Question

find the solution of :-
x'(t)= -1/2 *x(t)+2x(t-1)e^(-5x(t-1))
where x(t)=1 , t belongs to [-1,0]

Answer 1

$$x' = -x/2 + 2x(t-1)e^{-5x(t-1)}, \quad t \in [0,1]$$ $$x(t) = 1, \quad t \in [-1,0]$$ $$t \in [0,1] \Rightarrow t - 1 \in [-1,0] \Rightarrow x(t - 1) = 1, \quad t \in [0,1]$$ $$x(0) = 1$$ $$\left\{ \begin{array}{l} x' = -x/2 + 2e^{-5} \\ x(0) = 1 \end{array} \right.$$ $$1) x' = -x/2$$ $$\frac{dx}{x} = -(1/2)dt \Rightarrow x = Ce^{-t/2}$$ $$2) x = g(x)e^{-t/2} \Rightarrow x' = g'(x)e^{-t/2} - g(x)e^{-t/2}/2$$ $$g'(x)e^{-t/2} - g(x)e^{-t/2}/2 = -g(x)e^{-t/2}/2 + 2e^{-5}$$ $$g'(x) = 2e^{\frac{t}{2} - 5} \Rightarrow g(x) = 2e^{-5} \int e^{t/2}dt = 4e^{t/2 - 5}$$ $$x = g(x)e^{-t/2} = 4e^{-5}$$ $$3) x = Ce^{-t/2} + 4e^{-5}$$ $$x(0) = 1 \Rightarrow 1 = C + 4e^{-5} \Rightarrow C = 1 - 4e^{-5}$$ $$x = \left(1 - 4e^{-5}\right)e^{-t/2} + 4e^{-5}$$