Given a second order Differential Equation, how does one derive the Characteristic equation where one can evaluate and find the constants

Given a ODE of the 2nd order, Ay''+by'+cy = 0, we assume the general solution of the exponential form y=e^(mx).As we will see this leads to an easy simplification due to the properties of the exponential . From this we substitute in and we get Am^(2)(e^mx) +bm(e^mx) + c(e^mx) = 0 here we have a like term of e^mx and thus can be eliminated leaving a quadratic of the form Am^2 + Bm + C = 0 where for a particular ODE we can solve quadratically and will have two values of m for a well-defined solution of the ODE.

WP
Answered by William P. Maths tutor

2922 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The line AB has equation 5x+3y+3=0. The line AB is parallel to the line with equation y=mx+7 . Find the value of m.


Find the general solution, in degrees, of the equation 2sin(3x+45°)=1. Use your general solution to find the solution of 2sin(3x+45°)=1 that is closest to 200 °.


By consdering partial fractions find the integral of (1-x)/(5x-6-x^2) between x = 1 and x = 0, give your answer in an exact form.


Let X be a normally distributed random variable with mean 20 and standard deviation 6. Find: a) P(X < 27); and b) the value of x such that P(X < x) = 0.3015.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning