From Department of Mathematics at UTSA
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| − | With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.
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| − | Examples:
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| − | * <math> y' = 2x </math>, <math> y(2) = 0 </math>. With this point and the general solution <math>y = x^2 + C</math>, we can calculate the constant C to be -4. Thus the particular solution is <math>y = x^2 - 4</math>.
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| − | * <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>.
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| − | * <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.
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Revision as of 20:07, 17 September 2021
