MATH 308 Lecture 6

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Lecture Notes

Finding Domain

$\ln {x}\,y'+{\frac {1}{x}}\,y={\sqrt {5-x}}\quad y(3)=0$

Makes sense when $x>0$ and $5-x\geq 0$: $0<x\leq 5$

Now to solve it:

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

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

Theorem 2.4.1

Given a first-order linear differential equation

$y'+p(t)\,y=g(t)$

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

$y'-\theta \,y=\sin ^{2}{\theta }\quad y(\pi )=5$

• $\theta$ 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

${\begin{aligned}x\,y'+{\sqrt {x}}\,y&={\frac {3x}{x^{2}-3x+2}}\quad y(x_{0})=y_{0}\\y'+{\frac {1}{\sqrt {x}}}\,y&={\frac {3}{(x-2)(x-1)}}\end{aligned}}$

• $x_{0}>0$ or there is no solution
• $x\not \in \{2,1\}$

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

$y'=f(t,y)\quad y(t_{0})=y_{0}$

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

$y'=1+y^{2}$

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

Initial value solutions:

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

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

Example 2

$y'=x^{2}-x\,y^{3}\quad y(1)=2$

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

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

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