3.4
- Page ID
- 156609
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(w^{\prime}(t)=v^{\prime \prime}(t)\)
\[ \notag
\begin{array}{l}
w^{\prime}(t) \phi_1(t)+w(t)\left[2 \phi_1^{\prime}(t)+p(t) \phi_1(t)\right]=0 \\
w^{\prime}(t) \phi_1(t)=-w(t)\left[2 \phi_1^{\prime}(t)+p(t) \phi_1(t)\right] \\
\omega=\frac{d \omega}{d t} \quad \frac{w^{\prime}(t)}{w(t)}=\frac{2 \phi_1^{\prime}(t)+p(t) \phi_1(t)}{\phi_1(t)} \\
\int \frac{d w}{w}=\int \frac{2 \phi_1^{\prime}(t)+p(t) \phi_1(t)}{\phi_1(t)} d t \\
e^{\ln(|w|)} = e^{\int \frac{2 \phi_1^{\prime}(t)+p(t) \phi_1(t)}{\phi_1(t)} d t }\\
w=Ce^{A(t)}
\end{array}
\]