Date modified: Mon 2025-07-07 . 01:50 AM
Date created: Mon 2025-06-30 . 08:05 PM
# Week 3 Monday ## A simple motivation. Consider $y'+2y=0$. Solve it Consider $y'' + 3y' +2y=0$. Solve it ## Basic theory of linear ODEs. We are now concerned with linear ODEs of the form $$ a_{n}(x)y^{(n)}(x) + \cdots+ a_{0}(x) y(x) = F(x) $$If $a_{n}(x)=1$, then we call it **monic**. The $a_{k}$ are called the coefficient terms, and $F$ the driving term or forcing term. If $F =0$ then we say the linear ODE is **homogeneous**, otherwise **nonhomogeneous**. First, do solutions exists, and if so, are they unique? > **Theorem.** (Existence and uniqueness of solutions for linear ODEs) > Given an $n$-th order linear ODE $$ a_{n}(x) y^{(n)} +\cdots+a_{0}(x) y = F(x) $$Suppose $I$ is an interval where (1) $a_{n}(x)\neq0$ and (2) $a_{k}(x)$ and $F(x)$ all continuous, then there **exists** solutions to the DE on the interval $I$. Further more, if $x_{0} \in I$, and $b_{0},b_{1},\ldots,b_{n-1}$ are any $n$ real numbers, there exists a **unique** solution on $I$ satisfying the initial condition $$ y(x_{0}) = b_{0}, y'(x_{0})=b_{1},\ldots,y^{(n-1)}(x_{0})=b_{n-1}. $$ Also, we will say this interval $I$ where all the conditions are met as "nice". A full set of initial conditions determines a unique solution, on its interval of domain. Examples. Graphically specifying value and slope. Is it enough to guarantee uniqueness? Depends on the order! Now, what do solution sets look like? First the homogeneous case, we notice the following simple observation. > **Proposition.** (Linear combination of homogeneous solutions is a homogeneous solution) > Given an $n$-th order linear ODE that is homogeneous, $$ a_{n}(x)y^{(n)} + \cdots + a_{0}(x) y = 0 $$and let $y_{1}$ and $y_{2}$ be any two solution on some interval $I$. Then any linear combination $c_{1}y_{1} + c_{2}y_{2}$ is also a solution on $I$. Prove it. Write the DE as $\sum a_{k}(x)y^{(k)}=0$. Now plug in and factor. Note. The zero function is always a solution to a homogeneous linear ODE. And together with above shows that the set of all homogeneous solutions form a **linear subspace**, that is, it is a vector space. If the homogeneous solutions form a vector space, then it has a dimension. > **Theorem.** (The dimension of an $n$-th order homogeneous solution set is $n$) > Given an $n$-th order linear ODE that is homogeneous, $$ a_{n}(x)y^{(n)} + \cdots + a_{0}(x) y = 0 $$where it is "nice" on interval $I$, then the dimension of the solution set is $n$. In other words, there exists $n$ linearly independent solutions $y_{1},\ldots,y_{n}$ on $I$ such that every solution to this DE is a linear combination of $y_{1},\ldots,y_{n}$. This collection of $n$ solutions $y_{1},\ldots,y_{n}$ is called a **fundamental solution set**. So every solution is a linear combination $c_{1}y_{1}+\cdots+c_{n}y_{n}$. Now if it is nonhomogeneous, then it satisfies the following **structural theorem**. > **Theorem.** (Structural theorem of solutions to nonhomogeneous linear ODE) > Given an $n$-th order linear ODE that is nonhomogeneous, $$ a_{n}(x)y^{(n)} + \cdots + a_{0}(x) y = F(x) $$where it is "nice" on interval $I$, then every solution has the form $y_{H}+y_{P}$, where $y_{P}$ is any one particular solution to this nonhomogeneous problem, and $y_{H}$ is a solution to the associated homogeneous problem. In other words, every solution has the form $$ c_{1}y_{1}+\cdots c_{n}y_{n} + y_{P} $$where $y_{k}$ forms a fundamental solution set to the homogeneous problem. Prove it. Ok. So how do we find them? ## Constant coefficient homogeneous linear ODEs. - Auxiliary polynomial. - Repeated roots. - Complex roots. A constant coefficient homogeneous linear ODE has the form $$ a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_{0}y =0 $$where $a_{k}$ are all constants. How do we solve these? Let us reduce to simplest type of problem, which is first order constant coefficient. $$ a_{1}y'+a_{0}y=0, $$and without loss, $a_{1} \neq 0$, otherwise not first order. Let us solve this. Note this is separable, and we have $y' = -\frac{a_{0}}{a_{1}}y$, which is separable. Solving this gives $$ \frac{dy}{y} = -\frac{a_{0}}{a_{1}} \implies y=C e^{-(a_{0}/a_{1})x}. $$Note that this has a solution of the form $y=Ce^{rx}$ for some $r$. In this case $r = -\frac{a_{0}}{a_{1}}$.