4.6 #2: Find a particular solution to the second-order differential equation. y + 4y = sec 2t.

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1 4.6 #: Find a paricular soluion o he second-order differenial equaion y + 4y = sec. Soluion: To find he paricular soluion, we use variaion of parameers. Firs, we need o find a fundamenal se of soluions y and y o he associaed homogeneous equaion y + 4y = 0. The characerisic equaion is given by r + 4 = 0. Thus, r = 4 and so r = ± 4 = ±i. Thus, y () = cos and y () = sin form a fundamenal se of soluions o he homogeneous equaion. Now, we form y p = v y + v y = v cos + v sin. From here here are wo ways o proceed, as menioned in he summary in your book on pages The firs way is o use he formulas given in your book on p. 76 in (6.6). The second way is o go hrough he mehod which gave rise o hese equaions. The second way is more inuiive and doesn require you o memorize formulas; he firs way is probably faser. Le us go hrough boh mehods. Mehod : Using he formulas From he formulas given in (6.6) of your book, we see ha y ()g()d y ()g()d v () =, v () = W () W () where g() can be found by comparing he given differenial equaion wih he equaion in he general form y + p()y + q()y = g() and W () is he Wronskian of y and y. For us, we see ha ( ) ( ) y y W () = cos sin y y = = cos (4) + sin (4) = sin cos and g() = sec. Hence, we ge ha sin sec d v () = = sin cos d To inegrae, we le u = cos, du = sin d. Then v () = / du u = 4 ln u = ln(cos ). 4

2 Also, cos sec d v () = = cos cos d = d =. Therefore, he paricular soluion is given by y p () = v cos + v sin = 4 ln(cos ) cos + sin. We have ha y p = v cos + v sin. So Mehod : No using he formulas y p = v cos + v sin v sin + v cos. Now we se he erms wih derivaives of v and v equal o 0. So, se v cos + v sin = 0. This will form one of he main wo equaions for deermining v and v laer. Now, i follows ha Taking he second derivaive, we ge y p = v sin + v cos. y p = v sin + v cos 4v cos 4v sin. We plug y p ino our differenial equaion o ge sec = y p + 4y p = v sin + v cos 4v cos 4v sin + 4v cos + 4v sin = v sin + v cos. Thus, we ge he sysem of equaions v cos + v sin = 0 v sin + v cos = sec. We muliply he firs equaion by sin and he second equaion by cos o ge he equivalen sysem of equaions v cos sin + v sin = 0 v cos sin + v cos =

3 Adding he wo equaions gives v (sin + cos ) =, and so v =, which implies v =. Hence, v = d =. Now, from he equaion v cos + v sin = 0, we see ha v = v ge ha v = / sin. cos sin cos We inegrae by leing u = cos, du = sin d as in he previous mehod o ge Hence, we ge ha v = ln(cos ). 4 y p () = v cos + v sin = 4 ln(cos ) cos + sin.. So since v = /, we 4.6 #4: Verify ha y () = and y () = ln are soluions o he homogeneous equaion y () + 3y () + y() = 0. Use variaion of parameers o find he general soluion o y () + 3y () + y() =. Firs le us verify ha y () and y () are soluions o he homogeneous equaion. Noe y () =, y () = 3. Hence, we see ha y () + 3y () + y () = ( ) ( ) + = 3 + = 0. So y () is a soluion o he homogeneous equaion. Now we show y () = ln is a soluion. Noe y () = ln +, y () = 3 ln 3 3 = 3 ln 3 3.

4 Hence, y () + 3y + y () = ( 3 ln 3 ) ( ln + ) + ln = ln 3 3 ln ln = 0. So y () is also a soluion o he homogeneous equaion. So he homogeneous soluion is given by y h () = C y () + C y () = C + C ln. Now we use variaion of parameers o find he paricular soluion of y () + 3y () + y() =. We will use he formulas given in your book on p. 76. Thus, we need o calculae he Wronskian of y and y, and we need o know g(). Noe he Wronskian is given by ( ) y y W () = y y ( = ln ) ln + = 3 ln ln = 3. Now, noe g() is he righ-hand side when he equaion is in he form Our equaion is y + p()y + q()y = g(). y () + 3y () + y() = so we need o divide boh sides by. Doing so, we obain he equaion y () + 3 y () + y() = 3. Now, we see ha g() =. Thus, using he formulas given in he book, we ge 3 y ()g() d v () = W () = ln d 3 3 = ln d

5 To inegrae, le u = ln, du = d. Then v () = u du = u = (ln ). Nex, noe Therefore, our paricular soluion is given by Hence, he general soluion is given by y ()g() d v () = W () = d 3 = 3 d = ln. y p () = v ()y () + v ()y () = (ln ) + (ln ) = (ln ). y() = y h () + y p () = C + C ln + (ln ). 4.7 #9: Find a paricular soluion o he differenial equaion x + 4x + 5x = 3 sin, x(0) = 0, x (0) = 3 using undeermined coefficiens. Find and plo he soluion of he iniial value problem. Superimpose he plos of he ransien response and he seady-sae soluion. Use differen line syles or colors o differeniae he curves. Soluion: We ry a paricular soluion of he form We calculae he derivaives of x p : x p () = a cos + b sin. x p() = a sin + b cos, x p() = a cos b sin Plugging x p ino he differenial equaion gives 3 sin = x p() + 4x p() + 5x p () = a cos b sin 4a sin + 4b cos + 5a cos + 5b sin = (4a + 4b) cos + (4b 4a) sin.

6 Equaing coefficiens, we ge he sysem of equaions 4a + 4b = 0 4a + 4b = 3 Adding he wo equaions gives 8b = 3, so b = 3 8. From he firs equaion, we see ha a = b, and so a = 3 8. Thus, x p () = 3 8 cos + 3 sin. 8 Now we find he homogeneous soluion. The associaed homogeneous equaion is x + 4x + 5x = 0. The characerisic equaion is r + 4r + 5 = 0. So r = 4 ± 6 4()(5) = 4 ± 4 Hence, he homogeneous soluion is given by x h () = e ( ) C cos + C sin. Thus, he general soluion o our differenial equaion is = ± i. x() = x p () + x h () = 3 8 cos + 3 ) 8 sin + e ( C cos + C sin. Now we need o find C and C using our iniial condiions. Noe To find C, we ake he derivaive of x(): Thus, x(0) = 0 0 = C C = 3 8. x () = 3 8 sin + 3 ) 8 cos e ( C cos + C sin + e ( ) C sin + C cos. So he soluion is given x (0) = 3 3 = 3 8 C + C C = C = = 8. x() = x p () + x h () = 3 8 cos sin + e ( 3 8 cos 8 sin ).

7 The seady-sae soluion is he par of he soluion ha does no decay when. Noe So he seady-sae soluion is lim x() = 3 8 cos sin x seady-sae () = 3 8 cos + 3 sin. 8 The ransien response is he par of he soluion which quickly decays o 0 as. So he ransien response is given by x ransien () = e ( 3 ) cos 8 8 sin. Now we plo he soluion, he seady-sae soluion, and he ransien response.

1. y 5y + 6y = 2e t Solution: Characteristic equation is r 2 5r +6 = 0, therefore r 1 = 2, r 2 = 3, and y 1 (t) = e 2t,

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