Stuck trying to integrate dy/dx-y=cos(x)-2

  1. 1. The problem statement, all variables and given/known data

    integrate

    dy/dx-y=cos(x)-2

    2. Relevant equations
    3. The attempt at a solution

    dy/dx-y=cos(x)-2

    is in the form

    dy/dx+p(x)=q(x)

    take the I.F as e^int(p(x))dx=e^-x

    multiplying throughout by e^-x

    d(e^-x)z/dx=-(cos(x)e^-x)-(2e^-x)

    (e^-x)z=-int((cos(x)e^-x)-(2e^-x))dx

    so -int((cos(x)e^-x)-(2e^-x))dx=-int(cos(x)(e^-x)dx-2int(e^-x)dx

    2int(e^-x)dx=-2(e^-x)+c

    -int(cos(x)(e^-x)) this is where i have trouble im not sure if my method is correct using integration by parts multiple times

    1st integration
    let u=e^(-x) du/dx=-e^(-x) dv/dx=cos(x) v=-sin(x)

    -int(cos(x)(e^-x))dx = -e^(-x)sin(x)+int(-sin(x)e^(-x))dx

    2nd integration
    int(-sin(x)e^(-x))dx
    let u=e^(-x) du/dx=-e^(-x) dv/dx=-sin(x) v=cos(x)

    e^(-x)cos(x)+int(cos(x)(e^-x))dx

    so

    -int(cos(x)(e^-x)dx=-e^(-x)sin(x)+e^(-x)cos(x)+int(cos(x)(e^-x))dx

    this is where im not sure i take away from both sides int(cos(x)(e^-x))dx giving

    -2int(cos(x)(e^-x)dx=-e^(-x)sin(x)+e^(-x)cos(x)

    int(cos(x)(e^-x)dx=e^(-x)/2(sin(x)+cos(x))

    so
    (e^-x)z=-2(e^-x)+c+e^(-x)/2(sin(x)+cos(x))+c

    z=-2+(sin(x)+cos(x))/2+c/(e^-x)

    1/y=-2+(sin(x)+cos(x))/2+c/(e^-x)

    y=1/(-2+(sin(x)+cos(x))/2+c/(e^-x))

    thank you for your time to recap my problems is:

    im not sure if i am allowed to add integrals

    cheers

    dooogle
     
  2. jcsd
  3. fzero

    fzero 2,948
    Science Advisor
    Homework Helper
    Gold Member

    It's easier to do that integral by using the Euler formula

    [tex]e^{\pm ix} = \cos x \pm i \sin x[/tex]

    to obtain an expression for [tex]\cos x[/tex] in terms of exponentials. However, your integration is correct if you mean

    int(cos(x)(e^-x)dx= (e^(-x)/2 ) (sin(x)+cos(x)) +c
     
  4. HallsofIvy

    HallsofIvy 40,943
    Staff Emeritus
    Science Advisor

    Sorry but the use of two "/"s in
    is meaning less to me.
     
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