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Abstract:
This lecture (Vorlesung 13) by Prof. Karl-Heinz Lotze from the Friedrich-Schiller-Universität Jena introduces the "Variation of Constants" method as a second technique for finding a particular solution to inhomogeneous linear second-order ordinary differential equations with constant coefficients. The lecture systematically derives the formulas for the functions u'(x) and v'(x) used in the ansatz y_p(x) = u(x)y_1(x) + v(x)y_2(x), where y_1 and y_2 are fundamental solutions to the corresponding homogeneous equation. The method involves setting up a system of two algebraic equations for u' and v', solving them, and then integrating to find u and v. The lecture demonstrates the application of this method using the same example differential equation (y'' + y' - 2y = e^(2x)) that was solved in the previous lecture using a different technique, arriving at the same particular solution y_p = (1/4)e^(2x).
Summary of "Variation of Constants" for Solving Linear Second-Order ODEs
y_h = C_1y_1 + C_2y_2) and a particular solution (y_p) of the inhomogeneous equation. The goal is to find methods for determining y_p.y_p.y_1(x) and y_2(x) for the homogeneous equation ay'' + by' + cy = 0. The homogeneous solution is y_h(x) = C_1y_1(x) + C_2y_2(x).C_1 and C_2 by replacing them with functions u(x) and v(x). The ansatz for the particular solution is y_p(x) = u(x)y_1(x) + v(x)y_2(x).y_p and its derivatives y_p' and y_p'' are substituted into the original inhomogeneous differential equation ay'' + by' + cy = f(x).u' and v', an auxiliary condition is imposed: u'(x)y_1(x) + v'(x)y_2(x) = 0. This choice simplifies the expression for y_p'.u'(x) and v'(x):
u'(x)y_1(x) + v'(x)y_2(x) = 0u'(x)y_1'(x) + v'(x)y_2'(x) = f(x)/au'(x) and v'(x), yielding:
u'(x) = - (1/a) * f(x) * y_2(x) / W(y_1, y_2)v'(x) = (1/a) * f(x) * y_1(x) / W(y_1, y_2)W(y_1, y_2) = y_1y_2' - y_2y_1' is the Wronskian determinant of the fundamental solutions.u(x) and v(x) are found by integrating u'(x) and v'(x) respectively. Integration constants are typically omitted as we only need a particular solution.y'' + y' - 2y = e^(2x).
y_1 = e^x, y_2 = e^(-2x). (Here a=1, b=1, c=-2)y_1' = e^x, y_2' = -2e^(-2x).W(y_1, y_2) = -3e^(-x).f(x) = e^(2x).u'(x) = - (1/1) * e^(2x) * e^(-2x) / (-3e^(-x)) = (1/3)e^x.v'(x) = (1/1) * e^(2x) * e^x / (-3e^(-x)) = -(1/3)e^(4x).u(x) = (1/3)e^x.v(x) = -(1/12)e^(4x).y_p = u(x)y_1(x) + v(x)y_2(x) = (1/3)e^x * e^x + (-1/12)e^(4x) * e^(-2x)y_p = (1/3)e^(2x) - (1/12)e^(2x) = (1/4)e^(2x).y(x) = C_1e^x + C_2e^(-2x) + (1/4)e^(2x), which matches the result obtained by the previous method.f(x).