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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.

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  • $\begingroup$ 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. $\endgroup$ – John B Nov 3 '18 at 21:24

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