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Math Help - second order non-homogenous IVP

  1. #1
    Member Jones's Avatar
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    second order non-homogenous IVP

    Hi,

    Im having some trouble with an inhomogenous differential equation.

    I have:
    y'' + 3y' + 2y = sin(x)
     <br />
 \begin{cases}<br />
 y(0) = 1  \\<br />
 y'(0) = 1 <br />
\end{cases}<br />

    The solution to the homogenous equation r^2 + 3r + 2=0

    are \begin{aligned}r_{1} = -1 \\<br />
r_{2} = -2\end{aligned}

    And the general solution would therefore be: C_{1}e^{-2t} +C_{2}e^{-t}
    But since this whole equation is equal to sin x
    The solution is on the form A sin(x) + B cos(x)
    So let's try it:
     \begin{aligned}<br />
 y = A sin(x) + B cos(x)\\ <br />
 y'= A cos(x) -B sin(x) \\ <br />
 y''= -A sin(x) -B cos(x)\end{aligned}<br />

    Gives:

    -A sin(x) -B cos(x) + 3Acos(x) -3Bsin(x) +2Asin(x) +2Bcos(x) = 4Acos(x) -2Bsin(x) = sin(x)

    \begin{aligned} 4A = 0\\<br />
  A = 0 \\ <br />
-2B = 1\\<br />
  B = -\frac{1}{2}\end{aligned}

    This is the point where im stuck.
    The initial conditions give Both \begin{aligned}y(0)=Asin(x)+Bcos(x)~~ B=1\\<br />
y'(0) = Acos(x)-Bsin(x) ~~ A = 1 \end{aligned}

    This is where im stuck
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Conway AR
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    Quote Originally Posted by Jones View Post
    Hi,

    Im having some trouble with an inhomogenous differential equation.

    I have:
    y'' + 3y' + 2y = sin(x)
     <br />
\begin{cases}<br />
y(0) = 1 \\<br />
y'(0) = 1 <br />
\end{cases}<br />

    The solution to the homogenous equation r^2 + 3r + 2=0

    are \begin{aligned}r_{1} = -1 \\<br />
r_{2} = -2\end{aligned}

    And the general solution would therefore be: C_{1}e^{-2t} +C_{2}e^{-t}
    But since this whole equation is equal to sin x
    The solution is on the form A sin(x) + B cos(x)
    So let's try it:
     \begin{aligned}<br />
y = A sin(x) + B cos(x)\\ <br />
y'= A cos(x) -B sin(x) \\ <br />
y''= -A sin(x) -B cos(x)\end{aligned}<br />

    Gives:

    -A sin(x) -B cos(x) + 3Acos(x) -3Bsin(x) +2Asin(x) +2Bcos(x) = 4Acos(x) -2Bsin(x) = sin(x)

    \begin{aligned} 4A = 0\\<br />
A = 0 \\ <br />
-2B = 1\\<br />
B = -\frac{1}{2}\end{aligned}

    This is the point where im stuck.
    The initial conditions give Both \begin{aligned}y(0)=Asin(x)+Bcos(x)~~ B=1\\<br />
y'(0) = Acos(x)-Bsin(x) ~~ A = 1 \end{aligned}

    This is where im stuck
    Check on your particular solution, I got something different. So if the solution is

     <br />
y(x) = c_1 e^{-x} + c_2 e^{-2x} + \frac{1}{10} \sin x - \frac{3}{10} \cos x\;\;\;(*)<br />

    then use your initial conditions on (*).
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  3. #3
    Newbie
    Joined
    Jul 2009
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    Quote Originally Posted by Jones View Post
    Hi,

    Im having some trouble with an inhomogenous differential equation.

    I have:
    y'' + 3y' + 2y = sin(x)
     <br />
 \begin{cases}<br />
 y(0) = 1  \\<br />
 y'(0) = 1 <br />
\end{cases}<br />

    The solution to the homogenous equation r^2 + 3r + 2=0

    are \begin{aligned}r_{1} = -1 \\<br />
r_{2} = -2\end{aligned}

    And the general solution would therefore be: C_{1}e^{-2t} +C_{2}e^{-t}
    But since this whole equation is equal to sin x
    The solution is on the form A sin(x) + B cos(x)
    So let's try it:
     \begin{aligned}<br />
 y = A sin(x) + B cos(x)\\ <br />
 y'= A cos(x) -B sin(x) \\ <br />
 y''= -A sin(x) -B cos(x)\end{aligned}<br />

    Gives:

    -A sin(x) -B cos(x) + 3Acos(x) -3Bsin(x) +2Asin(x) +2Bcos(x) = 4Acos(x) -2Bsin(x) = sin(x)
    a)

    \begin{aligned} 4A = 0\\<br />
  A = 0 \\ <br />
-2B = 1\\<br />
  B = -\frac{1}{2}\end{aligned}

    This is the point where im stuck.
    The initial conditions give Both
    b)
    \begin{aligned}y(0)=Asin(x)+Bcos(x)~~ B=1\\<br />
y'(0) = Acos(x)-Bsin(x) ~~ A = 1 \end{aligned}

    This is where im stuck
    i know where you got stucked

    a) it is when you summed up, you should get
    = (-B+3A+2B)cos(x) +(-A-3B+2A)sin(x) = sin(x)

    b) to solve the initial value problem which is the particular solustion, you should include the general solution Ce-.... and the sin(x) ... al together
    and substitute the initial value
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  4. #4
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
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    Thanks!
    Last edited by mr fantastic; July 23rd 2009 at 06:42 PM. Reason: Moved new question to new thread.
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