# Solving a Delay Differential Equation with the Method of Steps

I am trying to verify if my solution is correct for the following Delayed Differentiable Equation of

$$y'(t) = 3y(t - 2)$$ with the history function $$h(t) = 1$$ for $$t \leq 0$$.

On the interval $$0 \leq t \leq 2$$, $$y(t -2)$$ has the known value of $$h(t - 2) = 1$$. This implies that on the interval $$0 \leq t \leq 2$$ the DDE reduces down to the ODE $$y'(t) = 3(h(t - 2)) = 3$$. It has initial value of $$y(0) = h(0) = 1$$. The solution to the DDE $$y'(t) = 3$$ over the interval $$0 \leq t \leq 2$$ is $$y(t) = 3t + C$$ and applying the initial conditions of $$y(0) = 1$$, $$C = 1$$ so $$y(t) = 3t + 1$$.

On the interval $$2 \leq t \leq 3$$ the DDE becomes the ODE $$y'(t) = 3y(t - 2) = 3(3(t - 2) + 1) = 3(3t - 5) = 9t - 15$$, with initial conditions $$y(2) = 3(2) + 1 = 7$$. Solving for the new solution for the ODE $$y'(t) = 9t - 15$$ on the interval $$2 \leq t \leq 3$$, we get $$y'(t) = 9t - 15 \Rightarrow \frac{dy}{dt} = (9t - 15) \Rightarrow \int dy = \int (9t - 15) dt =$$ $$y(t) = \frac{9t^2}{2} - 15t + C$$. Applying the initial conditions of $$y(2) = 7$$ the solution then becomes $$y(2) = \frac{9}{2}(2^2) - 15(2) + C = 18 - 30 + C = C - 12 = 7 \Rightarrow C = 19$$

$$y(t) = \frac{9t^2}{2} - 15t + 19$$, for $$2 \leq t \leq 3$$.

Putting all of these together my final result is then

$$y(t) = \left\{ \begin{array}{ll} 1, & t \leq 0\\ 3t + 1, & 0 \leq t \leq 2\\ \frac{9t^2}{2} - 15t + 19, & 2 \leq t \leq 3\\ \end{array} \right.$$

I just want to verify if my piecewise solution is correct, the instructor I am studying from does not provide any additional resources for this particular material regarding using the method of steps to solve delayed differential equations other than a simpler example covered in a short time. Thanks.

• All is quite fine. But you really don't need more examples since it is always the same. :) It would be different if you had a dependence on various former times instead of on a single one. – John B Nov 3 '18 at 21:24