**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.

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Tags: Differential Equation, Laplace Transform, Math, Monday Math

This entry was posted on April 26, 2010 at 12:06 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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May 3, 2010 at 4:24 am |

[…] 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 […]

June 10, 2010 at 3:39 am |

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?