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 .

2

step-by-step explanation:

1x2=2

hope this !

-payshence