Superposition 1. Superposition I We saw the principle of superposition already, for ﬁrst order equations. For example, we saw that if y1 is a solution to y� + 4y = sin(3t) and y2 a solution to y� + 4y = 2, then y1 + y2 is a solution to y� + 4y = sin(3t) + 2. Superposition will be useful for us again, though now we will use it in two slightly different ways. The ﬁrst version we already used in a previous session, but let’s state it carefully and explicitly: Superposition I: If y1 and y2 are solutions of a homogeneous linear equa tion, then so is any linear combination; that is, for any constants c1 and c2 , the function y3 = c1 y1 + c2 y2 will also be a solution. Example. Consider the ODE t2 y�� + ty� − 4y = 0. This is homogeneous, since the constant term (the one not involving y or any of its derivatives) is zero. You can easily check by substitution that y1 (t) = t2 and y2 (t) = 1/t2 are both solutions. Thus y(t) = c1 t2 + c2 /t2 is a solution for any c1 and c2 . Notice that we didn’t need the differential equation to have constant coefﬁcients: linearity and homogeneity is enough. If the equation is of second order with two solutions y1 and y2 such that neither is a multiple of the other, then c1 y1 + c2 y2 will be the general solution. It has the right number of parameters. The restriction on the solutions is to make sure that they are really “different” solutions, for instance, in the above example, it would be incorrect to take y1 = t2 and y2 = 3t2 , and then claim that y(t) = c1 t2 + c2 · 3t2 = (c1 + 3c2 )t2 is the general solution. 18.03SC 2. Superposition II Now consider the linear second order equation mx �� + bx � + kx = Fext (t), (1) and its associated homogeneous equation mx �� + bx � + kx = 0. (2) Superposition II: Suppose x p is any solution to (1). If xh is any solution to (2), then x = x p + xh is again a solution to (1). This is similar to the way we used superposition for ﬁrst order equa tions. To prove this, we just need to substitute x into (1) and check that it really is a solution: mx �� + bx � + kx = m( xh + x p )�� + b( xh + x p )� + k ( xh + x p ) = (mxh�� + mx �p� ) + (bxh� + bx �p ) + (kxh + kx p ) = (mxh�� + bxh� + kxh ) + = + 0 (mx �p� + bx �p + kx p ) Fext . So indeed, it is a solution. An important fact: if xh is the general solution to (2) (so it should have two parameters) then x p + xh is the general solution to (1). We’ll see an example of this shortly. This proof works for linear equations of any order. For example, we already saw it as a consequence of the method of integrating factors for ﬁrst order equations. We’ve already seen how to ﬁnd the general solution to the associated homogeneous equation (2) using the characteristic equation. Thus to ﬁnd the general solution to (1), we simply need to do is ﬁnd a single solution to this particular equation. This is what we’ll discuss next.