How would I solve the following equation d^2x/dt^2 + 5dx/dt + 6x = 0

Our given equation is d2x/dt2 + 5dx/dt + 6x = 0, which we need to recognise as a second order differential equation. Therefore we need to begin by solving the auxilary funtion m2+5m +6= 0. ( Side note: Most of the mathematical equations we solve are expressed in x and y, but in this equation it's expressed in terms of x and t, where x is the dependent variable). Solving the auxiliary funtion gives us values of -3&-2 for m. Because these are real values that are not equal to each other we can use the complimentary funtion y= Aect + Bedt where y is the dependent variable, t is our independent variable and A&B are constants of intergration. If we plug in our values the auxiliary funtion becaomes x = Ae-3t+Be-2t. Which is our final answer.

DM
Answered by Dimuthu M. Further Mathematics tutor

5657 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

write showing all working the following algebraic expression as a single fraction in its simplest form: 4-[(x+3)/ ((x^2 +5x +6)/(x-2))]


What is differentiation used for?


A circle has equation x^{2}-8x+y^{2}-6y=d. A line is tangent to this circle and passes through points A and B, (0,17) and (17,0) respectively. Find the radius of the circle.


x^3 + 2x^2 - 9x - 18 = (x^2 - a^2)(x + b) where a,b are integers. Work out the three linear factors of x^3 + 2x^2 - 9x - 18. (Note: x^3 indicates x cubed and x^2 indicates x squared).


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