Monday Math 116

The Laplace Transform
Part 17: Differential Equations with Non-constant Coefficients

The Laplace transform can also be used to solve some linear differential equations with non-constant coefficients. For example, let us consider the problem of solving, for t≥0, the differential equation  with initial conditions .

Taking the Laplace transform of both sides, again with ,
.

Now, recall our frequency differentiation formula:

Thus, we see
,
and
.
Thus, our transformed equation becomes:
,
and we have turned our second-order differential equation into a first order one. Further, it is a separable equation:
,
where C is a constant of integration. To find the value of this constant, we turn to the initial value theorem:
.
Here, we see
,
so we find that , and we have . Recalling that , we see that our solution is simply , t≥0.

2 Responses to “Monday Math 116”

1. Monday Math 117 « Twisted One 151's Weblog Says:

[…] Part 18: The Bessel Functions We can use the differential equation technique demonstrated in the previous post to find the Laplace transforms of the Bessel function of the first kind Jn(t) for nonnegative […]

2. so_stuck Says:

Dear Monday Math,

Your posts have just been the most wonderful resource. Thank you for showing every step – you have no idea how much help that has been.

I’ve been trying to solve a particular problem for a week now, and cannot come up with an answer. I was wondering if you could help me at all.

It is a second order differential equation, non-constant coefficients, and one of the differentials is squared. I’ve been trying to solve this using Laplace transforms.

For some x(t)…
x” = -[(x’^2)-2] / [2x]
And solve for x’ (not x!! we are to solve for dx/dt…) in terms of x, a, and b,
where x(0) = a and x'(0) = b.

The technique seems to be to define x’ = v(x) so that x” = v’v.
I came up with a half-baked solution: x’ = v = sqrt( [2x+C] / [x] ), but now I must get constant C in terms of a and b… and cannot figure out how.

Would you be able to help me?