Differential Equation Review In this course you will be expected to be able to solve two types of first order differential equations without Maple, just as you are expected to be able to do basic calculus without Maple. The first 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 of differential equations will not be difficult. 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 and ends at time t. ot Example 1: Consider the separable differential equation2()xtt=. We first rewrite the derivative as ()dxxtdt=, so we have2dxtdt=. Next we put all of the x’s on one side of the equation, and all of the t ’s on the other side of the equation, so we have . Now we want to integrate both sides of the equation. At time , 2dx t dt=otxhas value ()oxt , while at time t, xhas value ()xt. Hence we have ()2()ooxttxt tdx dλλ=∫∫ Note that we are using a dummy variable of integration (λ) so there is no confusion with the end points of the definite integral. Specifically, if we were to use as the dummy variable of integration and as one of the endpoints of the integral we would most likely make mistakes. t Performing the integration we get 33() ( )33oottxt xt−=− Finally we get the solution 33() ( )33oottxt xt=+− Example 2: Consider the separable differential equation() 2 ()yt tyt=−. We first rewrite the derivative as ()dyytdt=, so we have 2()dyty tdt=−. Next we put all of Fall 2006the ’s on one side of the equation, and all of the t’s on the other side of the equation, so we havey2dytdty=− . Now we want to integrate both sides of the equation. At time , has valueoty()oyt , while at time , has value . Hence we have ty ()yt()()2ooyttyt tdydyλλ=−∫∫ or 22 22()ln[ ( )] ln[ ( )] ln ( )()oooytyt yt t t t tyt⎡⎤−=o−+=−−⎢⎥⎣⎦= Finally we get the solution 22()() ( )ottoyt yt e−−= Example 3: Consider the separable differential equation() 3 ()yt yt=. We first rewrite the derivative as ()dyytdt=, so we have3()dyytdt=. Next we put all of they’s on one side of the equation, and all of the t ’s on the other side of the equation, so we have3dydty=. Now we want to integrate both sides of the equation. At time , has valueoty()oyt , while at time , has value . Hence we have ty ()yt()()3ooyttyt tdydyλ=∫∫ or 2()2()3(oo)ytyttt−=− Finally we get the solution 23() ( ) ( )2ooyt yt t t⎡⎤=+−⎢⎥⎣⎦ Fall 2006Differential Equations with Integrating Factors An integrating factor allows us to write one half of a first 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 and ends at time t. In the first example we go over all of the details, but in the final two examples we just use results from Example 4. ot Example 4: Consider the differential equation() () 2yt yt−=. We first rewrite the derivative as ()dyytdt=, so we have() 2dyytdt−=. Next we want to write the left hand side of the equation as()()atdytedt⎡⎤⎣⎦. Using basic properties from calculus we have () () () ()() () () ()() () ()at at at atd dyt dat dyt datyte e e yt e ytdt dt dt dt dt⎡⎤⎡⎤=+ = +⎣⎦⎢⎥⎣⎦ We want the term in the brackets to look like our original equation, that is, we want () () ()() ()dy t da t dy tyt ytdt dt dt⎡⎤+=−⎢⎥⎣⎦ Equating the two sides we get ()1da tdt=− which gives us ()at t=− At this point we have ()[() ] ()ttddytyte e ytdt dt−−⎡⎤=−⎢⎥⎣⎦ Now since we have (from our original differential equation) () 2dyytdt−= we can multiple both sides of this equation byte− to get ()() [2]ttdy teytedt−−⎡⎤−=⎢⎥⎣⎦ or Fall 2006()() () [2]ttdy t deytyteedt dt−−⎡⎤⎡⎤−= =⎣⎦⎢⎥⎣⎦t− At this point we have the left hand side as an exact derivative () 2ttdyte edt−−⎡⎤=⎣⎦ Now we want to integrate both sides of the equation () 2ootttttdyte dt e ddtλλ−−⎡⎤=⎣⎦∫∫ Integrating we have 0() ( ) 2 2ottttoyte yt e e e−−−−−=−+ Finally we get the solution () (() ( ) 2 2oott ttoyt yt e e−−=−+) Note: We can also solve this equation in the same way we solved the separable equation, by going through the following steps: 2dydty=+ ()()2ooyttyt tdydyλ=+∫∫ [][ ]2()ln 2 ( ) ln 2 ( ) ln2()oooytyt yt t tyt⎡⎤++−+ = =−⎢⎥+⎣⎦ ()2()2()ottoyteyt−+=+ []() () (2 () 2 () () 2oott tt ttooyt yt e yt e e−−+=+ = +)o− () ()() ( ) 2 2oott ttoyt yt e e−−=−+ Example 5: Consider the differential equation () 2 () ()yt tyt xt−=. From Example 4, we need ()2da ttdt=−, or 2()at t=− . We then have 22() ()ttdyte e xtdt−−⎡⎤=⎣⎦. Integrating both sides we have Fall 200622() ( )ootttttdyte dt x e ddtλλλ−−⎡⎤=⎣⎦∫∫ or 222() ( ) ( )ootttotyte yt e x e dλλλ−−−−=∫ Finally we have the solution 2222()() ( ) ( )oottttotyt yt e x e dλλλ−−=+∫ Note that we cannot go any further in the solution until we know ()xt. Example 6: Consider the differential equation3() () ()2tyt tyt ext+=. From Example 4, we need() 32da ttdt=, or.32()at t= We then have3322() ()tttdyte e extdt⎡⎤=⎢⎥⎣⎦. Integrating both sides we have 3322() ( )ootttttdyte dt x ee ddtλλλλ⎡⎤=⎢⎥⎣⎦∫∫ or 333222() ( ) ( )ootttotyte yt e x ee dλλλλ−=∫ Finally we have the solution 33332222()()() ( ) ( )oottttotyt yt e x ee dλλλλ−−−−=+∫ Note that we cannot go any further in the solution until we know()xt. Fall
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