Calculus Stewart ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
1. The auxiliary equation is r +3r+2=(r+2)(r+1)=0 , so the complementary solution is y (x)=c e
c 1 2 2x
2
+c e . We try the particular solution y (x)=Ax +Bx+C , so y =2Ax+B and y
2 p p 2 2 2
x
2
/
/ / p
=2A .
Substituting into the differential equation, we have (2A)+3(2Ax+B)+2(Ax +Bx+C)=x or 2Ax +(6A+2B)x+(2A+3B+2C)=x . Comparing coefficients gives 2A=1 , 6A+2B=0 , and 1 3 7 2A+3B+2C=0 , so A= , B= , and C= . Thus the general solution is 2 2 4 7 2x x 1 2 3 y(x)=y (x)+y (x)=c e +c e + x x+ . c p 1 2 2 2 4 2. The auxiliary equation is r +9=0 with roots r= 3i , so the complementary solution is y (x)=c cos (3x)+c sin (3x) . Try the particular solution y (x)=Ae
c 1 2 p 3x 3x 3x 3x 2
, so y =3Ae
p 3x
/
3x 3x
and y
/ / p
=9Ae
3x
.
Substitution into the differential equation gives 9Ae +9(Ae )=e the general solution is y(x)=y (x)+y (x)=c cos (3x)+c sin (3x)+
c p 1 2 2
or 18Ae =e
. Thus A=
1 and 18
1 3x e . 18
2x
3. The auxiliary equation is r 2r=r(r 2)=0 , so the complementary solution is y (x)=c +c e . Try
c 1 2
the particular solution y (x)=Acos 4x+Bsin 4x , so y = 4Asin 4x+4Bcos 4x and
p p
/
y
/ / p
= 16Acos 4x 16Bsin 4x . Substitution into the differential equation gives
( 16Acos 4x 16Bsin 4x) 2( 4Asin 4x+4Bcos 4x)=sin 4x ( 16A 8B)cos 4x+(8A 16B)sin 4x=sin 4x . 1 1 and B= . Thus the general solution is Then 16A 8B=0 and 8A 16B=1 A= 40 20 1 2x 1 y(x)=y (x)+y (x)=c +c e + cos 4x sin 4x . c p 1 2 40 20 4. The auxiliary equation is r +6r+9=(r+3) =0 , so the complementary solution is y (x)=c e
c 1 3x 2 2
+c xe
2
3x
. Try the particular solution y (x)=Ax+B , so y =A and y
p p
/
/ / p
=0 . Substitution
into the differential equation gives 0+6A+9(Ax+B)=1+x or (9A)x+(6A+9B)=1+x . Comparing 1 1 coefficients, we have 9A=1 and 6A+9B=1 , so A= and B= . Thus the general solution is 9 27 1 3x 3x 1 y(x)=c e +c xe + x+ . 1 2 9 27 5. The auxiliary equation is r 4r+5=0 with roots r=2 i , so the complementary solution is
1
2
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
y (x)=e (c cos x+c sin x) . Try y ( x ) =Ae , so y = Ae and y =Ae . Substitution gives c 1 2 p p p Ae
x
2x
x
/
x
/ /
x
4( Ae )+5(Ae )=e
2x 1 2
x
x
x
10Ae =e
x
x
x
A=
1 . Thus the general solution is 10
y(x)=e (c cos x+c sin x)+
x
1 e 10
.
2 x
6. y (x)=e (c x+c ) . Try y (x)=x ( Ax+B ) e so that no term in y is a solution of the complementary c 1 2 p p equation. Then y =[ Ax +(3A B)x +2Bx]e , y
p 3 2 / 3 2 x / / p
=[Ax +(B 6A)x +(6A 4B)x+2B]e and
3 2 3 2
3
2
x
substitution gives [Ax +(B 6A)x +(6A 4B)x+2B]+2[ Ax +(3A B)x +2Bx]+(Ax +Bx )=x 1 1 3 x 2 x x 6Ax+2B=x . So y ( x ) =x x e and the general solution is y(x)=e c x+c + x e . p 1 2 6 6
(
)
7. The auxiliary equation is r +1=0 with roots r= i , so the complementary solution is y (x)=c cos x+c sin x . For y
c 1 2 / /
2
+y=e try y (x)=Ae . Then y =y
p
1
x
x
/
/ / p
1
p
=Ae and substitution gives
2
x
1
Ae +Ae =e
/
x
x
x
A=
2
1 1 x , so y (x)= e . For y 2 2 p 1
/ / p
2
/ /
+y=x try y (x)=Ax +Bx +Cx+D .
p
2
3
3
Then y =3Ax +2Bx+C and y
p
2
=6Ax+2B . Substituting, we have C= 6 , and 2B+D=0 D=0 . Thus
6Ax+2B+Ax +Bx +Cx+D=x , so A=1 , B=0 , 6A+C=0 y (x)=x 6x and the general solution is
p
2
3
2
3
3
y(x)=y (x)+y (x)+y (x)=c cos x+c sin x+
c p
1
p
2
1
2
1 x 3 1 e +x 6x . But 2=y(0)=c + 1 2 2
c=
1
3 and 2
0=y (0)=c +
1 11 . Thus the solution to the initial value problem is 6 c= 2 2 2 2 3 11 1 x 3 y(x)= cos x+ sin x+ e +x 6x . 2 2 2
/ 2
8. The auxiliary equation is r 4=0 with roots r= 2 , so the complementary solution is y (x)=c e +c e
c 2x 2x / / 1 x 2
. Try y (x)=e (Acos x+Bsin x) , so y =e (Acos x+Bsin x+Bcos x Asin x) and
p p
x
/
x
y
e (2Bcos x 2Asin x) 4e (Acos x+Bsin x)=e cos x
p x
=e (2Bcos x 2Asin x) . Substitution gives
x x
(2B 4A)e cos x+( 2A 4B)e sin x=e cos x
2
x
x
x
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
1 1 1 1 2x 2x x , B= . Thus the general solution is y(x)=c e +c e +e cos x+ sin x . But 1 2 5 10 5 10 1 1 9 3 / 1=y(0)=c +c and 2=y (0)=2c 2c . Then c = , c = , and the solution to the initial 1 2 5 1 2 10 1 8 2 40 9 2x 3 2x x 1 1 value problem is y(x)= e + e +e cos x+ sin x . 8 40 5 10 A= 9. The auxiliary equation is r r=0 with roots r=0 , r=1 so the complementary solution is y (x)=c +c e . Try y (x)=x(Ax+B)e so that no term in y is a solution of the complementary
c 1 2 p p x x / 2 x / / p 2 x 2
equation. Then y =(Ax +(2A+B)x+B)e and y
p 2
=(Ax +(4A+B)x+(2A+2B))e . Substitution into the
x 2 x x
differential equation gives (Ax +(4A+B)x+(2A+2B))e (Ax +(2A+B)x+B)e =xe 1 1 2 x x x A= , B= 1 . Thus y (x)= x x e and the general solution is (2Ax+(2A+B))e =xe p 2 2 1 2 x x / y(x)=c +c e + x x e . But 2=y(0)=c +c and 1=y (0)=c 1 , so c =2 and c =0 . The solution 2 1 2 1 2 2 1 2 1 2 1 2 x x x to the initial value problem is y(x)=2e + x x e =e x x+2 . 2 2 10. y (x)=c e +c e
c 1 2 x 2x
. For y
/ /
+y
/
2y=x try y (x)=Ax+B . Then y =A , y
p
1
/
/ / p /
1
p
=0 , and substitution
1
gives 0+A 2(Ax+B)=x
1 1 1 1 A= , B= , so y (x)= x . For y 2 4 2 4 p 1
/ / / p
2
/ /
+y
2y=sin 2x try
y (x)=Acos 2x+Bsin 2x . Then y = 2Asin 2x+2Bcos 2x, y
p
2
p
= 4Acos 2x 4Bsin 2x , and substitution A= 1 3 , B= . 20 20
2
gives ( 4Acos 2x 4Bsin 2x)+( 2Asin 2x+2Bcos 2x) 2(Acos 2x+Bsin 2x)=sin 2x Thus y ( x ) =
1 3 cos 2x+ sin 2x and the general solution is 20 20 p 2 1 1 3 1 1 x 2x 1 y(x)=c e +c e x cos 2x sin 2x . But 1=y(0)=c +c and 1 2 1 2 4 20 2 4 20 20 1 3 17 1 / 0=y (0)=c 2c c= and c = . Thus the solution to the initial value problem is 1 2 2 10 1 15 2 6 17 x 1 2x 1 1 1 3 y(x)= e+ e x cos 2x sin 2x . 15 6 2 4 20 20 11. y (x)=c e
c 1 x/4
+c e
2
x
. Try y (x)=Ae . Then 10Ae =e , so A=
p
x
x
x
1 and the general solution is 10
3
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
1 x e . The solutions are all composed of exponential curves and with the 1 2 10 exception of the particular solution (which approaches 0 as x ), they all approach either 1 x as x . As x , all solutions are asymptotic to y = e . p 10 y(x)=c e
x/4
+c e +
x
or
12. The auxiliary equation is (2r+1)(r+1)=0 , so r= 1 , 2y
/ / / p p
1 x and y (x)=c e +c e c 1 2 2
/ / /
x/2
. For
+3y +y=1 , try y (x)=A ; substituting gives y (x)=1 . For 2y
1 1
+3y +y=cos 2x try
y =Acos 2x+Bsin 2x
p
2
y = 2Asin 2x+2Bcos 2x , y
p
2
/
/ / p
2
= 4Acos 2x 4Bsin 2x .
Substituting into the differential equation gives cos 2x= ( 6B 7A) cos 2x+ ( 7B 6A) sin 2x . 7 6 , B= . Thus, Then solving the equations 6B 7A=1 and 7B 6A=0 gives A= 85 85 7 6 7 6 x x/2 y ( x) = cos 2x+ sin and 2x the general solution is y ( x ) =c e +c e +1 cos 2x+ sin 2x 1 2 85 85 85 85 p 2 . The graph shows y =y +y and several other solutions. Notice that all solutions are asymptotic to y
p p
1
p
2
p
as x
.
13. Here y (x)=c cos 3x+c sin 3x . For y
c 1 2 2 2
/ /
+9y=e
2x
try y (x)=Ae
p
1
2x
and for y
/ /
+9y=x sin x try
2
y (x)=(Bx +Cx+D)cos x+(Ex +Fx+G)sin x . Thus a trial solution is
p p
2
y (x)=y (x)+y (x)=Ae +(Bx +Cx+D)cos x+(Ex +Fx+G)sin x .
p
1
2x
2
2
p
2
4
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
14. Since y (x)=c +c e
c 1 2
9x
, try y (x)=(Ax+B)e cos x+(Cx+D)e sin x .
p
x
x
15. Here y (x)=c +c e
c 1 2
9x
. For y
/ /
+9y =1 try y (x)=Ax (since y=A is a solution to the
p
1
/
complementary equation) and for y
x 4x
/ /
+9y =xe
3
/
9x
try y (x)=(Bx+C)e
p
2
9x
.
16. Since y (x)=c e +c e
c 1 2
try y (x)=x(Ax +Bx +Cx+D)e so that no term of y (x) satisfies the
p p
2
x
complementary equation. 17. Since y (x)=e (c cos 3x+c sin 3x) we try y (x)=x(Ax +Bx+C)e cos 3x+x(Dx +Ex+F)e sin 3x
c 1 2 p x 2 x 2 x
(so that no term of y is a solution of the complementary equation).
p
18. Here y (x)=c cos 2x+c sin 2x . For y
c 1 2 p
/ /
+4y=e
3x
try y (x)=Ae
p
1
3x
and for y
/ /
+4y=xsin 2x try
y (x)=x(Bx+C)cos 2x+x(Dx+E)sin 2x (so that no term of y is a solution of the complementary
2
p
2
equation). 19. (a) The complementary solution is y (x)=c cos 2x+c sin 2x . A particular solution is of the form
c 1 2
y (x)=Ax+B . Thus, 4Ax+4B=x
p
A=
1 and B=0 4
y (x)=
p
1 x . Thus, the general solution is 4
1 x. 4 (b) In (a), y (x)=c cos 2x+c sin 2x , so set y =cos 2x , y =sin 2x . Then y=y +y =c cos 2x+c sin 2x+
c p 1 2 c 1 2 2 2 1 2
y y
/
1 2
y y =2cos 2x+2sin 2x=2 so u =
2 1 1
/
/
1 xsin 2x 2
1 1 1 / 1 xsin 2xdx= xcos 2x+ sin 2x and u = xcos 2x 1 2 2 2 4 2 1 1 1 u (x)= xcos 2xdx= xsin 2x+ cos 2x . Hence 2 2 4 2 1 1 1 1 1 y (x)= xcos 2x+ sin 2x cos 2x+ xsin 2x+ cos 2x sin 2x= x . Thus p 4 2 4 2 4 1 y(x)=y (x)+y (x)=c cos 2x+c sin 2x+ x . c p 1 2 4 u (x)= 20. (a) Here r 3r+2=0
2
r=1 or 2 and
5
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
y (x)=c e +c e . We try a particular solution of the form y (x)=Acos x+Bsin x
c
2x
x
and y
1 / /
2
p
y = Asin x+Bcos x
p
/
p
= Acos x Bsin x . Then the equation y
/ /
3y +2y=sin x becomes A= 3 1 and B= . Thus, 10 10
/
(A 3B)cos x+(B+3A)sin x=sin x y (x)=
A 3B=0 and B+3A=1
3 1 cos x+ sin x . Therefore, the general solution is p 10 10 1 2x x 3 y(x)=y (x)+y (x)=c e +c e + cos x+ sin x . c p 1 2 10 10 (b) From (a) we know that y (x)=c e +c e . Setting y =e
c 1 2 1 2x x 2x
, y =e , we have
2 /
x
y y
/
1 2
y y =e
2 1 2x
/
3x
2e = e 1 e 5
3x
3x
. Thus u =
1
/
sin xe e
3x
x
=sin xe
2x
and u =
2
sin xe
2x
u (x)= e
1
sin xdx=
2x
( 2sin x cos x ) and u2(x)=
e 1 x x e sin xdx= e ( sin x cos x) . Thus 2
3x
= sin xe
x
. Then
1 1 1 3 ( 2sin x cos x)+ (sin x+cos x)= sin x+ cos x and the general solution is p 5 2 10 10 3 2x x 1 y(x)=y (x)+y (x)=c e +c e + sin x+ cos x . c p 1 2 10 10 y (x)= 21. (a) r r=r(r 1)=0
2
r=0 , 1 , so the complementary solution is y (x)=c e +c xe . A particular
c 2x p
x
x
solution is of the form y (x)=Ae
c
. Thus 4Ae
x 1
2x
4Ae +Ae =e
x 2x x
2x
2x
2x
Ae =e
2x
1 2x
2
A=1
y (x)=e
p
2x
. So
a general solution is y(x)=y (x)+y (x)=c e +c xe +e
p 2 x x x / c 1 2 1 2 x
.
/ x 1 2
(b) From (a), y (x)=c e +c xe , so set y =e , y =xe . Then, y y u = xe
1 / x
y y =e (1+x) xe =e
2 1
/
2x
2x
2x
and so
2x 2x
u ( x) = 1
xe dx= (x 1)e and u =e
2 c p
x
x
u (x)= e dx=e . Hence y ( x ) =(1 x)e +xe =e 2 p
x 2x 2
x
2x
and the general solution is y(x)=y (x)+y (x)=c e +c xe +e
1
x
.
22. (a) Here r 2r+1=(r+1) =0
x p
2
2
r= 1 and y (x)=c +c e and so we try a particular solution of the
c 1 2 / /
x
form y (x)=Axe . Thus, after calculating the necessary derivatives, we get y Ae (2+x) Ae (1+x)=e
x x x x p x
y =e
1
/
x x x
A=1 . Thus y (x)=xe and the general solution is y(x)=c +c e +xe .
x / c x x 1 2 1 2 1 2
(b) From (a) we know that y (x)=c +c e , so setting y =1 , y =e , then y y u = e /e = e and u =e /e =1 . Then u (x)=
1 2 1 / 2x x x / x x 2
y y =e 0=e . Thus
2 1 x x p
/
2 x
x
e dx= e and u (x)=x . Thus y (x)= e +xe and the
6
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
general solution is y(x)=c +c e e +xe =c +c e +xe .
1 2 1 3
x
x
x
x
x
23. As in Example 6, y (x)=c sin x+c cos x , so set y =sin x , y =cos x . Then
c 1
y y
2
/
1 2
sec xcos x =1 y y = sin x cos x= 1 , so u = 2 1 1 1
/ 2 2
2 /
1
2
u (x)=x and u =
1 2
/
sec xsin x = tan x 1
u (x)=
. Hence y (x)=xsin x+cos xln (cos x) and the p 2 general solution is y(x)=(c +x)sin x+ c +ln (cos x) cos x .
1 2
tan xdx=ln cos x =ln (cos x) on 0<x<
24. Setting y =sin x , y =cos x , then y y
1 2 /
/
1 2
cot xcos x cos x y y = sin x cos x= 1 . Thus u = = 2 1 1 1 sin x
/ 2 2 / 2
2
cot xsin x cos x = cos x . Then u (x)= dx= (csc x sin x)dx=ln (csc x cot x)+cos x and and u = 1 2 1 sin x u (x)= sin x . Thus y (x)=[cos x+ln (csc x cot x)]sin x+( sin x)(cos x) and the general solution is
2 p
y(x)=c sin x+c cos x+sin xln (csc x cot x) .
1 2
25. y =e , y =e
1 2
x
2x
and y y
/
1 2
y y =e
2 1 x /
/
3x
. So u =
1
/
e
2x x 3x
=
e
x x
and
(1+e )e =
x
1+e so
u (x)=
1
e
x x
dx=ln (1+e ) . u =
2
e
x x 3x
e
3x
x 2x
1+e e
3x x
(1+e )e
x
e +e
x
u (x)=
2
e +e
2x
dx=ln
e +1 e
x
x
e =ln (1+e ) e
x x
. Hence y (x)=e ln (1+e )+e [ln (1+e ) e ]
p x x 2x
x
x
2x
x
x
and the general solution is y(x)=[c +ln (1+e )]e +[c e +ln (1+e )]e
1 2 x 2x / / 3x /
.
x
26. y =e
1 /
, y =e
2 x x
and y y
x
1 2
y y = e
2 1
. So u =
1 x x
(sin e )e e
3x
x
2x
=e sin e and
x
x
u =
2
(sin e )e e
3x 2x
= e sin e . Hence u ( x ) = e sin e dx= cos e and 1
x x x x x x 2x p
2x
u (x)=
2
e sin e dx=e cos e sin e . Then y (x)= e cos e e
x x x x x 2x 1 2 /
[sin e e cos e ] and the general
x
x
x
solution is y(x)=(c cos e )e +[c sin e +e cos e ]e 27. y =e
1 x
.
, y =e and y y
2
x
/
1 2
y y =2 . So
2 1
7
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations
e / e u = ,u = and 1 2 2x 2x
/
x
x
y (x)= e
p
x
e x e dx+e dx . Hence the general solution is 2x 2x e dx 2x , y =xe
2 2x x
x
x
y(x)=
c
1
e +
x
e c+ dx 2 2x
/
x
e . e
2x 3
x
28. y =e
1 /
2x
and y y
1 2
y y =e
2 1
/
4x
. Then u =
1
/
xe
4x
2x
=
2x
1 x
2
so u (x)=x
1
1
and
xe xe
2x 2
u =
2
e
2x 3
e
2x 4x
=
1 x
3
so u (x)=
2
1 2x
2
. Thus y (x)=
p
e
2x
xe
2x
x
=
2x
e and the general solution is 2x
y(x)=e
[c +c x+1/(2x)] .
1 2
8