Answer to Question #163051 in Differential Equations for Leyvanhong

Question

solve the given system of differential equations:

dx/dt=2x+2y

dy/dt=x+3

Answer 1

Solution

d²x/dt²=2dx/dt+2dy/dt

Substitution here dy/dt from second equation gives

d²x/dt²=2dx/dt+2(x+3)  =>  x’’-2x’-2x = 6

We can solve homogeneous part of this ODE looking at the roots of the characteristic equation

r²-2r-2=0   => r_1,2=1±√3

So general solution of homogeneous equation is  x_h(t) = Aexp[(1+√3)t] + Bexp[(1-√3)t]

Some solution of nonhomogeneous equation can be found in the form x₀(t) = C

Substituting into equation -2C = 6

Therefore solution is x(t) = x_h(t) + x₀(t)  = Aexp[(1+√3)t] + Bexp[(1-√3)t] - 3

From the first equation of the system

y(t) = -x+x’/2 = -Aexp[(1+√3)t] - Bexp[(1-√3)t] + 3 + A(1+√3)exp[(1+√3)t]/2 + B(1-√3)exp[(1-√3)t]/2 = A(-1+√3)exp[(1+√3)t]/2 - B(1+√3)exp[(1-√3)t]/2 + 3

Finally

x(t) = Aexp[(1+√3)t] + Bexp[(1-√3)t] - 3

y(t) = -A(1-√3)exp[(1+√3)t]/2 - B(1+√3)exp[(1-√3)t]/2 + 3