MATH 308 Lecture 6

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Lecture Notes

Finding Domain

$\ln x{y}^{\prime }+\frac{1}{x}y=\sqrt{5-x}y(3)=0$

Makes sense when $x>0$ and $5-x\ge 0$: $0<x\le 5$

Now to solve it:

$\begin{array}{rl}\mu (x)& =\ln x\\ \ln xy& =-\frac{2(5-x)}{3}+C\\ y& =-\frac{2(5-x)}{3\ln x}+C\\ 0& =-\frac{2(5-3)}{3\ln x}+C\\ C& =\frac{2^{\frac{5}{2}}}{3}\\ y(x)& =-\frac{2(5-3)}{3\ln x}+\frac{2^{\frac{5}{2}}}{3}\end{array}$

The domain of the solution is $1<x\le 5$ (since $\ln 1=0$)

Theorem 2.4.1

Given a first-order linear differential equation

$y^{\prime }+p(t)y=g(t)$

where $p$ and $g$ are continuous functions on an open interval $I$ containing $t_0$,

For any real number $y_0$, there exists a unique solution to the initial value problem defined over $I$.

Example 1

$y^{\prime }-\theta y={\sin }^2\theta y(\pi )=5$

• $\theta$ is continuous for all real numbers
• ${\sin }^2\theta$ is also continuous for all real numbers

Therefore, for $x_0=\pi$, $y_0=5$, there exists a unique solution and it exists for all real numbers.

Example 2

$\begin{array}{rl}x{y}^{\prime }+\sqrt{x}y& =\frac{3x}{x^2-3x+2}y(x_0)=y_0\\ y^{\prime }+\frac{1}{\sqrt{x}}y& =\frac{3}{(x-2)(x-1)}\end{array}$

• $x_0>0$ or there is no solution
• $x\in ̸\{2,1\}$

For $x_0\in (0,1)$, $x_0\in (1,2)$, or $x_0>2$ there will be a unique solution that will be continuous on the rage of the selected solution

Theorem 2.4.2

Given a first order nonlinear initial value problem

$y^{\prime }=f(t,y)y(t_0)=y_0$

where $f$ and $\frac{\partial f}{\partial y}$ are continuous in some rectangle $(t,y)\in (\alpha ,\beta )\times (\gamma ,\delta )$ containing $(t_0,y_0)$,

There exists a unique solution to the initial value problem defined in a neigborhood $(a,b)\subset (\alpha ,\beta )$ of $t_0$.

Example 1

$y^{\prime }=1+y^2$

Separable general solution: ${\tan }^{-1}(y)=x+C$

Initial value solutions:

• $y(0)=0:y(x)=\tan xx\in (-\frac{\pi }{2})$
• $y(5)=0:y(x)=\tan (x-5)x\in (-\frac{\pi }{2}+5,\frac{\pi }{2}+5)$

Interval of solution is smaller than interval of differential equation, and the location of those intervals are dependent on the initial condition.

Example 2

$y^{\prime }=x^2-x{y}^{3}y(1)=2$

• $f(x,y)=x^2-x{y}^3$ is a polynomial function continuous on ${\mathbb{ℝ}}^2$.
• $\frac{\partial f}{\partial y}=-3x{y}^2$ is another polynomial function continuous on ${\mathbb{ℝ}}^2$

For the initial condition $y(1)=2$, there exists a unique solution whose domain is an interval that contains 1.

Example 3

$y^{\prime }=y^{\frac{1}{3}}y(1)=0$

The theorem does not apply since $\frac{\partial y}{\partial x}=\frac{1}{3}{y}^{-\frac{2}{3}}$ is not defined for $y=0$.