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Math Help - DE exam review questions help

  1. #1
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    DE exam review questions help

    Help ! Have 2 problems on tomorrow's exam similar to these 2 and I haven't been able to get anywhere with them in 2 days ! Please show me the steps so I don't bomb ! Thanks !

    1)
    Find the solution to the system:

    x'-2x-y'-y=6*exp^(3t)
    2x'-3x+y'-3y=6*exp^(3t)

    satisfying the condition x(0)=3 and y(0)=0

    2)
    solve
    y'+4y=4t^2-4t+10
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  2. #2
    Super Member Matt Westwood's Avatar
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    No. 2 is straightforward enough, it's in the form y' + py = q where p and q are functions of t only (p in fact being a constant here).

    So it's an integrating factor job - multiply through by e^(integral of p with respect to t).

    Then the LHS is d/dt (ye^(that integral)) and then it's just a matter of integrating both sides w.r.t. t and the job's done.
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  3. #3
    Senior Member
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    Problem 1

    On number one,
    First, you'll want to solve the equations for x' and y'. So adding the two equations together, we get:
    3x'-5x-4y=12e^{3t}
    x'-\frac{5}{3}x-\frac{4}{3}y=4e^{3t}
    x'=\frac{5}{3}x+\frac{4}{3}y+4e^{3t}
    And if we subtract twice the first equation from the second, we find:
    3y'+x-y=-6e^{3t}
    y'+\frac{1}{3}x-\frac{1}{3}y=-2e^{3t}
    y'=-\frac{1}{3}x+\frac{1}{3}y-2e^{3t}

    If we define the vectors \mathbf{x}(t)=\begin{pmatrix}x(t)\\y(t)\end{pmatri  x}, \mathbf{b}(t)=\begin{pmatrix}4e^{3t}\\-2e^{3t}\end{pmatrix}, and the matrix \mathbf{A}=\begin{pmatrix}\frac{5}{3}&\frac{4}{3}\  \-\frac{1}{3}&\frac{1}{3}\end{pmatrix}, then we can write the two equations above as:
    \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t)+\mathbf{b}(  t), which is a first order inhomogeneous linear system, and the general solution is a particular solution plus the general solution of the corresponding homogeneous equation \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t)
    (see here).

    As our matrix A has constant coefficients, we can find the solution to the homogeneous equation by finding the eigenvalues and eigenvectors; if he have an eigenvector \mathbf{v} corresponding to the eigenvalue \lambda, then \mathbf{x}(t)=Ce^{\lambda{t}}\mathbf{v} is a solution to \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t).

    Now, we have
    \det(\mathbf{A}-\lambda\mathbf{I})=0
    \begin{vmatrix}\frac{5}{3}-\lambda&\frac{4}{3}\\-\frac{1}{3}&\frac{1}{3}-\lambda\end{vmatrix}=0
    \left(\lambda-\frac{5}{3}\right)\left(\lambda-\frac{1}{3}\right)+\frac{4}{9}=0
    \lambda^2-2\lambda+1=0
    (\lambda-1)^2=0
    \lambda=1 is a double root of this equation. Looking for eigenvalues, we want \begin{pmatrix}x\\y\end{pmatrix} such that \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\f  rac{5}{3}&\frac{4}{3}\\-\frac{1}{3}&\frac{1}{3}\end{pmatrix}\begin{pmatrix  }x\\y\end{pmatrix}, which when multiplied out, both rows give x=-2y, so \mathbf{v}_1=\begin{pmatrix}2\\-1\end{pmatrix} is an eigenvalue.
    This gives one solution \mathbf{x}_1(t)=C_1e^{t}\begin{pmatrix}2\\-1\end{pmatrix}. There is, however, a second, linearly independent solution (see here). We want to find a vector \mathbf{v}_2 such that \mathbf{v}_1+\mathbf{v}_2=A\mathbf{v}_2. Letting \mathbf{v}_2=\begin{pmatrix}x\\y\end{pmatrix}, we get:
    \begin{pmatrix}x+2\\y-1\end{pmatrix}=\begin{pmatrix}\frac{5}{3}&\frac{4}  {3}\\-\frac{1}{3}&\frac{1}{3}\end{pmatrix}\begin{pmatrix  }x\\y\end{pmatrix}, which multiplies out to give x=-2y+3, so \mathbf{v}_2=\begin{pmatrix}1\\1\end{pmatrix} is one such vector. From this (see here), we have the second solution
    \mathbf{x}_2(t)=C_2e^{t}(t\mathbf{v}_1+\mathbf{v}_  2)
    \mathbf{x}_2(t)=C_2e^{t}\begin{pmatrix}2t+1\\1-t\end{pmatrix},
    and the general solution to the homegeneous equation is:
    \mathbf{x}_h(t)=C_1e^{t}\begin{pmatrix}2\\-1\end{pmatrix}+C_2e^{t}\begin{pmatrix}2t+1\\1-t\end{pmatrix}.

    Now, to find a specific solution to the inhomogeneous equation, try a solution of the form
    \mathbf{x}(t)=e^{3t}\begin{pmatrix}a\\b\end{pmatri  x}. Substituting into our equation \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t)+\mathbf{b}(  t), this gives:
    3e^{3t}\begin{pmatrix}a\\b\end{pmatrix}=e^{3t}\beg  in{pmatrix}\frac{5}{3}&\frac{4}{3}\\-\frac{1}{3}&\frac{1}{3}\end{pmatrix}\begin{pmatrix  }a\\b\end{pmatrix}+\begin{pmatrix}4e^{3t}\\-2e^{3t}\end{pmatrix}
    Multiplying the matrix, adding on the right hand side, and factoring e^{3t} from both sides,
    \begin{pmatrix}3a\\3b\end{pmatrix}=\begin{pmatrix}  \frac{5}{3}a+\frac{4}{3}b+4\\-\frac{1}{3}a+\frac{1}{3}b-2\end{pmatrix},
    so we get a-b=3 and a+8b=-6, which has solution a=2, b=-1. Thus, we have particular solution \mathbf{x}_p(t)=e^{3t}\begin{pmatrix}2\\-1\end{pmatrix}.

    Add this to the homogeneous solution, and we have:
    \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=C_1e^{t}\be  gin{pmatrix}2\\-1\end{pmatrix}+C_2e^{t}\begin{pmatrix}2t+1\\1-t\end{pmatrix}+e^{3t}\begin{pmatrix}2\\-1\end{pmatrix}. Our initial condition x(0)=3 and y(0)=0 gives us:
    \begin{pmatrix}3\\{0}\end{pmatrix}=C_1\begin{pmatr  ix}2\\-1\end{pmatrix}+C_2\begin{pmatrix}1\\1\end{pmatrix}  +\begin{pmatrix}2\\-1\end{pmatrix}, or
    2C_1+C_2=1 and C_2-C_1=1,
    so C_1=0, C_2=1, giving:
    \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=e^{t}\begin  {pmatrix}2t+1\\1-t\end{pmatrix}+e^{3t}\begin{pmatrix}2\\-1\end{pmatrix}
    Or separating,
    x(t)=(2t+1)e^t+2e^{3t}
    y(t)=(1-t)e^t-e^{3t}.

    --Kevin C.
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