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 .

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

what shape is