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Y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled out?

  1. Jul 9, 2008 #1
    1. The problem statement, all variables and given/known data
    I had to solve the 2nd order d.e
    y'' + 4y' + 5y=0
    Which I have done, then I need to find a solution for which y(0)=1
    and y'(0)=0



    3. The attempt at a solution

    My general soltuion for the d.e is y= e^(-2x) (c_1 *cos(x) + c_2*sin(x))

    so for y(0)=1= e^0 (c_1 * 1 + 0)
    so I end up with c_1=1 but I dont have an answer for c_2, I assume I just can't write c_2 =0? SO that is my first question. My second is how do I solve y'(0)=0 ... it might sound like a silly question but there is no examples in my text and I'm not sure, should I just do this

    y= e^(-2x) (c_1 *cos(x) + c_2*sin(x))
    y'= -e^(-2x)(c_2*cos(x) + 3*c_1*sin(x))
    y'(0)=0=-e^0 * (c_2*cos(0) + 0)
    0= c_2

    So If I'm even on the right track, this means the constant(s) equal zero?
     
  2. jcsd
  3. Jul 9, 2008 #2
    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    If you insert zero in x for your solution it comes down to a c_1 cos(0) = 1 so we get....

    If you derive your solution w.r.t. x and you again fill in zero you'll get: -2* c_2*cos(0) = 0 so we get...
     
  4. Jul 9, 2008 #3
    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    yes constant may out to be zero depend on conditions.

    your procedure was completely correct..
     
  5. Jul 9, 2008 #4

    Dick

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    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    Your derivative of y is way off. You need to use the product rule. If you correct that you'll find you don't get c2=0.
     
  6. Jul 10, 2008 #5
    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    Okay, I've re-done the derivative and ended up with

    y' = e^(-2x)(C_2*cos(x) - C_1*sin(x)) - 2e^(-2x)*(C_1*cos(x) + C_2*sin(x))
    so I had to find the solution for which y'(0)=0
    y'(0)=0=e^0(C_2) - 2e^0(C1)
    0=C_2 - 2C1
    so

    2C_1 = C_2

    Firstly, is that correct working for y' and if so, how do I answer the question when I end up with two variables? do I just sub one of them into the equation so they all have the same C for example

    y'=e^(-2x)(2C_1*cos(x) - C_1*sin(x)) - 2e^(-2x)*(C_1*cos(x) + 2C_1*sin(x))

    Would that be an acceptable answer?
     
  7. Jul 10, 2008 #6
    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    y(0) = 1 gives you the value of C_1 =1 .
     
  8. Jul 10, 2008 #7
    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    Does that mean I can simply plug that into my y' that would give me....


    y'=e^(-2x)(2cos(x) - sin(x)) - 2e^(-2x)*(cos(x) + 2sin(x))

    Can you just do that, I didn't know I could use a value I got from y(0) into y'
     
  9. Jul 10, 2008 #8
    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    In post 5 you correctly determined 2C_1 =C_2 and you know C_1....hence your y is solved for the constants. I don't know why you are struggling with y'.

    Just determine your constants and then you're done, m'kay?
     
  10. Jul 10, 2008 #9

    Dick

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    Homework Helper

    Re: y=e^(-2x)*(c1*cos(x) + c2*sin(x)) how do I solve y(0)=1 when c2 gets cancelled ou

    The conditions y(0)=1 and y'(0)=0 have to both hold. It's two equations in the two unknowns C1 and C2. So of course you can substitute one into the other.
     
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