Wolfram-Alfa liefert eine Schritt für Schritt
https://www.wolframalpha.com/input/?i=y%27%27-2.5y%27%2By%3De%5E%28-3x%29
Bei Wolfram-Alpha ist das allerdings schöner dargestellt als hier.
Solve ( d^2 y(x))/( dx^2)+y(x)-2.5 ( dy(x))/( dx) = e^{-3 x}:The general solution will be the sum of the complementary solution and particular solution.Find the complementary solution by solving ( d^2 y(x))/( dx^2)-2.5 ( dy(x))/( dx)+y(x) = 0:Assume a solution will be proportional to e^{lambda x} for some constant lambda.Substitute y(x) = e^{lambda x} into the differential equation:( d^2 )/( dx^2)(e^{lambda x})-2.5 ( d)/( dx)(e^{lambda x})+e^{lambda x} = 0Substitute ( d^2 )/( dx^2)(e^{lambda x}) = lambda^2 e^{lambda x} and ( d)/( dx)(e^{lambda x}) = lambda e^{lambda x}:1. lambda^2 e^{lambda x}-2.5 lambda e^{lambda x}+1. e^{lambda x} = 0Factor out e^{lambda x}:(1. lambda^2-2.5 lambda+1.) e^{lambda x} = 0Since e^{x lambda} !=0 for any finite lambda, the zeros must come from the polynomial:1. lambda^2-2.5 lambda+1. = 0Factor:1. (1. lambda-2.) (1. lambda-0.5) = 0Solve for lambda:lambda = 0.5 or lambda = 2.The root lambda = 0.5 gives y_1(x) = c_1 e^{0.5 x} as a solution, where c_1 is an arbitrary constant.The root lambda = 2. gives y_2(x) = c_2 e^{2. x} as a solution, where c_2 is an arbitrary constant.The general solution is the sum of the above solutions:y(x) = y_1(x)+y_2(x) = c_1 e^{0.5 x}+c_2 e^{2. x}Determine the particular solution to ( d^2 y(x))/( dx^2)+y(x)-2.5 ( dy(x))/( dx) = 0.+e^{-3 x} by the method of undetermined coefficients:The particular solution will be the sum of the particular solutions to ( d^2 y(x))/( dx^2)+y(x)-2.5 ( dy(x))/( dx) = 0. and ( d^2 y(x))/( dx^2)+y(x)-2.5 ( dy(x))/( dx) = e^{-3 x}.The particular solution to ( d^2 y(x))/( dx^2)+y(x)-2.5 ( dy(x))/( dx) = 0. is of the form:y_(p_1)(x) = a_1 x, where a_1 was multiplied by x to account for e^{0.5 x} and e^{2. x} in the complementary solution.The particular solution to ( d^2 y(x))/( dx^2)+y(x)-2.5 ( dy(x))/( dx) = e^{-3 x} is of the form:y_(p_2)(x) = a_2/e^{3 x}Sum y_(p_1)(x) and y_(p_2)(x) to obtain y_p(x):y_p(x) = y_(p_1)(x)+y_(p_2)(x) = a_1 x+a_2/e^{3 x}Solve for the unknown constants a_1 and a_2:Compute ( dy_p(x))/( dx):( dy_p(x))/( dx) = ( d)/( dx)(a_1 x+a_2/e^{3 x}) = a_1-(3 a_2)/e^{3 x}Compute ( d^2 y_p(x))/( dx^2):( d^2 y_p(x))/( dx^2) = ( d^2 )/( dx^2)(a_1 x+a_2/e^{3 x}) = (9 a_2)/e^{3 x}Substitute the particular solution y_p(x) into the differential equation:( d^2 y_p(x))/( dx^2)-2.5 ( dy_p(x))/( dx)+y_p(x) = e^{-3 x}+0.(9 a_2)/e^{3 x}-2.5 (a_1-(3 a_2)/e^{3 x})+(a_1 x+a_2/e^{3 x}) = e^{-3 x}+0.Simplify:-2.5 a_1+(17.5 a_2)/e^{3 x}+a_1 x = e^{-3 x}Equate the coefficients of 1 on both sides of the equation:-2.5 a_1 = 0.Equate the coefficients of e^{-3 x} on both sides of the equation:17.5 a_2 = 1Equate the coefficients of x on both sides of the equation:a_1 = 0Solve the system:a_1 = 0.a_2 = 0.0571429Substitute a_1 and a_2 into y_p(x) = a_1 x+a_2 e^{-3 x}:y_p(x) = 0.0571429/e^{3 x}+0.The general solution is:Answer: | | y(x) = y_c(x)+y_p(x) = 0.0571429/e^{3 x}+c_1 e^{0.5 x}+c_2 e^{2. x}+0.
