Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y′+p(t)y=g(t) have jump discontinuities. If t0 is such a point of discontinuity, then it is necessary to solve the equation separately for t < t0 and t > t0. Afterward, the two solutions are matched so that y is continuous at t0; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y′ continuous at t0.
Solve the initial value problem.

y' + 6y = g(t), y(0) = 0
where
g(t) = 1, 0 ≤ t ≤ 1,
= 0, t > 0.

For , we have

Given that , we have

so that

For (I think you mistakenly wrote , which overlaps with the first definition of ), we have

We want this to be a continuation of the previously found solution at , which means we need to pick such that

Then the solution to the IVP is

Alternatively, we can get around treating piecemeal and resorting to the Laplace transform. Write

where

is the unit step function.

Take the Laplace transform of both sides of the ODE:

where is the Laplace transform of .

We have

so that

Taking the inverse transform yields

which is equivalent to the same solution found earlier; for , , so that ; for , , and .

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: