1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Introductuin to differential equations

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    use the concept that y = c, -inf < x < inf is a constant function if and only if y' = 0 to determine whether the specified differential equation has any constant solutions:

    y'' + 4y' + 6y = 10

    3. The attempt at a solution

    What throws me off in this particular problem is y''. If y' = 0 then y'' = 0 as well.
    Only now when I integrate y'' to y' its y' = c1 so c1 = 0 and y' still = 0.

    then I basically substitute my results in the equation to get:

    (0) + 4(0) + 6(c) = 10

    6c = 10

    [c = 10/6, c1 = 0] are the constant solutions to the equation.

    Does this make sense?

  2. jcsd
  3. Oct 18, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    You are thinking too hard. You don't have to integrate anything. Set y''=y'=0 and solve the equation for y. y=0 is NOT a solution.
  4. Oct 18, 2007 #3
    ahh.. so

    y = (10 - 4y' - y'')/6

    y = (10 - 4(0) - 0)/6

    [y = 10/6] ?

    Thanks for the help! Please let me know if this is right.
  5. Oct 18, 2007 #4
    The great thing about ODE's is that you can usually quite readily check the answer by substituting your solution back into the differential equation. Give it a try and you'll find that your solution is correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Introductuin differential equations Date
Diffrential Equations Today at 2:02 AM
System of Differential Equations, Phase Plane Monday at 4:33 PM
Method of undetermined coefficients -- Help please Saturday at 5:04 PM
Substitution in a differential equation, independent variable Saturday at 7:41 AM
Prove that this function is holomorphic Mar 16, 2018