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3.4

  • Page ID
    156609
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    Next lets look at what happens if \(b^2-4 a c=0\), then \(r_1=r_2\). Hence one solution is \(y=e^{r_1 t}\) Need second solution. If \(y=e^{r t}\) is a solution, \(y=c e^{r t}\) is a solution. How about \(y=v(t) e^{r t}\) ?
    \[ \notag
    \begin{aligned}
    y^{\prime} & =v^{\prime}(t) e^{r t}+v(t) r e^{r t} \\
    y^{\prime \prime} & =v^{\prime \prime}(t) e^{r t}+v^{\prime}(t) r e^{r t}+v^{\prime}(t) r e^{r t}+v(t) r^2 e^{r t} \\
    & =v^{\prime \prime}(t) e^{r t}+2 v^{\prime}(t) r e^{r t}+v(t) r^2 e^{r t}
    \end{aligned}
    \]
    \[\notag
    \begin{array}{l}
    a y^{\prime \prime}+b y^{\prime}+c y=0 \\
    a\left(v^{\prime \prime} e^{r t}+2 v^{\prime} r e^{r t}+v r^2 e^{r t}\right)+b\left(v^{\prime} e^{r t}+v r e^{r t}\right)+c v e^{r t}=0 \\
    a\left(v^{\prime \prime}(t)+2 v^{\prime}(t) r+v(t) r^2\right)+b\left(v^{\prime}(t)+v(t) r\right)+c v(t)=0 \\
    a v^{\prime \prime}(t)+2 a v^{\prime}(t) r+a v(t) r^2+b v^{\prime}(t)+b v(t) r+c v(t)=0 \\
    a v^{\prime \prime}(t)+(2 a r+b) v^{\prime}(t)+\left(a r^2+b r+c\right) v(t)=0 \\
    a v^{\prime \prime}(t)+\left(2 a\left(\frac{-b}{2 a}\right)+b\right) v^{\prime}(t)+0=0
    \end{array}
    \]
    since \(a r^2+b r+c=0\) and \(r=\frac{-b}{2 a}\)
    \[ \notag
    a v^{\prime \prime}(t)+(-b+b) v^{\prime}(t)=0 . \quad \text { Thus } a v^{\prime \prime}(t)=0 \text {. }
    \]

    Hence \(v^{\prime \prime}(t)=0\) and \(v^{\prime}(t)=k_1\) and \(v(t)=k_1 t+k_2\).
    Hence \(v(t) e^{r_1 t}=\left(k_1 t+k_2\right) e^{r_1 t}\) is a soln Thus \(t e^{r_1 t}\) is a nice second solution.

    Hence general solution is \(y=c_1 e^{r_1 t}+c_2 t e^{r_1 t}\).

    Reduction of order


    Suppose \(y=\phi_1(t)\) is a solution to \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\)
    Guess \(y=v(t) \phi_1(t)\) is also a solution.
    Solve for unknown function \(v(t)\) by plugging in: \(v^{\prime \prime}(t) \phi_1(t)+v^{\prime}(t)\left[2 \phi_1^{\prime}(t)+p(t) \phi_1(t)\right]=0 \)

    Let \(w(t)=v^{\prime}(t)\), then \(w^{\prime}(t)=v^{\prime \prime}(t)\)
    \[ \notag
    \begin{array}{l}
    w^{\prime}(t) \phi_1(t)+w(t)\left[2 \phi_1^{\prime}(t)+p(t) \phi_1(t)\right]=0 \\
    w^{\prime}(t) \phi_1(t)=-w(t)\left[2 \phi_1^{\prime}(t)+p(t) \phi_1(t)\right] \\
    \omega=\frac{d \omega}{d t} \quad \frac{w^{\prime}(t)}{w(t)}=\frac{2 \phi_1^{\prime}(t)+p(t) \phi_1(t)}{\phi_1(t)} \\
    \int \frac{d w}{w}=\int \frac{2 \phi_1^{\prime}(t)+p(t) \phi_1(t)}{\phi_1(t)} d t \\
    e^{\ln(|w|)} = e^{\int \frac{2 \phi_1^{\prime}(t)+p(t) \phi_1(t)}{\phi_1(t)} d t }\\
    w=Ce^{A(t)}
    \end{array}
    \]


    This page titled 3.4 is shared under a not declared license and was authored, remixed, and/or curated by Isabel K. Darcy.

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