2-1 Integrating Factor $\mu (x)$

In general first-order linear differential equation in standard form:

$$ \frac{dy}{dt} + p(t) y = g(t) $$

In convenient ,we also write the equation in this form:

$$ P(t)\frac{dy}{dt} + Q(t) y = G(t) $$

then ,to solve this ODE, we tend to find a integrating factor $\mu (x)$ s.t. $(\mu (t) y)' = k(t)$

How to find the Integrating Factor

$$ P(t) \frac{dy}{dt} + Q(t)y = G(t) $$$$ \mu (x) P(t) \frac{dy}{dt} + \mu (x) Q(t)y = \mu (t)G(t) $$$$ \mu (t) \frac{dy}{dt} + \mu (t) \frac {Q(t)}{P(t)}y = \frac {\mu (t) G(t)}{P(t)} $$$$ [\mu (t) y]' = \frac{\mu (t) G(t)}{P(t)} $$

we can imply that $\mu '(t) = \frac{Q(t)}{P(t)} \mu (t)$ ,then

$$ \mu (t) = e^{\int_{}^{} \frac{Q(t)}{P(t)} dt} $$

2-2 Separable Differential Equation

The first general first-order differential equation is

$$ \frac{dy}{dt} = f(x ,y) $$

To identify this class of equation , we write in this form

$$ M(x, y) + N(x ,y)\frac{dy}{dx} = 0 $$

When$M$is a function o f $x$only and$N $is a function of $y$ ,the equation becomes

$$ M(x)dx+ N(y)dy = 0 $$

and we can $\int M(x)dx + N(y)dy = \int 0 dx$

2-4 Differences Between Linear and Nonlinear DEs

THM-1 (Existence and Unique for First-Order Linear ODE)

Statement

if p(t) is continuous on an open interval$I \quad \exist ! \ y =\phi (t) $satisfying $y' + p(t)y = q(t), \ \forall t \in I \ and \ y = y_0 $

Proof

Since P(t) is continuous on I ,we set $\mu(t) = e^{\int_{t_0}^{t}p(t)dt} \neq0 \ ,t \in I$($\mu '(t) = \mu(t) p(t)$)

We have $\mu(t) \in C'$($\mu \ is \ diff.)$

Multiplying (1) by $\mu(t)$, then we obtain $[\mu(t)y]' = \mu(t)q(t)$

Since q(t) is continuous on I ,,thus

$$ \mu(t) y(t) = \int _{t_0}^{t} \mu(t)q(t) dt +C $$$$ y(t) - \mu^{-1}(t) \int_{t_0}^{t} \mu(s)q(s)ds + C $$

$y = \phi(t)$ exists , is differential and satisfy (1). Also , from the mitial condition (2), we determine uniquely $c = y_0$. Hence ,there is only one solution of (1)-(2).

THM-2 (First-Order Nonlinear ODE)

Statement

Let $f (t ,y)$and $\frac{\delta f}{\delta y}$be continuous in some rectangle R ,$\alpha < t <\beta$ ,$\gamma < y < \delta$

$R = \{ (t, y)\ | \ \alpha < t < \beta ,\ \gamma < y < \delta \}$

containing $(t_0 ,y_0)$. Then , in some interval $(t_0 -h , t_0 + h) \subset (\alpha ,\beta)$for some $h > 0$ ,

$\exist! \ y = \phi(t) \ s.t. \ y'(t) = f(t ,y) \ and \ y(t_0) = y_0$

Remark

  • Thm-2 can imply Thm-1 (linear case)

    In the case of Thm-1 , $f(t ,y) = q(t) = p(t)y \in C$ and $\frac{\delta f}{\delta y}(t ,y) = -p(t) \in C$

    there is only one solution in some interval $(t_0 +h ,t_0 - h) \subset I$

  • Uniqueness

    We can make sure that the graph of two solution cannot have any intersection.

2-5 Autonomous DEs ,Population Dynamics

Autonmous

Ex1 (Exponential Growth)

$\phi (t)$is the popuation of the given species at time $t$

$$ \frac{dy}{dt} = f(y) $$

consider $\frac{dy}{dt} = ry$

  • r > 0 : increase
  • r < 0 : decline

Assume r > 0 ,and $y(0) = y_0$ ,$\phi(t) = y_0 e^{rt}$

Ex2 (Logistic DEs)

$$ r = h(y) $$

Condiser h(y) s.t. h(y) $\cong$r > 0,when y is small

  • h(y) decrease as y grows larger
  • h(y) < 0 when y is suff large

Assume that $h(y) = r-ay ,\quad a > 0$

Consider logistic equation:

$$ \frac{dy}{dt} = (r-ay)y $$$$ \frac{dy}{dt} = r(1- \frac{y}{k})y , \ where \ k =\frac{r}{a} $$
  • Find constant solution:
$$ \frac{dy}{dy} = 0 \Longleftrightarrow r(1-\frac{y}{k})y = 0 $$$$ the\ constant \ solutions \ are \quad y = \phi_1(t) \equiv0 ,\ y = \phi_2 \equiv k $$
  • Study for
$$ f( y) = r(1-\frac{y}{k})y = -\frac{r}{k}(y - \frac{k}{2})^2 + \frac{rk}{4} $$$$ \frac{dy}{dt} = \begin{cases} > 0 \quad if \ 0 < y < k \\ \quad \\ <0 \quad if \ y > k \\ \end{cases} $$

By Exsistence and Uniqueness Theorem ,no solution can intersect the equilibrium solution $y = k$

  • Find $\frac{d^2y}{dt^2} $
$$ \boxed{\frac{d^2y}{dt^2}} = \frac{d}{dt}(\frac{dy}{dt}) =\frac{d}{dt}f(y) = f'(y)\frac{dy}{dt} = \boxed{f'(y)f(y)} $$$$ \frac{d^2y}{dt^2} = \begin{cases} > 0 \quad if \ 0 < y < \frac{k}{2}\ or \ y > k \\ \quad \\ <0 \quad if \ \frac{k}{2}Note : $y = \frac{k}{2}$is an inflation points

2-6 Exact Differential Equations

THM

Let M ,N ,$M_y ,N_x$are continuous in $R = \{ (x ,y) |\ \alpha < x< \beta ,\ \gamma

Then the $equ(*)$ is exact in $R$ $\Longleftrightarrow \ M_y(x ,y) = N_x(x ,y)$at each point in $R$

proof

  • $\Longrightarrow$

Suppose (*) is an exact differential equation ,$\exist \ \psi \ s.t.\ \frac{\partial \psi}{\partial x} = M(x ,y) \ ,\ \frac{\partial \psi}{\partial y} = N(x ,y)$

then ,$M_y(x ,y) = \frac{\partial \psi}{\partial x\partial y} = \frac{\partial \psi}{\partial x \partial y}$, $M_y$and $N_x$are continuous ,$\psi_{xy} \ ,\ \psi_{yx}$are continuous.

Hence ,$\frac{\partial^2 \psi}{\partial y \partial x} = \frac{\partial^2 \psi}{\partial x \partial y}$ ,$M_y = N_x$

  • $\Longleftarrow$

Good Find a $\psi(x) \ s.t. \ \psi_x = M\ ,\ \psi_y = N$, Integrating $\psi_x = M \ w.r.t \ x$

$\psi(x ,y) = \int_{x_0}^{x}M(s ,y)\ ds + h(y) = Q(x ,y) + h(y)$ ,where $Q(x ,y) = \int_{x_0}^{x} M(s ,y)\ ds$and $x_0$is some specified constant with $\alpha < x < \beta$

Differentiating $Q(x ,y) + h(y) \quad w.r.t \quad y$

$\psi_y(x ,y) = \frac{\partial Q}{\partial y}(x ,y) + h'(y) = N(x ,y)$, Then $h'(y) = N(x ,y) - \frac{\partial Q}{\partial y}(x ,y)$

To show $N(x ,y) - \frac{\partial Q}{\partial y}(x ,y)$only depends on y

$$ \frac{\partial}{\partial x}(N(x ,y) - \frac{\partial Q}{\partial y}(x ,y)) = N_x - \frac{\partial^2Q}{\partial y \partial x} = N_x - M_y = 0 $$

2-9 First-Order Difference Equations

Definition

$$ y_{n+1} = f(n ,y_n) ,\quad n = 0 ,1 ,2... $$

It is called first-order difference equation. if f is a function of $y_n$, the difference equation is linear, otherwise, it is nonlinear. The solutions of difference equation is a sequence of numbers $y_0 ,y_1 ,y_2$ that satisfy the equation for each n.

General case

$$ y_{n+1} = f(y_n), \quad n = 0 ,1, 2... $$

then$y_1 = f(y_0) ,\quad y_2 = f(y_1) = f(f(y_0) = f^2(y_0)$

In general , the $n^{th}$ iterate$y_n$is $y_n = f^n(y_0)$

  • Equilibrium solutions

    Solutions for which $y_n$has the same value for all n

Linear Equations

Case-1

$$ y_{n+1} = \rho_ny_n,\quad n= 0 ,1 ,2... $$

In general ,$y_n = \rho_{n-1}...\rho_0y_0 ,\quad n = 1 ,2 ,...$

if $\rho_n = \rho \ \ \forall n \text{ ,the equations becomes } y_n = \rho^n y_0$

$$ \lim_{n \to \infty}{y_n} = \begin{cases} 0, \quad if \ |\rho| < 1; \\ \quad \\ y_0, \quad if \ \rho = 1; \\ \\ \text{doesn't exist , otherwise.} \end{cases} $$

Case-2

$$ y_{n+1} = \rho y_n + b_n,\quad n = 0,1,2 ... $$$$ y_1 = \rho y_0 + b_0 $$$$ y_2 = \rho(\rho y_0 + b_0) + b_1 = \rho^2 y_0 + \rho b_0 + b_1 $$$$ y_n = \rho^n y_0 + \sum_{j = 0}^{n-1} \rho^{n-1-j} b_j $$

$\text{In special case where }b_n = b \neq 0\text{for all n ,the difference equations is } $

$$ y_{n+1} = \rho y_n + b $$

$\text{If }\rho \neq 1, \text{we can write the solution in the more compact form}$

$$ y_n = \rho^n (y_0 - \frac{b}{1-\rho})+\frac{b}{1 - \rho} $$

3-1 Homogeneous DEs with constant coefficients

Definition

Many second-order ordinary differential equations have the form

$$ \frac{d^2y}{dt^2} = f(t ,y ,\frac{dy}{dt}) $$

It is linear if the function has the form

$$ \frac{d^2 y}{dt^2} = g(t)- p(t)\frac{dy}{dt} - q(t) y $$

In general,

$$ y'' + p(t)y' + q(t)y = g(t) $$$$ P(t)y'' + Q(t)y' + R(t)y = 0 $$

if $g(t) = 0 \Longrightarrow \text{homogeneous}$

Initial Value Problem(IVP)

we consider that

$$ \begin{cases} ay'' + by' +cy = 0\\ \quad \\ y(0) = y_0 ,\quad y'(0) = y_0'\\ \end{cases} $$

We start by seeking exponential solutions of the form$y = e^{rt}$,where r is a parameter to be determined. Then it follows that $y' = re^{rt}$,$y' = r^2 e^{rt}$Substituting these expressions.

$$ (a r^2 + b r + c) e^{rt} = 0 $$

Since $e^{rt} \neq 0$

$$ a r^2 + b r + c = 0 $$

This equation is called the characteristic equation for the differential equation

Assuming that the roots of the characteristic equation are real and different, let them be denoted by $r_1$ and $r_2$, where $r_1 \neq r_2$. Then $y_1(t) = e^{r_1 t} \ and \ y_2(t) = e ^{r_2 t}$

$$ then ,\exist \ c_1 ,c_2 \in R \ s.t.\ y = c_1 e^{r_1t} + c_2 e ^{r_2t} $$$$ y' = c_1 r_1 e^{r_1 t} + c_2 r_2 e^{r_2 t} $$$$ y'' = c_1 r_1^2 e^{r_1 t} + c_2 r_2^2 e^{r_2 t} $$

Substituting these expression,

$$ ay'' + by' + c = c_1 (ar_1^2 + b r_1 + c)e^{r_1 t} + c_2 (ar_2^2 + b r_2 + c)e^{r_2 t} = 0 $$

Consider the initial value $y(t_0) = y_0 ,\ y'(t_0) = y_0'$

$$ \begin{cases}c_1 e^{r_1t_0} + c_2 e ^{r_2t_0} = y_0\\ \\ c_1 r_1 e^{r_1 t_0} + c_2 r_2 e^{r_2 t_0} = y_0'\\ \end{cases} $$$$ c_1 = \frac{y_0'-y_0 r_2}{r_1 - r_2}e^{-r_1t_0},\ c_2 = \frac{y_0r_1 - y_0'}{r_2 - r_1} e^{-r_2 t_0} $$

3-2 Solution of Linear Homogeneous Equations; the Wronskian

Differential Operator

Let p and q be continuous functions on an open interval I , for $\alpha < t < \beta$. The cases for $\alpha = -\infty$or $\beta = \infty$or both , are included.Then for any function $\phi$ that is twice differential on I , we define the differential operator L by the equation

$$ L[\phi] = \phi'' + p \phi' + q \phi $$

for example ,if $p(t) = t^2 ,\ q(t) = 1+t,\ \phi(t) = sin3t$

$$ L[\phi](t) = (sin3t)'' + t^2(sin3t)' + (1+t)sin3t $$

we usually write this equation in the form

$$ L[y] = y''+p(t)y'+q(t)y = 0 $$

with the initial value $y(t_0) = y_0,\quad y'(t_0) = y_0'$

Theorem 3.2.1 Existence and uniqueness Theorem

Consider the initial value problem

$$ y'' + p(t) y' + q(t) y = g(t),\quad y(t_0) = y_0, \quad y'(t_0) = y_0' $$

where p, q ,and g are continuous on an open interval I that contains the point $t_0$. This problem has exactly one solution $y = \phi (t)$,and the solution exists throughout the interval I.

  • Note : the initial problem has an unique solution on an interval I

Theorem 3.2.2 Principle of Superposition

if $y_1,\ y_2$are two solutions of the differential equation

$$ L[y] = y'' + p(t) y' + q(t)y = 0 $$

then the linear combination $c_1y_1 + c_2 y_2$is also a solution for any value of the constants $c_1 ,\ c_2$

proof

$$ \begin{aligned} L[c_1y_1 + c_2 y_2] &= [c_1y_1 + c_2 y_2]'' + p[c_1y_1 + c_2 y_2]' + q[c_1y_1 + c_2 y_2] \\\\ &= c_1y_1'' + c_2y_2'' + c_1py_1' + c_2py_2' + c_1qy_1 + c_2q y_2 \\\\ &= c_1[y_1'' + py_1' + q y_1] + c_2[y_2'' + py_2' + q y_2] \\\\ &= c_1L[y_1] + c_2L[y_2] \end{aligned} $$

Note

  • $W[y_1 ,y_2](x) $ is not everywhere zero iff $c_1 y_1 + c_2 y_2$ contains all solutions of equation.
  • Therefore, $y = c_1 y_1(t) + c_2 y_2(t)$is called general solution
  • $y_1$and $y_2$ are said to form a fundamental set of solutions iff $Wy_1 ,y_2 \neq 0$

3-4 Solution of Linear Homogeneous Equations; the Wronskian

Solution of Repeated Roots

In this section, we are going to solve what if $r_1 = r_2$

$$ ay'' + by' +cy = 0 \quad --(a) $$

if $r_1 = r_2$, then

$$ y_1(t) = c_1 e^{\frac{-b}{2a}t} $$

To find second solution, we assume that

$$ y = v(t)y_1(t) = v(t)e^{\frac{-b}{2a}t} $$$$ y' = v'(t)e^{\frac{-b}{2a}t} - \frac{b}{2a}v(t)e^{\frac{-b}{2a}t} $$$$ y'' = v''(t)e^{\frac{-b}{2a}t} - \frac{b}{a}v'(t)e^{\frac{-b}{2a}t} + \frac{b^2}{4a^2}v(t)e^{\frac{-b}{2a}t} $$

By substitude equation (a) ,cancelling the factor $e^{\frac{-b}{2a}t}$, rearranging the items, we find that

$$ v'' = 0 $$$$ v = c_1 + c_2 t $$

then we have

$$ y = c_1 e^{\frac{-b}{2a}t} + c_2 t e^{\frac{-b}{2a}t} $$

Thus $y$ is a linear combination of the two solutions

$$ y_1(t) = e^{\frac{-b}{2a}t}, \quad y_2(t) = t e^{\frac{-b}{2a}t} $$$$ W[y_1 ,y_2](t) = e^{\frac{-b}{2a}} $$

Since ,$W[y_1 ,y_2](t)$ never zero, $y_1$and $y_2$form a fundamental set of solutions.

Reduction of Order

Suppose that we know one solution $y_1(t)$ , not everywhere zero, of

$$ y'' + p(t)y' + q(t)y = 0 \quad --(b) $$

To find the second solution, let

$$ y = v(t) y_1(t) $$$$ y' = v'(t)y_1(t) + v(t)y_1'(t) $$$$ y'' = v''(t)y_1(t) + v'(t)y_1'(t) + v'(t)y_1'(t) + v(t)y_1''(t) $$

Substitude $y$, $y'$and $y''$to the equation (b)

$$ y_1 v'' + (2y_1' + py_1)v' + (y_1'' + p(t)y_1' + q(t)y_1)v = 0 $$

Since $y_1$is a solution of equation (b).

$$ y_1 v'' + (2y_1' + py_1)v' = 0 $$

Here becomes a first-order differential equation for the function $v'$

3-6 Variation of Parameters

In this section，we will intoduce the gerneral method in principle at least, it can be applied

to any equation, and it requires no detailed assumptions about the form of the solution.

Method of Variation of Parameters

In gerneral, we consider

$$ y'' + p(t)y' + q(t)y = g(t) $$

where $p$ , $q$ and $g$ are continuous functions. We assume that we know

$$ y_h = c_1y_1(t) + c_2 y_2(t) $$

In this idea, we replace the constant $c_1$,$c_2$ by $u_1(t)$,$u_2(t)$ $\text{resp.}$

$$ y = u_1(t) y_1(t) + u_2(t) y_2(t) \quad --(a) $$$$ y' = u_1'(t) y_1(t) + u_1(t)y_1'(t) + u'_2(t) y_2(t) + u_2(t)y_2'(t) $$

Assume that $u_1'(t) y_1(t) + u'_2(t) y_2(t) = 0$, we have

$$ y' = u_1(t)y_1'(t) + u_2(t)y_2'(t) $$$$ y'' = u_1'(t)y_1'(t) + u_1(t) y_1''(t) + u_2'(t)y_2'(t) + u_2(t)y''_2(t) $$

Now, we substitute$y$,$y'$and $y''$to the orginal equation. We find that

$$ \begin{aligned} &u_1(t) \left( y_1''(t) + p(t)y_1'(t) + q(t)y_1(t) \right) \\\\ +&u_2(t) \left( y_2''(t) + p(t)y_2'(t) + q(t)y_2(t) \right) \\\\ +&u_1'(t)y_1'(t) + u_2'(t)y_2'(t) = g(t) \end{aligned} $$

where $y_1$and $y_2$are the homogeneous solutions ,hence the equation above can be reduced to

$$ u_1'(t)y_1'(t) +u_2'(t)y_2'(t) = g(t) \quad --(b) $$

Since (a) and (b) forms a system of two linear algebraic equations for derivatives $u_1'(t)$ and $u_2'(t)$of the unknown function. By Carmer’s rule, we have:

$$ u_1'(t) = -\frac{y_2(t)g(t)}{W[y_1 ,y_2](t)} ,\quad u_2'(t) = -\frac{y_1(t)g(t)}{W[y_1 ,y_2](t)} $$

Integrate both of them, and substitude into (a) ,you can get the answer.