Second order differential equations - sine forcing

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8 years, 11 months ago

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Daniel Nucinkis 8 years, 11 months ago

Gave some feedback: Ready to use

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Daniel Nucinkis 9 years ago

Created this as a copy of Second order differential equations 2.

Name	Status	Author	Last Modified
Julie's copy of Second order differential equations 5	draft	Julie Crowley	08/10/2016 20:26
Second order differential equations - sine forcing	Ready to use	Daniel Nucinkis	01/06/2016 09:47
Simon's copy of Julie's copy of Second order differential equations 5	draft	Simon Thomas	20/03/2019 11:58

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Question statement

Find the complete general solution of the equation:
$\displaystyle\simplify{y''+{2*p}*(y')+{p^2-q^2}y={A}sin({f}x)}$
in the form $\displaystyle{y_{CF}(x)+y_{PI}(x)}$ where $y_{CF}$ is the complementary function of the form $\displaystyle{Ae^{ax}+Be^{bx}}$ , $A$ and $B$ are arbitrary constants, and $y_{PI}$ is a particular integral.

Calculate $y_{CF}$ and $y_{PI}$ and input the values of $a$ , $b$ and $y_{PI}$ below.

Note that we require that $a \gt b$ .

Finally, obtain the values of $A$ and $B$ by using the initial conditions $\simplify{y(0)={y0}}$ and $\simplify{ y'(0)={yd0}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

q1	integer	7
p	integer	-1
q	integer	7
A	integer	5
f	integer	3
yd0	integer	-3
y0	integer	1

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$a=\;\;$ Gap 0. $\;\;\;b=\;\;$ Gap 1.. Remember that we require $a \gt b$

Advice

First we find the Complementary Function (CF) i.e. the solution of:

$\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y=0}$

The auxillary equation is $\simplify[std]{lambda^2+{2*p}lambda+{p^2-q^2}=0} \Rightarrow \simplify[std]{(lambda+{p-q})(lambda+{p+q})=0}\Rightarrow \lambda = \var{q-p}\textrm{ or }\lambda = \var{-q-p}$

Hence the CF is $\displaystyle{y_{CF}(x)=\simplify[std]{A*e^({q-p}x)+ B*e^({-q-p}x)}}$ for $A$ , $B$ arbitrary constants.

So $a=\var{q-p}$ and $b=\var{-q-p}$ as we required $a \gt b$ .

To find the Particular Integral (PI) for $\displaystyle{\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y={A}sin({f}x)}}$ , we observe that $y_{PI}(x)=\simplify[std]{C*sin({f}x)+D*cos({f}x)}$ is a possible PI for constants $C,\;D$ to be found.

Substituting $y_{PI}(x)=\simplify[std]{C*sin({f}x)+D*cos({f}x)}$ in to the equation gives:

$\displaystyle \simplify[std]{-{f^2}C*sin({f}x) -{f^2}D*cos({f}x)+{2*p*f}C*cos({f}x)-{2*p*f}D*sin({f}x)+{p^2-q^2}C*sin({f}x)+{p^2-q^2}D*cos({f}x) ={A} sin({f}x)}$ . Now collect terms to obtain

$\displaystyle \simplify[std]{sin({f}x)*({-f^2+p^2-q^2}C-{2*p*f}D)+cos({f}x)*({-f^2+p^2-q^2}D+{2*p*f}C)={A} sin({f}x)}$

so that

$\simplify[std]{{-f^2+p^2-q^2}C-{2*p*f}D={A}}$ and $\simplify[std]{{-f^2+p^2-q^2}D+{2*p*f}C=0}$

We solve the two equations to obtain $\displaystyle\simplify[std]{C={(A*p^2-A*f^2-A*q^2)/((p^2+2*p*q+q^2+f^2)*(p^2-2*p*q+q^2+f^2))}}$ and $\displaystyle\simplify[std]{D={-2*A*f*p/((p^2+2*p*q+q^2+f^2)*(p^2-2*p*q+q^2+f^2))}}$ .

and obtain the particular integral:

$\displaystyle\simplify[std]{{(A*p^2-A*f^2-A*q^2)/((p^2+2*p*q+q^2+f^2)*(p^2-2*p*q+q^2+f^2))}sin({f}x)+{-2*A*f*p/((p^2+2*p*q+q^2+f^2)*(p^2-2*p*q+q^2+f^2))}cos({f}x)}$

Now differentiate $y=y_{CF}+y_{PI}$ and substitute the initial conditions to obtain $A$ and $B$ .

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Second order differential equations - sine forcing

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Second order differential equations - sine forcing

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Taxonomy: mathcentre

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Feedback

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Comment

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Daniel Nucinkis 8 years, 11 months ago

Daniel Nucinkis 9 years ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`