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Solutions homework 2
1 Exercise 1
The differential equations given in this exercise are homogeneous, second-order differential equations with constant coefficients. Hence they can besolve with the help of the corresponding auxiliary equation. The solutionsare:
1. For 2y 13y 7y = 0, the auxiliary equation is 2r2 13r 7 = 0which has two real solutions r = 1
2and r = 7. Therefore the solutions
are y(x) = C1ex
2 + C2e7x
.
2. For z 6z + 10z = 0, the auxiliary equation is r2 6r + 10 = 0which has two complex (conjugate) solutions r = 3 + i and r = 3 i.Therefore the solutions are z(x) = C1e
3x cos(x) + C2e3x sin(x).
3. For u 6u + 9u = 0, the auxiliary equation is r2 6r + 9 = 0which has one double solution r = 3. Therefore the solutions areu(x) = C1e
3x + C2xe3x.
4. For 2u + 12u + 18u, the auxiliary equation is 2r2 + 12r + 18 = 0
which has one double solution r = 3. Therefore the solutions areu(x) = C1e3x + C2xe3x.5. For 3x 6x + 6x = 0, the auxiliary equation is 3r2 6r + 6 = 0
which has two complex solutions r = 1 + i and r = 1 i. Thereforethe solutions are x(t) = C1e
t cos(t) + C2et sin(t).
2 Exercise 2
In this exercise, the differential equation are second-order linear differentialequations with constant coefficients with a forcing term (the r.h.s. is nolonger zero).
1. For y + 2y + 2y = 8t2et , y(0) = 0 , y(0) = 1, the auxiliary equationis r2 + 2r + 2 = 0 that has two complex roots: r = 1 i. Thereforethe solutions of the homogeneous equations are y0(x) = C1e
x cos(x) +C2e
x sin(x). We need to find a particular solution to the equation. The
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form of the r.h.s. suggests that we look for y(x) = (Ax2 + Bx + C)ex.
Inserting this into the equation leads to A = 8 , B = 0 , C = 16.Therefore the general solution is:
y(x) = C1ex cos(x) + C2e
x sin(x) + (8x2 16)ex
The initial conditions give C1 = 16 and C2 = 1.
2. For y4y+4y = t sin(t), y(0) = 1 , y(0) = 1, the auxiliary equation isr24r+4 = 0 that has a double roots: r = 2. Therefore the solutions ofthe homogeneous equations are y0(t) = C1e
2t + C2te2t. We need to find
a particular solution to the equation. The form of the r.h.s. suggeststhat we look for y(t) = (At + B)cos(t) + (Ct + D) sin(t). Inserting
this into the equation leads to A = 20125 , B =22125 , C =
15125 , D =
4125 .
Therefore the general solution is:
y(t) = C1e2t + C2te
2t + (20
125t +
22
125)cos(t) + (
15
125t +
4
125)sin(t)
The initial conditions give C1 =103125
and C2 =2125
.
3. For x + x + x = t2, x(0) = 2 , x(0) = 0, the auxiliary equation is
r2 + r +1 = 0 that has two complex roots: r = 12 i
32
. Therefore the
solutions of the homogeneous equations are x0(t) = C1e 1
2t cos(
32
t) +
C2e1
2t
sin(
3
2 t). We need to find a particular solution to the equation.The form of the r.h.s. suggests that we look for x(t) = At2 + Bt + C.Inserting this into the equation leads to A = 1 , B = 2 , C = 0.Therefore the general solution is:
x(t) = C1e 1
2t cos(
3
2t) + C2e
12t sin(
3
2t) + t2 2t
The initial conditions give C1 = 2 and C2 = 2
3.
4. For 2u 12u + 18u = e3x, u(0) = 3 , u(0) = 1, the auxiliary equationis 2r2
12r + 18 = 0 that has one double root: r = 3. Therefore the
solutions of the homogeneous equations are u0(x) = C1e3x
+ C2xe3x
.We need to find a particular solution to the equation. The form of ther.h.s. suggests that we look for u(x) = Ax2e3x. Inserting this into theequation leads to A = 1
4. Therefore the general solution is:
u(x) = C1e3x + C2xe
3x +1
4x2e3x
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The initial conditions give C1 = 3 and C2 =
8.
5. For u+2u+u = xex+x, u(0) = 2 , u(0) = 1, the auxiliary equationis r2 + 2r + 1 = 0 that has one double root: r = 1. Therefore thesolutions of the homogeneous equations are u0(x) = C1e
x + C2xex.
We need to find a particular solution to the equation. The form of ther.h.s. suggests that we look for u(x) = (Ax3 + Dx2)ex + Bx + C.Inserting this into the equation leads to A = 1
6, D = 0 , B = 1 , C =
2. Therefore the general solution is:
u(x) = C1ex + C2xe
x +1
6x3ex + x 2
The initial conditions give C1 = 0 and C2 = 0 so that: u(x) =16
x3ex+x 2.
3 Exercise 3
The differential equations are of the same type as in the previous exerciseexcept that we need to use the superposition theorem to find a particularsolution.
1. For y + y = (et
+ t)2
, the auxiliary equation is r2
+ 1 = 0 having thetwo complex roots r = i. Therefore the solutions of the homogeneousequations are y0(t) = C1 cos(t) + C2 sin(t). We need to find a particularsolution to the equation. The r.h.s. can be expanded as e2t + 2tet + t2
and therefore we look for y(t) = Ae2t + (B1t + B2)et + Ct2 + Dt + E.
Inserting this into the equation leads to A = 15
, B1 = 1 , B2 = 1 , C =1 , D = 0 , E = 2. Therefore the general solution is:
y(t) = C1 cos(t) + C2 sin(t) +1
5e2t + (t 1)et + t2 2
2. For y+3y4y = 2t+sin2 t+3, the auxiliary equation is r2+3r4 = 0having the two real roots r = 1 and r = 4. Therefore the solutionsof the homogeneous equations are y0(t) = C1e
(t) + C2e( 4t). We
need to find a particular solution to the equation. The r.h.s. can bewritten as 2t + 1
2(1 cos(2t)) + 3 and therefore we look for y(t) =
At + B + Ccos(2t) + D sin(2t). Inserting this into the equation leads to
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A =
12
, B1 =
54
, C = 125
, D =
3
100. Therefore the general solution
is:
y(t) = C1e(t) + C2e
( 4t) + 12
t +54
+1
25cos(2t) 3
100sin(2t)
3. For y 2y + 3y = cosh t, the auxiliary equation is r2 2r + 3 = 0having the two complex roots r = 1 i
2 . Therefore the solutions of
the homogeneous equations are y0(t) = C1et cos(
2t) + C2e
t sin(
2t).We need to find a particular solution to the equation. The r.h.s. canbe written as 1
2et + 1
2et and therefore we look for y(t) = Aet + Bet.
Inserting this into the equation leads to A = 14
, B1 =112
. Thereforethe general solution is:
y(t) = C1et cos(
2t) + C2e
t sin(
2t) +1
4et +
1
12et
4 Exercise 4
In this exercise it is asked to generalize the standard procedure obtainedfor a linear second-order differential equation with constant coefficientsto a third order one. We associated to the equation:
y 2y y + 2y = sin(2t) cos(2t)an auxiliary equation: r3 2r2 r + 2 = 0 which has three real rootsr = 1, r = 1 and r = 2. Therefore we have that y0(t) = C1et +C2e
t + C3e2t is a solution of the differential equation, a fact that can
be checked easily as soon as you have the formula. Now we need todetermine a particular solution. The r.h.s. suggests that we look fory(t) = A cos(2t) + B sin(2t). Inserting this back into the equation leadsto A = 0 and B = 1
10. Therefore the solutions can be written as:
y(t) = C1et + C2e
t + C3e2t +
1
10sin(2t)
We see that we have now three degrees of freedom (three unknown con-stants C1, C2, C3. Therefore it is unlikely that only two initial condi-tions will be enough to make the solution unique. Indeed the conditionsimposed requires:
C1 + C2 + C3 = 0 , C1 + 2C2 + C3 = 1
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which admits an infinite number of solution (affine space of dimension
1):C2 =
1
3 1
2C1 , C3 =
1
3 3
2C1
for any values of C1.
Remark: In general a linear differential equation or order n with con-stant coefficients can be solved by the same method except that theauxiliary equation is a polynomial equation of order n that is oftenimpossible to solve explicitly. However determining the particular so-lution when the r.h.s. is nice is still possible. When the r.h.s. is notnice, then variation of parameters method can be used but in general
is hard to solve.
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