Solution to Review Problems for Midterm II

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1 Soluion o Review Problems for Miderm II MATH Correcion: (i) should be y () ( + )y () + ( + )y() = e (1 + ). Given ha () = e is a soluion of y () ( + )y () + ( + )y() = 0. You should do problems involving he exac equaion. Here is one example. Solve (xy 3 + y + 1) + (3x y + xy 3 + 3y ) dy = 0. dx Soluion: Le M(x, y) = xy 3 + y + 1 and N(x, y) = 3x y + xy 3 + 3y, We have have M y = 6xy + y 3 and N x = 6xy + y 3. So M y = N x and he given equaion is exac. Thus here is a funcion φ(x, y) such ha φ x = M = xy 3 +y +1 and φ y = N = 3x y +xy 3 +3y. Inegraing he firs equaion, we have φ(x, y) = (xy 3 + y + 1)dx = x y 3 + xy + x + h(y). Using φ y = N = 3x y + xy 3 + 3y, we have (x y 3 +xy +x+h(y)) = 3x y + xy 3 + 3y, y 3x y + xy 3 + h (y) = 3x y + xy 3 + 3y, h (y) = 3y and h(y) = (3y )dy = y 3 + c. Hence φ(x, y) = x y 3 + xy + x + h(y) = x y 3 + xy + x + y 3 and he soluion saisfies φ(x, y) = x y 3 + xy + x + y 3 = c. (1) Find he general soluion of he following differenial equaions. (a) y () + 6y () + 9y = 0. The characerisic equaion of y () + 6y () + 9y() = 0 is r + 6r + 9 = (r + 3) = 0. We have repeaed roos r = 3. Thus he general soluion is y() = c 1 e 3 + c e 3. (b) y () + 5y () + y = 0. The characerisic equaion of y () + 5y () + y = 0 is r + 5r + = (r + 1)(r + ) = 0. We have r = 1 or r =. Thus he general soluion is y() = c 1 e + c e. (c) y () + y () + 5y = 0. The characerisic equaion of y () + y () + 5y = 0 is r + r + 5 = 0. We have r = ± i. Noe ha e ( +i) = e e i = e cos() + ie sin(). Thus he general soluion is y() = c 1 e cos() + c e sin(). (d) y () + 7y () + 8y() = 0. Suppose y() = r, we have y () = r r 1 and y () = r(r 1) r. Thus y () + 7y () + 8y() = (r(r 1) + 7r + 8) r = (r + 6r + 8) r. Thus y = r is a soluion of y () + 7y () + 8y() = 0 if r + 6r + 8 = (r + )(r + ) = 0. The roos of r + 6r + 8 = 0 are and. Therefore he general soluion is y() = c 1 + c. (e) y () + 7y () + 10y() = 0. Suppose y() = r, we have y () = r r 1 and y () = r(r 1) r. Thus y () + 7y () + 10y() = (r(r 1) + 7r + 10) r = (r + 6r + 10) r. Thus y = r is a soluion of y () + 7y () + 10y() = 0 if r + 6r + 10 = 0. The roos of r + 6r + 10 = 0 are 3 + i and 3 i. Noe ha = e ln and 3+i = 3 e i ln = 3 cos(ln ) + i 3 sin(ln ). Therefore he general soluion is c 1 3 cos(ln ) + c 3 sin(ln ). page 1 of 8

2 MAT ODE Review: page of 8 (f) y () + 5y () + y() = 0. Suppose y() = r, we have y () = r r 1 and y () = r(r 1) r. Thus y () + 5y () + y() = (r(r 1) + 5r + ) r = (r + r + ) r. Thus y = r is a soluion of y () + 5y () + y() = 0 if r + r + = (r + ) = 0. The roos of r + r + = 0 are. Therefore he general soluion is c 1 + c ln ). (g) y () + y () + 9y = 0. Suppose y() = r, we have y () = r r 1 and y () = r(r 1) r. Thus y () + y () + 9y() = (r(r 1) + r + 9) r = (r + 9) r. Thus y = r is a soluion of y () + αy () + βy() = 0 if r + 9 = 0. The roos of r + 9 = 0 are 3i and 3i. Noe ha = e ln and 3i = e i3 ln = cos(3 ln ) + i sin(3 ln ). Therefore he general soluion is c 1 cos(3 ln ) + c sin(3 ln ). (h) Le p() = y (). Then y () + y () = can be rewrien as p () + p () = Thus ( p()) = Thus p() = C and 5 y() = p()d = C + D. 0 1 (i) Le p() = y (). Then y () + (y ()) 3 = 0 can be wrien as p () + p 3 () = 0. We have p() = ± 1 +C and y() = p()d = ± 1 +C d = ± + C + D. (j) Le p() = y (). Then y () = ( + y ()) 1 can be wrien as p () = ( + p()) 1. Le v = + p(). We have v () = 1 + p () and p () = v () 1. Thus p () = ( + p()) 1 is equivalen o v () 1 = v () 1, ha is v () = v (). We ge v() = 1 and +C y () = p() = v() = 1. Hence +C y() = 1 ( )d = ln + C + D. +C () Find he soluion of he following iniial value problems. (a) y () + y () + 5y = 0, y(0) = 1 and y (0) = 3. From (1C), we have y() = c 1 e cos()+c e sin() and y () = c 1 e cos() c 1 e sin() c e sin()+ c e cos() = ( c 1 + c )e cos() + ( c 1 c )e sin(). Using y(0) = 1 and y (0) = 3, we have c 1 = 1 and c 1 + c = 3. So c 1 = 1 and c = 5. Hence y() = e cos() + 5e sin(). (b) y () + 7y () + 10y() = 0, y(1) = and y (1) = 5. From (1C), we have y() = c 1 3 cos(ln )+c 3 sin(ln ) and y () = 3c 1 cos(ln ) c 1 3 sin(ln ) 3c sin(ln ) + c 3 cos(ln ) = ( 3c 1 + c ) cos(ln ) + ( c 1 3c ) sin(ln ). Using y(1) = and y (1) = 5., we have c 1 = and 3c 1 + c = 5. So c 1 = and c = 1. Hence y() = 3 cos(ln ) + 3 sin(ln ). (3) In he following problems, a differenial and one soluion are given. Use he mehod of reducion of order o find he general soluion soluion. (a) y () (+)y ()+(+)y() = 0; () =. Rewrie he equaion y () (+ )y ()+(+)y() = 0 as y + y + + y = 0. So p() = +. Le y be anoher

3 ODE Review: page 3 of 8 MAT soluion of y () ( + )y () + ( + )y() = 0. We have ( y ) = y y 1 y = W () = Ce R p()d R + = Ce d R = Ce (1+ )d Ce(+ ln()) = = Ce e ln = Ce = Ce. = Ce d = Ce + D and y = (Ce + D) = (Ce + D) = Ce + D. So y So he general soluion is y = Ce + D. (b) ( + 1)y () ( + )y () + y() = 0; () = e. Rewrie he equaion ( + 1)y () ( + )y () + y() = 0 as y + +1 y y = 0. So p() =. Le y be anoher soluion of ( + 1)y () ( + )y () + y() = 0. We have ( y ) = = fracce R R + p()d e = Ce +1 d y y 1 y y1 Ce e ln(+1) e = W () R = Ce ( )d e = Ce(+ln(+1)) = e = Ce (+1) = Ce ( + 1) = C(e + e ). So y e = C(e + e )d = C( e e ) + D and y = (C( e e ) + D) = e (C( e e ) + D) = C( + ) + De. So he general soluion is y = c( + ) + de. () Find he general soluion of he following differenial equaions. (a) y () + 5y () + 6y() = e + sin(). Solving r + 5r + 6 = (r + )(r + 3) = 0, we know ha he soluion of y () + 5y () + 6y() = 0 is y() = c 1 e + c e 3. We ry y p = ce + d sin() + e cos() o be a paricular soluion of y () + 5y () + 6y() = e + sin(). We have y p = ce +d sin()+e cos(), y p = ce +d cos() e sin(), y p = ce d sin() e cos() and y p()+5y p()+6y p () = (ce d sin() e cos())+5(ce +d cos() e sin())+6(ce + d sin()+e cos()) = (c+5c+6c)e +( d 5e+6d) sin()+( e+5d+6e) cos() = 1ce + (5d 5e) sin() + (5d + 5e) cos() = e + sin() if 1c = 1, 5d 5e = 1 and 5d + 5e = 0. So c = 1, d = 1 and e = 1. Thus he general soluion of y () + 5y () + 6y() = e + sin() is y() = 1 1 e + 1 sin() + 1 cos() + c e + c e 3. (b) y () + y = sin() + 3 cos() Solving r + = 0, we know ha he soluion of y ()+y = 0 is y() = c 1 sin()+c cos(). We ry y p = c sin()+d cos()+ e sin() + f cos() o be a paricular soluion of y () + y = sin() + 3 cos(). We have y p = c sin() + d cos() + e sin() + f cos(), y p = c sin ( ) + c cos ( ) + d cos ( ) d sin ( ) + e cos () f sin (), y p = c cos ( ) c sin ( ) d sin ( ) d cos ( ) e sin () f cos () and y p()+y p () = c cos ( ) d sin ( )+3e sin ()+3f cos () = sin()+3 cos() if c = 0, d = 1, e = 0 and f = 1. Thus he general soluion of y () + y = sin() + 3 cos() is y() = 1 cos ( ) + cos () + c 1 sin() + c cos(). (c) y () + y = e We ry y p () = ce. Then y p = ce, y p = 16ce. So y p() + y p = 0ce = e if c = 1. The general soluion is y() = e + c 1 sin() + c cos().

4 MAT ODE Review: page of 8 (d) y () + y + y() = e + e Solving r + r + = (r + ) = 0, we know ha he soluion of y () + y + y() = 0 is y() = c 1 e + c e. We ry y p = c e + de o be a paricular soluion of y () + y + y() = e + e. We have y p = c e + de, y p = ce c e + de, y p = ce ce ce + c e + de = ce 8ce + c e + de and y p()+y p()+y p () = (ce 8ce +c e +de )+(ce c e + de ) + (c e + de ) = ce + 16de. So y p() + y p() + y p () = e + e if c = 1 and 16d = 1, c = 1 and d = 1. Thus he general soluion of 16 y () + y + y() = e + e is y() = 1 e e + c 1 e + c e. (e) y () + 5y () + 6y() = + 1. Solving r + 5r + 6 = (r + )(r + 3) = 0, we know ha he soluion of y () + 5y + 6y() = 0 is y() = c 1 e + c e 3. We ry y p () = a +b+c o be a paricular soluion of y ()+5y ()+6y() = +1. So y p = a+b, y p = a and y p()+5y p()+6y p () = a+5(a+b)+6(a +b+c) = 6a + (10a + 6b) + a + 5b + 6c. So y p() + 5y p() + 6y p () = + 1 if 6a=1, 10a+6b = 0 and a+5b+6c = 1. Thus a = 1, b = 5a = 5 1 a 5b and c = = = =. The general soluion of 108 y () + 5y () + 6y() = + 1 is y() = c e + c e 3. (variaion of parameer) Suppose () and y () are independen soluions of y () + p()y () + q()y() = 0. Then a paricular soluion is given by y p () = () y ()g() d + y W () () y ()g() d W () where W () = W (, y )() = ()y () y ()() is he Wronskian of and y. (f) y () y () 3y() =. Firs, we solve y () y () 3y() = 0. Suppose y() = r, we have y () = r r 1 and y () = r(r 1) r. Thus y () y () 3y() = (r(r 1) r 3) r = (r r 3) r. Thus y = r is a soluion of y () y () 3y() = 0 if r r 3 = (r 3)(r + 1) = 0. The roos of r r 3 = 0 are 1 and 3. Therefore he general soluion is c c 3. Le = 1 and y = 3 o be he soluions of y () y () 3y() = 0. W (, y )() = ()y () y ()() = 1 (3 ) 3 ( 1 ) =. Now g() =. We have y g() d = 3 W (,y d = d = 1 )() c and g() d = 1 W (,y d = )() 1d = + c. So y() = y1 () y ()g() W () 3 ( + c) = 5 + c 3 d 1. (g) y () + y = sec() We will use he variaion of parameer formula. y () = cos(), d + y () y ()g() W () d = 1 ( d) + We have () = sin(),

5 ODE Review: page 5 of 8 MAT W (, y )() = ()y () y ()() = sin() ( sin()) cos() ( cos()) =, y g() d = cos() sec() W (,y d = cos() d = 1 d = + c and )() cos() g() d = sin() sec() W (,y d = sin() ln cos() d = + d. We have used )() cos() subsiuion u = cos() and du = sin()d. Thus y() = sin() ( sin() + cos() ln cos(). + c) + cos()( ln cos() + d) = c sin() + d cos() + (h) y () + y = an() We will use he variaion of parameer formula again. From previous example, we have () = sin(), y () = cos() and W (, y )() =. y g() d = cos() an() W (,y d = cos() sin() d = sin() d = cos() + c. )() cos() g() d = sin() an() W (,y )() sec() )d = sin() Thus y() = sin() ( cos() d = sin()sin() ln sec()+an() + d. cos() + c) cos()( sin() d == 1 cos () cos() d = ( cos() + ln sec()+an() + d) = ln sec()+an() cos() c sin() + d cos(). (i) y () ( + )y () + ( + )y() = e (1 + ). Given ha () = e is a soluion of y () ( + )y () + ( + )y() = 0. This equaion of his problem should be y () (+)y ()+y() = e (1+) Rewrie y () ( + )y () + y() = e (1 + ) as y () (+) y () + y() = e (1 + ). Firs, we find he soluion of y () (+) Le p() = + y () + y() = 0.. Le y be anoher soluion of y () (+)y ()+(+)y() = 0. We have ( y ) = y y 1 y Ce (+ ln()) e = W () = Ce R p()d (e ) = Ce e ln e = Ce e = Ce. So y R + = Ce d e R = Ce (1+ )d = e = Ce d = Ce + D and y = ( Ce + D) = e ( Ce + D) = C + De. So he general soluion is y = C + De. We may choose he second independen soluion o be y =. Now = e, y =, = e + e,y = 1 and W (, y )() = y y = e 1 (e + e ) = e. Recall ha g() = e (1 + ). y g() d = e (1+) W (,y )() d = ( )d = 3 + c. e 3 g() d = e e (1+) W (,y d = ( e e )d = e e e + d. )() e Thus y() = e ( 3 + c) + 3 (e e e + d) = ( )e ce + d. (j) (1 )y () + y () y() = ( 1) e. Given ha () = is a soluion of (1 )y () + y () y() = 0.

6 MAT ODE Review: page 6 of 8 Rewrie (1 )y () + y () y() = ( 1) e as y () + 1 y () 1 y() = ( 1) e = (1 )e. 1 1 Firs, we find he soluion of y () + 1 y () 1 y() = 0. 1 Le p() =. Le be anoher soluion of y () + 1 y () 1 y() = 0. We have ( y ) = y y 1 y R (1+ 1 Ce 1 )d e = Ce(+ln( 1)) = W () = Ce ( 1) 1 )d = C e + D and y = (C e = Ce R p()d R = Ce 1 d = Ce ( 1) = Ce ( 1 1 ). So y 1 R +1 1 = Ce 1 d = = Ce ( 1 + D) = (C e + D) = Ce + D. So he general soluion is y = Ce + D. We may choose he second independen soluion o be y = e. Now =, y = e, = 1,y = e and W (, y )() = y y = e e = e ( 1). Recall ha g() = (1 )e. y g() d = e (1 )e W (,y )() d = ( e )d = e + c. e ( 1) g() d = (1 )e W (,y d = ( e )d = 1 )() e ( 1) e ( + 1) + d. Thus y() = (e + c) + e ( 1 e ( + 1) + d) = e + 1 e c + de. (5) Find he soluion of he following iniial value problems. (a) y () + y = sin() + 3 cos(), y(0) = 3 and y (0) = 5. You may use he resul in b. From (b), we have y() = 1 cos( ) + cos() + c 1 sin() + c cos(). So y () = 1 cos( ) + sin( ) sin() + c 1 cos() c sin(). Using y(0) = 3 and y (0) = 5, we have 1 + c = 3 and 1 + c 1 = 5. Thus c 1 = 11, c = and y() = 1 cos( ) + cos() + 11 sin() + cos(). (b) y () y () + sin(y()) = 0, y(1) = 0 and y (1) = 0. One can check ha y() = 0 is a soluion of y () y () + sin(y()) = 0. By he uniqueness and exisence Theorem, we have y() = 0. (6) Wha is he form of he paricular soluion of he following equaions? (You don have o find he paricular soluion. For example, he form of he paricular soluion of y + y = sin() is y p () = c sin() + d sin().) (a) y () + y = e Solving r + = 0, we have r = ±i. The soluion of y () + y = 0 is y() = c 1 sin() + c cos(). So he paricular soluion of y ()+y = e is y p () = c e +de +fe. (b) y () + y + y() = e + e Solving r + r + = 0, we have r =. The soluion of y () + y + y() = 0 is y() = c 1 e + c e.

7 ODE Review: page 7 of 8 MAT Noe ha e is a soluion of y () + y + y() = 0 and e is no a soluion of y ()+y+y() = 0. So he paricular soluion of y ()+y+y() = e +e is y p () = c e + de. (c) y () + y () + y() = e cos(). Solving r + r + = 0, we have r = 1 ± i. The soluion of y () + y () + y() = 0 is y() = c 1 e cos() + c e cos(). Noe ha e cos() is a soluion of y () + y () + y() = 0 So he paricular soluion of y ()+y ()+y() = e cos() is y p () = c e cos()+d e sin()+ fe cos() + ge sin(). (d) y () + y () + y() = e sin(). Solving r + r + = 0, we have r = 1 ± i. The soluion of y () + y () + y() = 0 is y() = c 1 e cos() + c e cos(). Noe ha e sin() is no a soluion of y ()+y ()+y() = 0 So he paricular soluion of y ()+y ()+y() = e sin() is y p () = ce sin()+de cos()+ fe sin() + ge cos(). (e) y () + y = sin() + 3 cos(). Solving r + = 0, we have r = ±i. The soluion of y () + y = 0 is y() = c 1 sin() + c cos(). Noe ha sin() and cos() are soluions of y ()+y = 0. So he paricular soluion of y ()+y() = sin() + 3 cos() is y p () = c sin() + d cos() + f sin() + g cos(). (7) Express he soluion of he following equaion in he form of y = Ae B cos(c D). (a) y () + y () + y() = 0, y(0) = and y (0) = 3. Solving r + r + = 0, we have r = 1 ± i. So he general soluion of y () + y () + y() = 0 is y() = c 1 e sin() + c e cos(). We have y () = c 1 e sin() + c 1 e cos() c e cos() c e sin() = ( c 1 c )e sin()+(c 1 c )e cos(). Using y(0) = and y (0) = 3, we have c = and c 1 c = 3. This gives c 1 = c + 3 = 5 and c =. Thus y() = 5e sin() + e cos() = e (5 sin() + cos()) = e 9( 5 9 sin() + 9 cos()) = 9e cos( θ) where θ is deermined by cos(θ) = 9 and sin(θ) = 5 9. (b) y () + y () + 5y() = 0, y(0) = and y (0) = 3. Solving r + r + 5 = 0, we have r = ± i. So he general soluion of y () + y () + 5y() = 0 is y() = c 1 e sin() + c e cos(). We have y () = c 1 e sin() + c 1 e cos() c e cos() c e sin() = ( c 1 c )e sin() + (c 1 c )e cos(). Using y(0) = and y (0) = 3, we have c = and c 1 c = 3. This gives c 1 = 7 and c =. Thus y() = 7e sin() + e cos() = e (7 sin() + cos()) = e 53( 7 53 sin() + 53 cos()) = 53e cos( θ) where θ is deermined by cos(θ) = 53 and sin(θ) = (8) Solve he following problems and describe he behavior of he soluions.

8 MAT ODE Review: page 8 of 8 (a) y () + y() = A cos(w) if w. The soluion of y () + y() = 0 is y() = c 1 cos() + c sin() If w, we can ry y p () = c sin(w) + d cos(w). Then y p = cw cos(w) dw sin(w) and y p = cw sin(w) dw cos(w). So y p + y p = cw sin(w) dw cos(w) + (c sin(w) + d cos(w)) = c( w ) sin(w) + d( w ) cos(w) = A cos(w) if d( w ) = A and c( w ) = 0. Using w, we have c = 0 and d = A. w Thus y() = A cos(w) + c w 1 cos() + c sin(). (b) y () + y() = A cos(). The soluion of y ()+y() = 0 is y() = c 1 cos()+c sin() We can ry y p () = c sin() + d cos(). Then y p = c sin() + c cos() + d cos() d sin(), y p = c cos() + c cos() c sin() d sin() d sin() d cos() and y p +y p = c cos()+c cos() c sin() d sin() d sin() d cos()+ c sin() + d cos() = c cos() d sin(). Thus y p + y p = A cos() if c = A and d = 0. Hence y() = A sin() + c 1 cos() + c sin().

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