Linear Independence and the Wronskian Recall from linear algebra that two vectors v and w are called linearly dependent if there are nonzero constants c_{1} and c_{2} with. c_{1}v + c_{2}w = 0 We can think of differentiable functions f(t) and g(t) as being vectors in the vector space of differentiable functions. The analogous definition is Let f(t) and g(t) be differentiable functions. Then they are called linearly dependent if there are nonzero constants c_{1} and c_{2} with c_{1}f(t) + c_{2}g(t) = 0 for all t. Otherwise they are called linearly independent.
Example The functions f(t) = 2sin^{2 }t and g(t) = 1  cos^{2}(t) are linearly dependent since (1)(2sin^{2 }t) + (2)(1  cos^{2}(t)) = 0
Example The functions f(t) = t and g(t) = t^{2} are linearly independent since otherwise there would be nonzero constants c_{1} and c_{2} such that c_{1}t + c_{2}t^{2} = 0 for all t. First let t = 1. Then c_{1} + c_{2} = 0 Now let t = 2. Then 2c_{1} + 4c_{2} = 0 This is a system of 2 equations and two unknowns. The determinant of the corresponding matrix is 4  2 = 2 Since the determinant is nonzero, the only solution is the trivial solution. That is c_{1} = c_{2} = 0 The two functions are linearly independent.
In the above example, we arbitrarily selected two values for t. It turns out that there is a systematic way to check for linear dependence. The following theorem states this way.
Example Show that the functions f(t) = t and g(t) = e^{2t} are linearly independent. Solution We compute the Wronskian. f '(t) = 1 g '(t) = 2e^{2t} The Wronskian is (t)(2e^{2t})  (e^{2t})(1) Now plug in 0 to get W(f,g)(0) = 1 which is nonzero. We can conclude that f and g are linearly independent.
Proof If c_{1}f(t) + c_{2}g(t) = 0 Then we can take derivatives of both sides to get c_{1}f '(t) + c_{2}g '(t) = 0 This is a system of two equations with two unknowns. The determinant of the corresponding matrix is the Wronskian. Hence, if the Wronskian is nonzero at some t_{0}, only the trivial solution exists. Hence they are linearly independent. There is a fascinating relationship between second order linear differential equations and the Wronskian. This relationship is stated below.
Proof First the Wronskian W = y_{1}y_{2}'  y_{1}'y_{2} has derivative W' = y_{1}'y_{2}' + y_{1}y_{2}''  y_{1}''y_{2}  y_{1}'y_{2}' = y_{1}y_{2}''  y_{1}''y_{2} Since y_{1} and y_{2} are solutions to the differential equation, we have y_{1}'' + p(t)y_{1}' + q(t)y_{1} = 0 y_{2}'' + p(t)y_{2}' + q(t)y_{2} = 0 Multiplying the first equation by y_{2} and the second by y_{1} and adding gives (y_{1}y_{2}''  y_{1}''y_{2}) + p(t)(y_{1}y_{2}'  y_{1}'y_{2}) = 0 This can be written as W' + p(t)W = 0 This is a separable differential equation with dW/W = p(t)dt Now integrate and Abel's theorem appears. Example Find the Wronskian (up to a constant) of the differential equations y'' + cos(t) y = 0 Solution We just use Abel's theorem, the integral of cos t is sin t hence the Wronskian is W(t) = ce^{sin t}
A corollary of Abel's theorem is the following Corollary Let y_{1} and y_{2} be solutions to the differential equation L(y) = y'' + p(t)y' + q(t)y = 0 Then either W(y_{1},y_{2}) is zero for all t or never zero.
Example Prove that y_{1}(t) = 1  t and y_{2}(t) = t^{3} cannot both be solutions to a differential equation y'' + p(t)y + q(t) = 0 for p(t) and q(t) continuous on [1,5]. Solution We compute the Wronskian y_{1}' = 1 y_{2}' = 3t^{2} W(y_{1},y_{2}) = (1  t)(3t^{2})  (t^{3})(1) = 3t^{2}  2t^{3} Notice that the Wronskian is zero at t = 0 but nonzero at t = 1. By the above corollary, y_{1} and y_{2} cannot both be solutions.
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