Continuous ECE 300
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This 5 page Class Notes was uploaded by Dr. Reina Hane on Monday October 19, 2015. The Class Notes belongs to ECE 300 at Rose-Hulman Institute of Technology taught by Robert Throne in Fall. Since its upload, it has received 16 views. For similar materials see /class/225083/ece-300-rose-hulman-institute-of-technology in Electrical Engineering & Computer Science at Rose-Hulman Institute of Technology.
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Date Created: 10/19/15
Differential Equation Review In this course you will be expected to be able to solve two types of rst order differential equations without Maple just as you are expected to be able to do basic calculus without Maple The rst type of differential equation to be reviewed is a separable differential equation and the second is a differential equation that can be solved using an integrating factor Once you understand how to deal with these simple types of equations and how they affect system properties generalizing to other types ofdifferential equations will not be dif cult Separable Equations A separable differential equation is generally one where we can rewrite the equation so all of the variables on one side of the equation are the same In what follows we will assume the system starts at time In and ends at time t Example 1 Consider the separable differential equation x0 t2 We rst rewrite the derivative as xt so we have Next we put all of the x s on one side ofthe equation and all of the t s on the other side ofthe equation so we have dx tzdt Now we want to integrate both sides ofthe equation At time In xhas value xt While at time t xhas value xt Hence we have Note that we are using a dummy variable ofintegration 1 so there is no confusion with the end points ofthe de nite integral Speci cally if we were to use I as the dummy variable of integration and as one of the endpoints ofthe integral we would most likely make mistakes Performing the integration we get t3 t3 t t i x x7 3 3 Finally we get the solution I3 t3 I t i x x7 3 3 Example 2 Consider the separable differential equation ya 2tyt We rst rewrite the derivative as ya so we have 21yt Next we put all of Fall 2006 they s on one side of the equation and all ofthe t s on the other side ofthe equation so we have 2tdt Now we want to integrate both sides ofthe y equation At time In y has valueyt0 while at time t y has value yt Hence we have N t j d y j 4sz mo y to or yt 0 1nyt 1nyr 1 1n y r2 r r2 r5 Finally we get the solution ya ya e w f Examgle 3 Consider the separable differential equation yt 3 yt We first rewrite the derivative as yt so we have 31 yt Next we put all of they s on one side of the equation and all ofthe t s on the other side ofthe equation so we haved f3dt Now we want to integrate both sides of the y equation At time In y has valueyt while at time t y has value yt Hence we have N 61 yin7 2 Jim 2 Jim 3t tg 33511 to or Finally we get the solution ya Jym r 2 Fall 2006 Differential Equations with Integrating Factors An integrating factor allows us to write one half of a rst order differential equation as an exact derivative something easy to integrate and the other part as a function with no derivatives In what follows we will assume the system starts at time In and ends at time t In the rst example we go over all of the details but in the nal two examples we just use results from Example 4 Example 4 Consider the differential equation yt yt 2 We rst rewrite the derivative as yt so we haveg ya 2 Next we want to write the left hand side of the equation asyte Using basic properties from calculus we have d a a dyt 61610 a a dyt 61610 t 2 2 m I m t whoa 8 air dt 8 yo 8 air air yo We want the term in the brackets to look like our original equation that is we want dt 7 0 dyU 61610 dyU dt y t Equating the two sides we get dat dr 1 which gives us at t At this point we have d 2 2 dya t t dtye e dt y Now since we have from our original differential equation dy 2 dt y we can multiple both sides ofthis equation by equot to get m 2 e dt m e 2 Of Fall 2006 m dt d it it t t 2 yo dime e 1 At this point we have the left hand side as an exact derivative lytequot l 26quot Now we want to integrate both sides of the equation 2 d t t quot dt Z ldi dime e Integrating we have ytequot yt7 e39tquot Zequot 26quot Finally we get the solution yt We e H 2 26W Note We can also solve this equation in the same way we solved the separable equation by going through the following steps idt 2y N t fizj39dg ylo2y o 2 ln2yt ln2yt01n2 t t0 80720 2 We 2yt 2yt eHo ytueHo 28040 ya yam e Ho 2 2e Examgle 5 Consider the differential equation yt Ztyt xt From Example dat dt Integrating both sides we have 4 we need 2t or at t2 We then have ytequotz equot2xt Fall 2006 jytequot2 dt ixM 12 d1 of 2 ya yt0e z lx1equotzd1 to Finally we have the solution 2 yt yt7 e 2quot 191sz all to Note that we cannot go any further in the solution until we know xt Examgle 6 Consider the differential equation ytyt e xt From 3 3 3 Example 4 we needx oratt2 We then haveyte e e xt Integrating both sides we have 261 3 t 3 t air 1 dl imbue M W or 3 3 t 3 yte yt0 e g have d1 to Finally we have the solution 3 3 t 3 3 yt yt7 e w m Jx eie39 z39md to Note that we cannot go any further in the solution until we knOWxt Fall 2006
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