Introductory Differential Equations
Introductory Differential Equations
5th Edition
ISBN: 9780128149485
Author: Abell, Martha L. L.
Publisher: Elsevier Science
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Chapter 7, Problem 1RE

(a)

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=106C, I(0)=0 A and E(t)=0 V.

(a)

Expert Solution
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Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e23tet) and I(t)=1500000(e23t+et)_.

Explanation of Solution

Result used:

From Kirchhoff’s voltage law, RI+LdIdt+1CQE(t)=0.

From the above equation, the system of differential equations {dQdt=IdIdt=1LCQRLI+1LE(t).

The initial condition Q(0)=Q0 and I(0)=I0 on charge and current, respectively.

Here, I(t)=current(dQ(t)dt), Q(t)=charge, R is resistance, C is capacitance, E is voltage and L is inductance.

Calculation:

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=0 V.

The initial condition is Q(0)=106C  and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=1LCQRLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=1(1)(32)Q(53)1I+11(0)

Simplify the system of equations as follows.

{dQdt=IdIdt=23Q53I

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(012353)(QI)+(00)

Consider the vector [012353], compute eigenvalues of the matrix as follows.

|012353|λ|1001|=|012353||λ00λ|=|λ12353λ|=λ(53λ)(23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further as,

(3λ+2)(λ+1)=0λ=23 and λ=1

Thus, the corresponding eigenvectors are v1=(321) and v2=(11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(32e23tete23tet).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|32e23tete23tet|=(32e23t)(et)(et)(e23t)=32e53t+e53t=12e53t

Now compute inverse matrix as follows.

Φ1(t)=1(12e53t)adj(32e23tete23tet)=2e53t(etete23t32e23t)=(2e23t2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ1(0)X(0)+Φ(t)0tΦ1(u)F(u)du.

Φ1(0)=(2e23(0)2e23(0)2e(0)3e(0))=(2223)

Note that,

X(0)=(Q(0)I(0))=(1060)

X(t)=Φ(t)Φ1(0)X(0)+Φ(t)0tΦ1(u)F(u)du={(32e23tete23tet)(2223)(1060)+(32e23tete23tet)0t(2e23u2e23u2eu3eu)(00)du}={(32e23tete23tet)(2223)(1060)+0}=(3e23t2et3e23t3et2e23t+2et2e23t+3et)(1060)

Simplify further as,

X(t)=(3e23t2et3e23t3et2e23t+2et2e23t+3et)(1060)=((3e23t2et)106+0(2e23t+2et)106+0)=(1500000(32e23tet)1500000(e23t+et))

The solution of one-loop L-R-C circuit is, Q(t)=1500000(32e23tet) and I(t)=1500000(e23t+et)_.

b.

To determine

The solution of one loop circuit with L=1 H, R=53Ω, C=32 F, Q=106C, I(0)=0 A and E(t)=et V.

b.

Expert Solution
Check Mark

Answer to Problem 1RE

The solution of one-loop L-R-C circuit is, Q(t)=(3e23t2et)10695e23t(e53t1)3tet_ and I(t)=(2e23t+2et)10615e23t(6e53t6)3tet_.

Explanation of Solution

Given that, the condition L=1 H, R=53Ω, C=32 F and E(t)=et V.

The initial condition is Q(0)=106C  and I(0)=0 A.

Substitute the respective values in the system of equation {dQdt=IdIdt=1LCQRLI+1LE(t) and obtain the homogeneous equation as follows.

{dQdt=IdIdt=1(1)(32)Q(53)1I+11et

On further simplification.

{dQdt=IdIdt=23Q53I+et

From above system of equation, the solution in matrix form as follows.

(dQdtdIdt)=(012353)(QI)+(0et)

Consider the vector [012353], compute eigenvalues of the matrix as follows.

|012353|λ|1001|=|012353||λ00λ|=|λ12353λ|=λ(53λ)(23)(1)=53λ+λ2+23

Let λ2+53λ+23=0. Compute the values of λ as follows.

λ2+53λ+23=03λ2+5λ+2=03λ2+3λ+2λ+2=03λ(λ+1)+2(λ+1)=0

Simplify further,

(3λ+2)(λ+1)=0λ=23 and λ=1

Thus, the corresponding eigenvectors are v1=(321) and v2=(11).

Use the eigenvalues and eigenvectors and compute fundamental matrix as, Φ(t)=(32e23tete23tet).

Compute the inverse matrix of Φ(t) as follows.

|Φ(t)|=|32e23tete23tet|=(32e23t)(et)(et)(e23t)=32e53t+e53t=12e53t

Now compute inverse matrix as follows.

Φ1(t)=1(12e53t)adj(32e23tete23tet)=2e53t(etete23t32e23t)=(2e23t2e23t2et3et)

Thus, the value of X(t)=Φ(t)Φ1(0)X(0)+Φ(t)0tΦ1(u)F(u)du.

Φ1(0)=(2e23(0)2e23(0)2e(0)3e(0))=(2223)

Here, the matrix X(0)=(1060).

X(t)=Φ(t)Φ1(0)X(0)+Φ(t)0tΦ1(u)F(u)du={(32e23tete23tet)(2223)(1060)+(32e23tete23tet)0t(2e23u2e23u2eu3eu)(0eu)du}={(32e23tete23tet)(2223)(1060)+(32e23tete23tet)0t(2e53u3)du}={(32e23tete23tet)(2223)(1060)+(32e23tete23tet)(65e53t653t)}

Simplify further as,

X(t)=(3e23t2et3e23t3et2e23t+2et2e23t+3et)(1060)+(32e23tete23tet)(65e53t653t)=((3e23t2et)106+0(2e23t+2et)106+0)+(95e23t(e53t1)3tet15e23t(6e53t6)3tet)=((3e23t2et)10695e23t(e53t1)3tet(2e23t+2et)10615e23t(6e53t6)3tet)

The solution of one-loop L-R-C circuit is, Q(t)=(3e23t2et)10695e23t(e53t1)3tet_ and I(t)=(2e23t+2et)10615e23t(6e53t6)3tet_.

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Chapter 7 Solutions

Introductory Differential Equations

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