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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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