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:

$X'(t)=c_{1}'(t)\,X_{1}+c_{2}'(t)\,X_{2}+c_{1}(t)\,X_{1}'+c_{2}(t)\,X_{2}'$

By the differential equation, this has to be equal to

$A\,\left(c_{1}\,X_{1}+c_{2}\,X_{2}\right)+G(t)$

Notice that we have an equation for $X_{1}'=A\,X_{1}$ and $X_{2}'=A\,X_{2}$

Therefore,

${\begin{aligned}c_{1}'\,X_{1}+c_{2}'\,X_{2}&=G(t)\\X\,{\frac {\mathrm {d} {\vec {c}}}{\mathrm {d} t}}&=G(t)\\{\frac {\mathrm {d} {\vec {c}}}{\mathrm {d} t}}&=X^{-1}\,G(t)\\{\vec {c}}&=\int X^{-1}\,G(t)\,\mathrm {d} t\end{aligned}}$

Exercise 3

$X'(t)={\begin{bmatrix}4&-2\\8&-4\end{bmatrix}}X(t)+{\begin{bmatrix}t^{-3}\\-t^{-2}\end{bmatrix}}$

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

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

Eigenvectors: ${\vec {x}}_{1}=\left\langle \alpha ,2\alpha \right\rangle$

Generalized Eigenvector: ${\vec {x}}_{2}=\left\langle 0,-{\frac {1}{2}}\right\rangle$

Homogeneous solutions: ${\begin{cases}X_{1}=\left\langle 1,2\right\rangle \\X_{2}=t\,\left\langle 1,2\right\rangle -\left\langle 0,{\frac {1}{2}}\right\rangle \end{cases}}$

Particular solution is of form $X_{p}=c_{1}(t)\,X_{1}+c_{2}(t)\,X_{2}$, where $c_{1}'\,X_{1}+c_{2}'\,X_{2}=G(t)$.

$c_{1}'{\begin{bmatrix}1\\2\end{bmatrix}}+c_{2}'{\begin{bmatrix}t\\2t-{\frac {1}{2}}\end{bmatrix}}={\begin{bmatrix}t^{-3}\\-t^{-2}\end{bmatrix}}$

From this, we find ${\begin{cases}c_{1}'=t^{-3}+4t^{-2}+2t^{-1}\\c_{2}'=4t^{-3}-2t^{-2}\end{cases}}$

Therefore,

${\begin{aligned}c_{1}&=-{\frac {t^{-2}}{2}}+4t^{-1}-2\ln {t}\\c_{2}&=-{\frac {4t^{-2}}{2}}-2t^{-1}\end{aligned}}$

Extra Material: Chapter 5

not on final

Taylor series:

${\begin{aligned}f(t)&=\sum _{i=0}^{\infty }{\frac {f^{(i)}(0)}{i!}}\,x^{i}\\&={\frac {f(0)}{0!}}\,x^{0}+{\frac {f'(0)}{1!}}\,x+{\frac {f''(0)}{2!}}\,x^{2}+\dots +{\frac {f^{(n)}(c)}{n!}}\end{aligned}}$

For $c\in (0,x)$

Suppose we had a differential equation $y''-y=0$ for $y=a_{0}+a_{1}\,x+a_{2}\,x^{2}+\dots +a_{n}\,x^{n}$

${\begin{aligned}y'&=a_{1}+2a_{2}\,x+3a_{3}\,x^{2}+\dots +n\,a_{n}\,x^{n-1}y''&=2a_{2}+3\cdot 2\,a_{3}\,x+4\cdot 3\,a_{4}\,x^{2}+\dots +n(n-1)a_{n}\,x^{n-2}\end{aligned}}$

in other words

Suppose we had a differential equation $y''-y=0$ for $y=\sum _{k=0}^{\infty }a_{k}\,x^{k}$

${\begin{aligned}y'&=\sum _{k=0}^{\infty }(k+1)\,a_{k+1}\,x^{k}\\y''&=\sum _{k=0}^{\infty }(k+1)(k+2)\,a_{k+2}\,x^{k}\end{aligned}}$

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

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