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Thread: DE exam review questions help

  1. #1
    Jul 2008

    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 !

    Find the solution to the system:


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

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  2. #2
    MHF Contributor Matt Westwood's Avatar
    Jul 2008
    Reading, UK
    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
    Dec 2007
    Anchorage, AK

    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:
    $\displaystyle 3x'-5x-4y=12e^{3t}$
    $\displaystyle x'-\frac{5}{3}x-\frac{4}{3}y=4e^{3t}$
    $\displaystyle x'=\frac{5}{3}x+\frac{4}{3}y+4e^{3t}$
    And if we subtract twice the first equation from the second, we find:
    $\displaystyle 3y'+x-y=-6e^{3t}$
    $\displaystyle y'+\frac{1}{3}x-\frac{1}{3}y=-2e^{3t}$
    $\displaystyle y'=-\frac{1}{3}x+\frac{1}{3}y-2e^{3t}$

    If we define the vectors $\displaystyle \mathbf{x}(t)=\begin{pmatrix}x(t)\\y(t)\end{pmatri x}$, $\displaystyle \mathbf{b}(t)=\begin{pmatrix}4e^{3t}\\-2e^{3t}\end{pmatrix}$, and the matrix $\displaystyle \mathbf{A}=\begin{pmatrix}\frac{5}{3}&\frac{4}{3}\ \-\frac{1}{3}&\frac{1}{3}\end{pmatrix}$, then we can write the two equations above as:
    $\displaystyle \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t)+\mathbf{b}( t)$, 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 $\displaystyle \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t)$
    (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 $\displaystyle \mathbf{v}$ corresponding to the eigenvalue $\displaystyle \lambda$, then $\displaystyle \mathbf{x}(t)=Ce^{\lambda{t}}\mathbf{v}$ is a solution to $\displaystyle \mathbf{x}'(t)=\mathbf{A}\mathbf{x}(t)$.

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

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

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

    --Kevin C.
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