Solving system of differential equations...?

Derive second equation:

d²R/dt = dJ/dt

You can convert this system of two first order differential equations to a second order differential equation of either R or J.

To obtain a second differential equation of R derive first DE.

d²R7dt = dJ/dt

then substitute dJ/dt by second equation

d²R/dt = - R + J

substitute J by first equation

d²R/dt = - R + dR/dt

<=>

d²R/dt - dR/dt + R = 0

That is constant coefficient ODE with characteristic equation:

λ² - λ + 1 =

It has two conjugate complex roots:

λ = 1/2 ± √( (1/2)² - 1) = (1/2) ± i∙√(3/4)

So the general solution for R is:

R(t) = e^((1/2)∙t) ∙ [ C₁∙sin(√(3/4)∙t) + C₂∙cos(√(3/4)∙t) ]

General solution for J can be found from 1st DE:

J(t) = dR/dt

= (1/2)∙e^((1/2)∙t) ∙ [ C₁∙sin(√(3/4)∙t) + C₂∙cos(√(3/4)∙t) ]

+ e^((1/2)∙t) ∙ [ √(3/4)∙C₁∙cos(√(3/4)∙t) - √(3/4)∙C₂∙sin(√(3/4)∙t) ]

= e^((1/2)∙t) ∙ [ ((1/2)∙C₁ - √(3/4)∙C₂)∙sin(√(3/4)∙t) + (√(3/4)∙C₁ + (1/2)∙C₂)∙cos(√(3/4)∙t) ]

Apply initial conditions to find the constants:

(i)

R(0) = 1

<=>

e^((1/2)∙0) ∙ [ C₁∙sin(√(3/4)∙0) + C₂∙cos(√(3/4)∙0) ] = 1

<=>

<=>

1 ∙ [ C₁∙0 + C₂∙1 ] = 1

<=>

C₂ = 1

(ii)

J(0) = 0

<=>

e^((1/2)∙0) ∙ [ ((1/2)∙C₁ - √(3/4)∙C₂)∙sin(√(3/4)∙0) + (√(3/4)∙C₁ + (1/2)∙C₂)∙cos(√(3/4)∙0) ] = 0

<=>

1∙[ ((1/2)∙C₁ - √(3/4)∙C₂)∙0 + (√(3/4)∙C₁ + (1/2)∙C₂)1 ] = 0

<=>

(√(3/4)∙C₁ + (1/2)∙C₂) = 0

<=>

C₁ = -(1/√3)∙C₂ = -(1/√3)

So the solution to this IVP is:

R(t) = e^((1/2)∙t) ∙ [ cos(√(3/4)∙t) - (1/√3)∙sin(√(3/4)∙t) ]

J(t) = e^((1/2)∙t) ∙ [ (- (1/2)∙(1/√3) - √(3/4)∙C₂)∙sin(√(3/4)∙t) + -(√(3/4)∙(1/√3) + (1/2)∙C₂)∙cos(√(3/4)∙t) ]

= - (2/√3) ∙ e^((1/2)∙t) ∙sin(√(3/4)∙t)