Let's review the outline of how to use the Laplace transform to solve differential equations. We have an ODE. For us, it will always have constant coefficients. We use the Laplace transform on the entire equation. And we get an algebraic equation for the transforms of x and f. Now that it's just algebra, we can easily solve for the transform of the solution that we want. Then we take the inverse Laplace transform, which is the hard step to get the solution to the original ODE. Here's our table of transforms. There's a new entries here. Everything starts with taking the transforms of derivatives. Then once we've solved for capital X, we have to match terms in capital X to these things. Over here are the standard forms that we then know how to take the inverse of. For our first example, let's find a particular solution of this equation. When I transform both sides, x double prime gives me S squared times capital X minus S times x at 0 minus x prime at 0. My transform x, I just get capital X. When I transform e to the 40, I get 1 over S minus 4. We're only seeking a particular solution. We'll take things as easy as we can on ourselves by setting the initial values to 0. Then it's easy to solve for capital X. Both terms on the left, multiply it. So all we have to do is divide both sides by S squared minus 1. Our greatest weapon. In finding an inverse Laplace transform is partial fraction decomposition. I need to factor the denominator. And then we can write out what the terms must be and the decomposition. If we clear the denominators on both sides. So we multiply through by that denominator on the left. We have two choices. We can expand the write into a quadratic polynomial and set two polynomials equal to each other. Or we can just choose each of those routes in turn. And we get simple equations. So S equals 4. The a term survives, but the other two terms are 0. So that tells us a. If we choose x equals 1, then the a term drops out. B survives, and C also drops out. So that tells us p. And finally, if we let S be the third root in the denominator, a and B will drop out. And we'll be able to find c. Now that we have values for a, b, and c, we have the Partial Fraction Decomposition. Each term in this decomposition is easy to invert. In fact, each one leads to an exponential function. 1 over S minus 4 gives us e to the 40. To 1 over S minus one gives us e to the t. And 1 over S minus one gives us e to the negative t. This is a particular solution of the original ODE. I want to point out though that if we looked at the characteristic polynomial, it has roots one and negative one. And those show up in our particular solution. This isn't like the method of undetermined coefficients where we ignore that part because we chose our initial conditions in a particular way. For general second-order linear equation with constant coefficients will transform both sides. And again, if we're looking for our particular solution and we don't care which one we get. Then we should choose the one where the initial values are 0. Everything left and the left multiplies capital X. We use a P now to remind us this particular solution. And it has this form F of S divided by this polynomial ins. And if you look back, this is exactly what we call it the characteristic polynomial. Each of those characteristic roots is a place where the transform of capital X actually goes to infinity. In the Laplace context, we call them poles. Each pole corresponds to an exponential solution, right? One over S minus lambda gives us e to the lambda t. Here's our next example. We get the transform of the forcing function divided by the characteristic polynomial. Now we have to transform a cosine function. The transform of cosine Omega t is S over S squared plus Omega squared. So here that gives us S squared plus 4. To use partial fractions on this, I'm going to use the factorization into real quadratics because if I factored further, they'd be complex numbers. We could do it that way, but the arithmetic isn't any easier. Now be clear denominators in this equation. And this time we'll expand the right-hand side as a polynomial and S. We wanted to do it the way as in the last example. We could use imaginary poles, thigh and to why it would work out. But again, it's not really an easier than doing this. Now I want to equate this cubic on the right side to the 10 asks that we have on the left, which means that the coefficient of x cubed is 0. The coefficient of x squared is 0. The coefficient of S is ten, and the coefficient of one is 0. So I have four linear equations for four unknowns. Fortunately, they come in pairs of two. So I can solve those separately. And what we find is that a is 2, c is negative two, and b and d are both 0. Now we're ready to invert this expression to S over S squared plus four. And negative 2, S over S squared plus nine. Well, S over S squared plus 4 is cosine of two t and S over S squared plus 9 cosine of three t. There's our solution.
V.1-2 Laplace for 2nd order problems
From Tobin Driscoll March 31, 2021
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