MATH 308 Lecture 38

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Lecture Notes

Variation of Parameters in Systems

• $X_1$ and $X_2$ are two linearly independent solutions to homogeneous equation
• Find a particular solution to nonhomogeneous system $X(t)=c_1(t)X_1+c_2(t)X_2$, where $c_1$ and $c_2$ are real-valued functions.

Taking the derivative of the equation for $X(t)$, we get:

$X^{\prime }(t)={c_1}^{\prime }(t)X_1+{c_2}^{\prime }(t)X_2+c_1(t){X_1}^{\prime }+c_2(t){X_2}^{\prime }$

By the differential equation, this has to be equal to

$A(c_1{X}_1+c_2{X}_2)+G(t)$

Notice that we have an equation for ${X_1}^{\prime }=A{X}_1$ and ${X_2}^{\prime }=A{X}_2$

Therefore,

$\begin{array}{rl}{c_1}^{\prime }{X}_1+{c_2}^{\prime }{X}_2& =G(t)\\ X\frac{\mathrm{d}\overrightarrow{c}}{\mathrm{d}t}& =G(t)\\ \frac{\mathrm{d}\overrightarrow{c}}{\mathrm{d}t}& =X^{-1}G(t)\\ \overrightarrow{c}& =\int X^{-1}G(t)\mathrm{d}t\end{array}$

Exercise 3

$X^{\prime }(t)=\left[\begin{array}{cc}4& -2\\ 8& -4\end{array}\right]X(t)+\left[\begin{array}{c}{t}^{-3}\\ -t^{-2}\end{array}\right]$

Use variation of parameters, but first solve homogeneous equation first:

Eigenvalues: $\lambda \in \{0,0\}$

Eigenvectors: ${\overrightarrow{x}}_1=⟨\alpha ,2\alpha ⟩$

Generalized Eigenvector: ${\overrightarrow{x}}_2=⟨0,-\frac{1}{2}⟩$

Homogeneous solutions: $\{\begin{array}{l}{X}_1=⟨1,2⟩\\ X_2=t⟨1,2⟩-⟨0,\frac{1}{2}⟩\end{array}$

Particular solution is of form $X_p=c_1(t)X_1+c_2(t)X_2$, where ${c_1}^{\prime }{X}_1+{c_2}^{\prime }{X}_2=G(t)$.

${c_1}^{\prime }\left[\begin{array}{c}1\\ 2\end{array}\right]+{c_2}^{\prime }\left[\begin{array}{c}t\\ 2t-\frac{1}{2}\end{array}\right]=\left[\begin{array}{c}{t}^{-3}\\ -t^{-2}\end{array}\right]$

From this, we find $\{\begin{array}{l}{c_1}^{\prime }=t^{-3}+4t^{-2}+2t^{-1}\\ {c_2}^{\prime }=4t^{-3}-2t^{-2}\end{array}$

Therefore,

$\begin{array}{rl}{c}_1& =-\frac{t^{-2}}{2}+4t^{-1}-2\ln t\\ c_2& =-\frac{4t^{-2}}{2}-2t^{-1}\end{array}$

Extra Material: Chapter 5

not on final

Taylor series:

$\begin{array}{rl}f(t)& =\sum _{i=0}^{\mathrm{\infty }}\frac{f^{(i)}(0)}{i!}{x}^i\\ & =\frac{f(0)}{0!}{x}^0+\frac{f^{\prime }(0)}{1!}x+\frac{f^{″}(0)}{2!}{x}^2+\dots +\frac{f^{(n)}(c)}{n!}\end{array}$

For $c\in (0,x)$

Suppose we had a differential equation $y^{″}-y=0$ for $y=a_0+a_{1}x+a_2{x}^2+\dots +a_n{x}^n$

$\begin{array}{rlr}{y}^{\prime }& =a_1+2a_{2}x+3a_3{x}^2+\dots +n{a}_n{x}^{n-1}{y}^{″}& =2a_2+3\cdot 2a_{3}x+4\cdot 3a_4{x}^2+\dots +n(n-1)a_n{x}^{n-2}\end{array}$

in other words

Suppose we had a differential equation $y^{″}-y=0$ for $y=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k$

$\begin{array}{rl}{y}^{\prime }& =\sum _{k=0}^{\mathrm{\infty }}(k+1)a_{k+1}{x}^k\\ y^{″}& =\sum _{k=0}^{\mathrm{\infty }}(k+1)(k+2)a_{k+2}{x}^k\end{array}$

By the differential equation, for any $k\ge 0$,

$(k+2)(k+1)a_{k+2}=-a_k$