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.

Related Further Mathematics GCSE answers

All answers ▸

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).


A curve has equation: y = x^3 - 3x^2 + 5. Show that the curve has a minimum point when x = 2.


Can you explain rationalising surds?


Find the coordinates of the minimum/maximum of the curve: Y = 8X - 2X^2 - 9, and determine whether it is a maximum or a minimum.


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 2013–2024

Terms & Conditions|Privacy Policy