2}*x(t)}=0$

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Sample Solution

$\class{jme-subvar}{\left\{\operatorname{simplify} \left ( \operatorname{expression} \left ( \texttt{sol} \right ), \textrm{all} \right )\right\}}$

A possible solution, to be shown to the student on 'Reveal Answer'.

Differential Equation Coefficients

A list of strings corresponding to the (possible function-valued) coefficients of the various terms in the differential equation.
The entries $1,0,4$ gives the differential equation $\ddot{y}+4y=0$.

Right-hand Side

$0$

The right-hand side of the differential equation. Will be zero if the differential equation is homogeneous.

Argument

The argument with respect to whom the quantity is being differentiated.

Variable

The function quantity being differentiated.

Advice

The characteristic equation is given by trying the solution $x(t)=e^{\lambda t}$: here this gives

$\simplify{lambda^(2)*e^(lambda*t)+{-root1-root2}*lambda*e^(lambda t)+{root1*root2}*e^(lambda t)}=0.$

Since $e^{\lambda t}>0$ for all $t\in(-\infty,\infty)$, to solve this we must have

$\simplify{lambda^(2)+{-root1-root2}lambda+{root1*root2}}=0$

which means that we have a repeated root, $\lambda=\var{root1}$. To have two linearly independent solutions, we must therefore take $x_{1}(t)=e^{\var{root1}t}$ and $x_{2}(t)=te^{\var{root1}t}$. Then the general solution is given by

$x(t)=\simplify{(A+B*t)*e^({root1}*t)},$

which means that $\lambda=\var{root1}$ or $\lambda=\var{root2}$. This gives two linearly independent solutions: $x_{1}(t)=e^{\var{root1}t}$ and $x_{2}(t)=e^{\var{root2}t}$. So the general solution is

$x(t)=\simplify{A*e^({root1}*t)+B*e^({root2}*t)},$

where $A$ and $B$ are undetermined constants of integration.

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This question is used in the following exam:

Differentiation Extension Examples by Andrew Iskauskas in Andrew's workspace.

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Taxonomy: mathcentre

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Checkpoint description

Andrew Iskauskas 6 years, 1 month ago

Andrew Iskauskas 6 years, 1 month ago

Error in variable testing condition

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Generated value: `integer`

Parts

Differential Equation

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Differential Equation Question Type

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Andrew Iskauskas 6 years, 1 month ago

Andrew Iskauskas 6 years, 1 month ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Differential Equation

Generated value: `integer`