First integrals of the Lotka-Volterra Model

Integrating this by parts yields, I₁ = 1

2e^2x(3x²−2x−1)−1 2

e^2x(6x−2)dx, (4.32) integrating the second half of I₁ by parts, we get

I₂ = 1

4e^2x(6x−2)− Z 6

4e^2xdx = 1

4e^2x(6x−2)− 3

4e^2x+G. (4.33) and thus,I₁ results in

I1 = 1

2e^2x(3x²−2x−1)− 1

4e^2x(6x−2) + 3

4e^2x+G, (4.34) where Gis the integration constant.

Integrating equation (4.36) leads to the generalised form of the first integral, I = ˙xR₁[1 + tan(S₁)]¹² + 1

A[ ˙x+Aln( ˙x−A)] +ψ(x, t). (4.37) In order to solve for the general functions in (4.37), we first differentiate then equate like terms in the corresponding coefficients resulting with the following equations,

∂I

∂x = x˙

R_1x[1 + tan(S₁)]¹² +R₁

2 [1 + tan(S₁)]⁻¹² sec²(S₁)·S_1x

+ψ_x, (4.38)

∂I

∂t = xR˙ ₁

2 [1 + tan(S₁)]⁻¹² sec²(S₁)S_1t+ψ_t, (4.39) where the subscript xor t refers to partial differentiation with respect to the variable.

Substituting the derivatives of I(x, t) into the next part of the equation, we separate by powers ofx,˙

˙ x² :

R_1x[1 + tan(S₁)]¹² + R₁

2 [1 + tan(S₁)]⁻¹² sec²(S₁)S_1x

= be^x+a

A(A−x)˙ , (4.40)

x : −(be^x+a)R₁[1 + tan(S₁)]¹² − be^x+a

A(A−x)˙ = R₁

2 [1 + tan(S₁)]⁻¹² sec²(S₁)S_1t

+ψ_x. (4.41)

Integrating with respect to x, ψ = −

Z "

(be^x+a)R₁[1 + tan(S₁)]¹² + R₁

2 [1 + tan(S₁)]⁻¹² sec²(S₁)S_1t + be^x+a

A(A−x)˙

∂x+F(t). (4.42)

Equating coefficients by x˙⁰,

A R₁ [1 + tan(S₁)]¹² (be^x+a) = ψ_t, (4.43)

=⇒ F(t) = AR₁[1 + tan(S₁)]¹² (be^x+a), (4.44) +

Z "

(be^x+a)R₁[1 + tan(S₁)]¹² +R1

2 [1 + tan(S₁)]⁻¹² sec²(S₁)S_1t + be^x+a

A(A−x)˙

∂x

+G. (4.45)

Thus, the first integral is given by I = ˙xR1[1 + tan(S1)]¹² + 1

A[ ˙x+Aln( ˙x−A)] +AR1(be^x+a) Z

[1 + tan(S1)]¹² ∂t+G.

(4.46) where Gis the constant of integration.

Case 1.2: To determine the first integral we apply the equation (3.9) to (3.6) to obtain the relation,

R₂[1 + tan(S₂)]¹² [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.47) Equating coefficients ofx,¨

∂I

∂x˙ =R₂[1 + tan(S₂)]¹² . (4.48) Integrating equation (4.48) results in,

I = ˙xR₂[1 + tan(S₂)]¹² +ψ(x, t). (4.49) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = x˙

R_2x[1 + tan(S₂)]¹² + R₂

2 [1 + tan(S₂)]⁻¹² sec²(S₂)S_2x

+ψ_x (4.50)

∂I

∂t = xR˙ ₂

2 [1 + tan(S₂)]⁻¹² sec²(S₂)S_2t+ψ_t. (4.51)

Substituting the derivatives of I(x, t) into the next part of the equation, we separate by powers ofx,˙

˙ x² :

R_2x[1 + tan(S₂)]¹² +R₂

2 [1 + tan(S₂)]⁻¹² sec²(S₂)S_2x

= 0, (4.52)

x : −(be^x+a)R2[1 + tan(S2)]¹² = R₂

2 [1 + tan(S2)]⁻¹² sec²(S2)S2t

+ψ_x. (4.53)

Integrating with respect to x we obtain, ψ = −

(be^x+a)R₂[1 + tan(S₂)]¹² + R₂

2 [1 + tan(S₂)]⁻¹² sec²(S₂)S_2t

∂x

+ F(t). (4.54)

Equating coefficients by x˙⁰,

AR₂ [1 + tan(S₂)]¹² (be^x+a) =ψ_t,

=⇒ F(t) =AR₂[1 + tan(S₂)]¹² (be^x+a) +

Z "

(be^x+a)R₂[1 + tan(S₂)]¹²

+ R₂

2 [1 + tan(S₂)]⁻¹² sec²(S₂)S_2t

∂x

+G. (4.55)

Thus, the first integral is given by

I = ˙xR₂[1 + tan(S₂)]¹² +AR₂(be^x+a) Z

[1 + tan(S₂)]¹² ∂t+G. (4.56) where Gis the constant of integration.

Case 1.3: To determine the first integral we apply the equation (3.10) to (3.6) to obtain the relation,

R₃[1 + tan(S₃)]¹² [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.57)

Equating coefficients ofx,¨

∂I

∂x˙ =R₃[1 + tan(S₃)]¹² . (4.58) Integrating equation (4.58) results in,

I = ˙xR₃[1 + tan(S₃)]¹² +ψ(x, t). (4.59) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = x˙

R_3x[1 + tan(S₃)]¹² +R₃

2 [1 + tan(S₃)]⁻¹² sec²(S₃)S_3x

+ψ_x, (4.60)

∂I

∂t = xR˙ ₃

2 [1 + tan(S3)]⁻¹² sec²(S3)S3t+ψt. (4.61) Substituting the derivatives of I(x, t) into the next part of the equation, we separate by powers ofx,˙

˙ x² :

R_3x[1 + tan(S₃)]¹² +R3

2 [1 + tan(S₃)]⁻¹² sec²(S₃)S_3x

= 0, (4.62)

x : −(be^x+a)R₃[1 + tan(S₃)]¹² = R₃

2 [1 + tan(S₃)]⁻¹² sec²(S₃)S_3t

+ψ_x. (4.63)

Integrating with respect to x we obtain, ψ = −

(be^x+a)R₃[1 + tan(S₃)]¹² + R3

2 [1 + tan(S₃)]⁻¹² sec²(S₃)S_3t

∂x

+ F(t). (4.64)

Equating coefficients by x˙⁰,

A R₃ [1 + tan(S₃)]¹² (be^x+a) = ψ_t,

=⇒ F(t) = AR[1 + tan(S₃)]¹² (be^x+a) +

Z "

(be^x+a)R₃[1 + tan(S₃)]¹²

+ R₃

2 [1 + tan(S₃)]⁻¹² sec²(S₃)S_3t

∂x

+G. (4.65)

Thus, the first integral is given by

I = ˙xR3[1 + tan(S3)]¹² +AR3(be^x+a) Z

[1 + tan(S3)]¹² ∂t+G. (4.66) where Gis the constant of integration.

Case 2.1: To determine the first integral for this case, we apply equation (3.12) to (3.6) to obtain the relation,

˙ x A−x˙

x+ (be^x+a)(A−x)˙ i

= ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.67) By equating coefficients of x¨ , we get

∂I

∂x˙ = x˙

A−x˙ (4.68)

Integrating equation (4.68) yields, I =−x˙

A −ln( ˙x−A) +ψ(x, t). (4.69) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = ψx, (4.70)

∂I

∂t = ψ_t. (4.71)

Substituting the derivatives of I(x, t), we have

x(be^at+a) = ∂I

∂t + ˙x∂I

∂x, (4.72)

x(be^at+a) = ψ_t+ ˙xψ_x, (4.73)

and equating by powers of x˙ we get,

x : be^x+a=ψ_x, (4.74)

x⁰ : 0 =ψ_t. (4.75)

Integrating x˙⁰,yields

ψ = C(x), (4.76)

ψ_x = C(x),˙

=⇒ C(x) =˙ be^x+a,

=⇒ C(x) = be^x+ax+G. (4.77) Therefore, our first integral becomes

I =−x˙

A −ln( ˙x−A) +be^x+ax+G. (4.78) where Gis the constant of integration.

Case 2.2.1: To determine the first integral we apply equation (3.14) to (3.6) which results in,

Acos(S₁)− x˙₁ A(A−x˙1)

[¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.79) Equating coefficients ofx,¨

∂I

∂x˙ = 1

Acos(S₁)− x˙₁

A(A−x˙1) (4.80)

Integrating equation (4.80) yields, I = x˙

Acos(S₁)− 1

A[ ˙x+Aln( ˙x−A)] +ψ(x, t). (4.81) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = −x˙

Asin(S₁)S_1x+ψ_x, (4.82)

∂I

∂t = −x˙

Asin(S₁)S_1t+ψ_t. (4.83)

Substituting the derivatives of I(x, t) into the next part of the equation, we separate by powers ofx,˙

x² : (be^x+a)

A(A−x˙₁) =−1

Asin(S)S_1x, (4.84)

x : −(be^x+a)

A cos(S₁)− (be^x+a)

(A−x)˙ =−1

Asin(S₁)S_1t+ψ_x. (4.85) Integrating with respect to x yields,

ψ = Z

−(be^x+a)

cos(S1)

A + 1

A−x˙

+ 1

Asin(S1)S1t

∂x+F(t), (4.86) ψt = ∂

∂t Z

−(be^x+a)

cos(S1)

A + 1

A−x˙

+ 1

Asin(S1)S1t

∂x

+ ˙F(4.87) Equating coefficients by x˙⁰,

(be^x+a) cos(S₁) = ψ_t, (4.88)

=⇒ F(t) = − Z

−(be^x+a)

cos(S₁)

A + 1

A−x˙

+ 1

Asin(S1)S1t

∂x + (be^x+a)

cos(S₁)∂t+G. (4.89)

Thus, the first integral is given by I = x˙

Acos(S₁) + 1

A[ ˙x+Aln( ˙x−A)] + (be^x+a) Z

cos(S₁)∂t+G. (4.90) where Gis the constant of integration.

Case 2.2.2: To acquire the first integral for this case, we apply equation (3.15) to (3.6) which yields,

R₂cos(S₂) [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.91)

Equating coefficients ofx,¨

∂I

∂x˙ =R₂cos(S₂). (4.92)

Integrating equation (4.92) results in,

I = ˙xR2cos(S2) +ψ(x, t). (4.93) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = −xR˙ ₂sin(S₂)S_2x+ψ_x, (4.94)

∂I

∂t = x˙h

R˙₂cos(S₂)−R₂sin(S₂)S_2ti

+ψ_t. (4.95)

Substituting the derivatives of I(x, t), we separate by powers ofx,˙

x² : −R₂sin(S₂)S_2x = 0, (4.96)

x : −(be^x+a)R₂cos(S₂) = ˙R₂cos(S₂)−R₂sin(S₂)S_2t+ψ_x. (4.97) Integrating with respect to x,

ψ = − Z h

(be^x+a)R₂cos(S₂) +

R˙₂cos(S₂)−R₂sin(S₂)S_2ti

∂x

+ F(t). (4.98)

Equating coefficients by x˙⁰,

A(be^x+a)R₂cos(S₂) = ψ_t, (4.99)

=⇒ F(t) = Z h

(be^x+a)R₂cos(S₂) +

R˙₂cos(S₂)−R₂sin(S₂)S_2ti

∂x + A(be^x+a)

R₂cos(S₂)∂t+G. (4.100) Thus, the first integral is given by

I = ˙xR₂cos(S₂) +A(be^x+a) Z

R₂cos(S₂)∂t+G. (4.101)

where Gis the constant of integration.

Case 2.2.3 : In order to obtain the first integral, we apply equation (3.16) to (3.6) which gives us,

Acos(S₃) [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.102) Equating coefficients ofx,¨

∂I

∂x˙ = 1

Acos(S₃). (4.103)

I = x˙

Acos(S3) +ψ(x, t). (4.104) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = −x˙

Asin(S3)S3x+ψx, (4.105)

∂I

∂t = −x˙

Asin(S₃)S_3t+ψ_t. (4.106) Substituting the derivatives of I(x, t) into the next part of the equation, we separate by powers ofx,˙

x² : −1

Asin(S3)S3x= 0, (4.107)

x : −(be^x+a)

A cos(S₃) = −1

Asin(S₃)S_3t+ψ_x. (4.108) Integrating with respect to x,

ψ = 1 A

Z 1

Asin(S₃)S_3t−(be^x+a)

A cos(S₃)

∂x+F(t), (4.109) ψ_t = ∂

∂t 1

A Z

Asin(S₃)S_3t− (be^x+a)

A cos(S₃)

∂x

+ ˙F (4.110)

Equating coefficients by x˙⁰,

(be^x+a) cos(S₃) = ψ_t, (4.111)

=⇒ F(t) = −1 A

Z 1

Asin(S3)S3t−(be^x+a)

A cos(S3)

∂x + (be^x+a)

cos(S₃)∂t+G. (4.112)

Thus, the first integral is given by I = x˙

Acos(S₃) + (be^x+a) Z

cos(S₃)∂t+G. (4.113) where Gis the constant of integration.

Case 2.2.4: To determine the first integral we apply equation (3.17) to (3.6) which yields,

e^A

A cos(S₄) [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.114) Equating coefficients ofx,¨

∂I

∂x˙ = e^A

A cos(S₄). (4.115)

Integrating equation (4.115) results in, I = xe˙ ^A

A cos(S₄) +ψ(x, t). (4.116) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = −xe˙ ^A+x

A sin(S₄)S_4x+ψ_x, (4.117)

∂I

∂t = −xe˙ ^A

A sin(S₄)S_4t+ψ_t. (4.118)

Substituting the derivatives of I(x, t) into the next part of the equation, we separate by powers ofx,˙

x² : −e^A+x

A sin(S₄)S_4x = 0, (4.119)

x : −(be^x+a)e^A

A cos(S₄) =−e^A

A sin(S₄)S_4t+ψ_x. (4.120) Integrating with respect to x,

ψ = e^A A

Z e^A

A sin(S₄)S_4t− (be^x+a)e^A

A cos(S₄)

∂x+F(t), (4.121) ψ_t = ∂

∂t e^A

A Z

e^A

A sin(S₄)S_4t−(be^x+a)e^A

A cos(S₄)

∂x

+ ˙F . (4.122) Equating coefficients by x˙⁰,

e^A(be^x+a) cos(S₄) = ψ_t, (4.123)

=⇒ F(t) = −e^A A

Z e^A

A sin(S4)S4t− (be^x+a)e^A

A cos(S4)

∂x + e^A(be^x+a)

cos(S₄)∂t+G. (4.124)

Thus, the first integral is given by I = xe˙ ^A

A cos(S₄) +e^A(be^x+a) Z

cos(S₄)∂t+G. (4.125) where Gis the constant of integration.

Case 3.1: To obtain the first integral, we apply equation (3.19) to (3.6) which yields,

˙ x

A−x˙ [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.126) By equating coefficients of x,¨

∂I

∂x˙ = x˙

A−x˙ (4.127)

Integrating equation (4.127) results in, I =−x˙

A −ln( ˙x−A) +ψ(x, t). (4.128) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = ψ_x, (4.129)

∂I

∂t = ψ_t. (4.130)

Substituting the derivatives of I(x, t), we have

x(be^at+a) = ∂I

∂t + ˙x∂I

∂x, (4.131)

x(be^at+a) = ψ_t+ ˙xψ_x. (4.132) So by equating coefficients we get,

x : be^x+a=ψx, (4.133)

x⁰ : 0 =ψ_t. (4.134)

Integrating x˙⁰,yields

ψ = C(x), (4.135)

ψ_x = C(x),˙

=⇒ C(x) =˙ be^x+a,

=⇒ C(x) = be^x+ax+G. (4.136) Therefore our first integral becomes

I =−x˙

A −ln( ˙x−A) +be^x+ax+G. (4.137)

where Gis the constant of integration.

Case 3.2.1 : To determine the first integral we apply equation (3.21) to (3.6) which results in,

R₁cos(S₁)− x˙₁ A(A−x˙₁)

[¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.138) Equating coefficients ofx,¨

∂I

∂x˙ =R₁cos(S₁)− x˙₁

A(A−x˙₁). (4.139)

Integrating equation (4.139) yields, I = ˙xR₁cos(S₁)− 1

A[ ˙x+Aln( ˙x−A)] +ψ(x, t). (4.140) Then differentiating and equating like terms in the corresponding coefficients leads to

∂I

∂x = ψx, (4.141)

∂I

∂t = x˙h

R˙₁cos(S₁)−R₁sin(S₁)S_1ti

+ψ_t. (4.142)

Substituting the derivatives of I(x, t) into the next part of the equation, we can then separate by powers of x,˙

x² : (be^x+a)

A(A−x)˙ = 0 (4.143)

x : −R₁(be^x+a) cos(S₁)− (be^x+a)

(A−x)˙ = ˙R₁cos(S₁)−R₁sin(S₁)S_1t+ψ_x. (4.144) Integrating with respect to x we obtain,

ψ = − Z

R₁(be^x+a) cos(S₁) + (be^x+a) (A−x)˙ +

R˙₁cos(S₁)−R₁sin(S₁)S_1t

∂x

+ F(t). (4.145)

Equating coefficients by x˙⁰,

A (be^x+a)R₁cos(S₁) =ψ_t,

=⇒ F(t) = Z "

R₁(be^x+a) cos(S₁) + (be^x+a) (A−x)˙ +

R˙₁cos(S₁)−R₁sin(S₁)S_1t

∂x + A(be^x+a)

Rcos(S)∂t+G. (4.146)

Therefore, the first integral is given by I = ˙xR₁cos(S₁)− 1

A[ ˙x+Aln( ˙x−A)] +A(be^x+a) Z

R₁cos(S₁)∂t+G. (4.147) where Gis the constant of integration.

Case 3.2.2 : To determine the first integral of this case, we apply equation (3.22) to (3.6)

R₂cos(S₂) [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.148) Equating coefficients ofx,¨

∂I

∂x˙ =R₂cos(S₂). (4.149)

Integrating equation (4.149) results in,

I = ˙xR₂cos(S₂) +ψ(x, t). (4.150) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = ψx, (4.151)

∂I

∂t = x˙h

R˙₂cos(S₂)−R₂sin(S₂)S_2ti

+ψ_t. (4.152)

Substituting the derivatives of I(x, t), we then separate by powers ofx,˙

x : −R₂(be^x+a) cos(S₂) = ˙R₂cos(S₂)−R₂sin(S₂)S_2t+ψ_x. (4.153)

Integrating with respect to x,

ψ = −R2(be^x+ax) cos(S2)−x

hR˙2cos(S2)−R2sin(S2)S2t

+F(t). (4.154) Equating coefficients by x˙⁰,

e^A(be^x+a) cos(S₂) = ψ_t, (4.155)

=⇒ F(t) = R(be^x+ax) cos(S₂) +xh

R˙₂cos(S₂)−R₂sin(S₂)S_2ti + A(be^x+a)

R₂cos(S₂)∂t+G. (4.156) Thus, the first integral is given by

I = ˙xR₂cos(S₂) +A(be^x+a) Z

R₂cos(S₂)∂t+G. (4.157) where Gis the constant of integration.

Case 3.2.3 : To determine the first integral, we apply equation (3.23) to (3.6) which results in,

R₃cos(S₃) [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.158) Equating coefficients ofx,¨

∂I

∂x˙ =R₃cos(S₃). (4.159)

Integrating equation (4.159) yields,

I = ˙xR₃cos(S₃) +ψ(x, t). (4.160) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = ψ_x, (4.161)

∂I

∂t = x˙h

R˙₃cos(S₃)−R₃sin(S₃)S_3ti

+ψ_t. (4.162)

Substituting the derivatives of I(x, t), we can separate by powers of x,˙

x : −R(be^x+a) cos(S₃) = ˙R₃cos(S₃)−R₃sin(S₃)S_3t+ψ_x. (4.163) Integrating with respect to x,

ψ = −R₃(be^x+ax) cos(S₃)−xh

R˙₃cos(S₃)−R₃sin(S₃)S_3ti

+F(t). (4.164) Equating coefficients by x˙⁰,

e^A(be^x+a) cos(S₃) = ψ_t, (4.165)

=⇒ F(t) = R₃(be^x+ax) cos(S₃) +xh

R˙₃cos(S₃)−R₃sin(S₃)S_3ti + A(be^x+a)

R3cos(S3)∂t+G. (4.166) Thus, the first integral is given by

I = ˙xR₃cos(S₃) +A(be^x+a) Z

R₃cos(S₃)∂t+G. (4.167) where Gis the constant of integration.

Case 3.2.4 : To determine the first integral, we can apply (3.24) to (3.6) which yields, R₄cos(S₄) [¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.168) Equating coefficients ofx,¨

∂I

∂x˙ =R4cos(S4). (4.169)

Integrating equation (4.169) results in,

I = ˙xR₄cos(S₄) +ψ(x, t). (4.170)

The differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = ψ_x, (4.171)

∂I

∂t = x˙

hR˙4cos(S4)−R4sin(S4)S4t

+ψt. (4.172)

Substituting the derivatives ofI(x, t)into the next part of the equation, we can separate by powers ofx,˙

x : −R₄(be^x+a) cos(S₄) = ˙R₄cos(S₄)−R₄sin(S₄)S_4t+ψ_x. (4.173) Integrating with respect to x,

ψ = −R₄(be^x+ax) cos(S₄)−xh

R˙₄cos(S₄)−R₄sin(S₄)S_4ti

+F(t). (4.174) Equating coefficients by x˙⁰,

e^A(be^x+a) cos(S₄) = ψ_t,

=⇒ F(t) = R₄(be^x+ax) cos(S₄) +xh

R˙₄cos(S₄)−R₄sin(S₄)S_4ti + A(be^x+a)

R₄cos(S₄)∂t+G. (4.175) Thus, the first integral is given by

I = ˙xR₄cos(S₄) +A(be^x+a) Z

R₄cos(S₄)∂t+G. (4.176) where Gis the constant of integration.

Case 3.3.1 : To determine the first integral for this case, we apply equation (3.26) to (3.6) which results in,

[¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.177)

By equating coefficients x,¨

∂I

∂x˙ = 1. (4.178)

Integrating equation (4.178) yields,

I = ˙x+ψ(x, t). (4.179)

Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = ψ_x, (4.180)

∂I

∂t = ψ_t. (4.181)

Substituting the derivatives of I(x, t), we can then separate by powers of x,˙

x:−(be^x+a) = ψx, (4.182a)

=⇒ ψ = −(be^x+ax) +C(t), (4.182b) ψ_t = C(t).˙

x⁰ :A(be^x+a) = ψ_t, (4.182c)

A(be^x+a) = C(t),˙

=⇒ C(t) = At(be^x+a) +G. (4.182d) Therefore, our first integral becomes

I = ˙x−(be^x+a) (At−1) +G. (4.183) where Gis the constant of integration.

Case 3.3.2 : To determine the first integral, we apply equation (3.27) to (3.6) which results in,

e^−at[¨x+ (be^x+a)(A−x)] =˙ ∂I

∂t + ˙x∂I

∂x + ¨x∂I

∂x˙. (4.184)

By equating coefficients x,¨

∂I

∂x˙ =e^−at. (4.185)

Integrating equation (4.185) yields,

I =e^−atx˙ +ψ(x, t). (4.186) Then differentiating and equating like terms in the corresponding coefficients leads to,

∂I

∂x = ψ_x, (4.187)

∂I

∂t = −ae^−atx˙ +ψt. (4.188)

Substituting the derivatives of I(x, t), we have e^−at(be^x+a)(A−x) =˙ ∂I

∂t + ˙x∂I

∂x (4.189)

e^−at(be^x+a)(A−x) =˙ −ae^−atx˙ +ψt+ ˙xψx. (4.190) We can now separate by powers of x,˙

x : −e^−at(be^x+a) = −ae^−at+ψ_x, (4.191a) ψ_x = −e^−atbe^x,

ψ = −be^x−at+C(t), (4.191b)

ψ_t = abe^x−at+ ˙C(t).

x⁰ : e^−at(be^x+a)A = ψ_t, (4.191c)

e^−at(be^x+a)A = abe^x−at+ ˙C(t), C(t) =˙ e^−at(be^x−abe^x),

=⇒ C(t) = e^−atbe^x− e^−at

a (be^x+a)A+G. (4.191d)

Therefore, our first integral becomes I =e^−atx˙ − e^−at

a (be^x+a)A+G. (4.192)

where Gis the constant of integration.

Dalam dokumen The characteristic approach in determining first integrals of a predator-prey system. (Halaman 47-67)