Consider the DEd²y dydx²which is linear with constant coefficients.8 + 16y = xdxFirst we will work on solving the correspondinghomogeneous equation. The auxiliary equation (using mas your variable) is= 0which has rootBecause this is a repeated root, we don't have muchchoice but to use the exponential functioncorresponding to this root:reduction of order.Y2 =ue4xThen (using the prime notation for the derivatives)Y/₂YSo, plugging y2 into the left side of the differentialequation, and reducing, we gety2-8y2 + 16y2 =U =So now our equation is e4u" = x. To solve for u weneed only integrate ce-4 twice, using a as our firstconstant of integration and b as the second we getTherefore y2 =general solution.to dotheWe knew from the beginning that e4 was a solution. Wehave worked out is that xe4 is another solution to thehomogeneous equation, which is generally the casewhen we have multiple roots. Then 2+4 is the particular64solution to the nonhomogeneous equation, and thegeneral solution we derived is pieced together usingsuperposition.

Given: The linear differential equation with constant coefficients is given by,…

Consider the DE which is linear with constant coefficients. d²y dx² First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is = 0 which has root Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: 3/2 dy 8 + 16y = x dx reduction of order. Y2 = ue4x Then (using the prime notation for the derivatives) = y₁ = So, plugging y2 into the left side of the differential equation, and reducing, we get y8y2 + 16y2 = U= So now our equation is exu" =x. To solve for u we need only integrate xe twice, using a as our first constant of integration and b as the second we get -4x Therefore y2 = general solution. to do the We knew from the beginning that e4 was a solution. We have worked out is that xe4 is another solution to the homogeneous equation, which is generally the case 2+4x when we have multiple roots. Then 24 is the particular 64 solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.

Consider the DE which is linear with constant coefficients. d²y dx² First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is = 0 which has root Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: 3/2 dy 8 + 16y = x dx reduction of order. Y2 = ue4x Then (using the prime notation for the derivatives) = y₁ = So, plugging y2 into the left side of the differential equation, and reducing, we get y8y2 + 16y2 = U= So now our equation is exu" =x. To solve for u we need only integrate xe twice, using a as our first constant of integration and b as the second we get -4x Therefore y2 = general solution. to do the We knew from the beginning that e4 was a solution. We have worked out is that xe4 is another solution to the homogeneous equation, which is generally the case 2+4x when we have multiple roots. Then 24 is the particular 64 solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter7: Exponents And Exponential Functions

Section7.8: Transforming Exponential Expressions

Problem 13PFA

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