Find the general solution to the differential equation y'' + 4y' + 3y = 6e^(2x) [where y' is dy/dx and y'' is d^2 y/ dx^2]

First find the general solution to the differential equation y'' + 4y' + 3y = 0 as an arbitrary number of the solution to this differential equation can be added to the solution of the differential equation in the question. Equations of this form have the solution y = Ae^(mx) (where A is an arbitrary constant) so y ' = my and y'' = (m^2)y. Cancelling y this gives m^2 + 4m + 3 = 0, solved by m = -3, -1. The solution of this is equation (the complementary function) is y = Ae^(-3x) + Be^(-x) [where A and B are arbitrary constants]Then find the particular integral, the solution to the differential equation in the question. This is found by trying a solution of the form of the right hand side of the equation, y = ce^(2x). This gives y' = 2y, y'' = 4y. Substituting y, y' and y'' in and cancelling y gives 15c = 6, so c = 0.4. The general solution is the sum of the particular integral and the complementary function, y = Ae^(-3x) + Be^(-x) + 0.4e^(2x)

ES
Answered by Elliot S. Further Mathematics tutor

4519 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Integrate ln(x) with respect to x.


The cubic equation 27(z^3) + k(z^2) + 4 = 0 has roots α, β and γ. In the case where β=γ, find the roots of the equation and determine the value of k


Find the general solution of y'' - 3y' + 2y = 2e^x


Using your knowledge of complex numbers, such as De Moivre's and Euler's formulae, verify the trigonometric identities for the double angle.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences