文章目錄

• 1. Separable Differentiable Equations
• 2. Linear Equations
• • First-order linear equation
  • • Solution of the homogeneous equation
    • Solution of the inhomogeneous equation
    • • An alternate solution -- Variation of parameters
    • Theorem -- Structure of the Solution
• 3. Exact Differential Equations
• • Definition
  • Theorem
  • Solving Exact Differential Equation
  • • Integrating Factors
    • Separable Equations
  • Homogenous Equations
  • • Solve Homogenous Equations

1. Separable Differentiable Equations

$$\begin{gathered} \frac{dy}{dt}=\frac{g(t)}{h(y)} \\ or \\ \frac{dy}{dt}=g(t)f(y) \end{gathered}$$

Separate the variables
$$\begin{gathered} h(y)dy=g(t)dt \\ or \\ \frac{dy}{f(y)}=g(t)dt \end{gathered}$$

Integrate both sides
$$\begin{gathered} \int h(y)dy=\int g(t)dt \\ or \\ \int \frac{dy}{f(y)}=\int g(t)dt \end{gathered}$$

Solve for the solution y(t), if possible

PS: For $\frac{dy}{f(y)}$, if $f(y)=0$, then if $f(y_0)=0$, $y(t)=y_0$ is a solution.

2. Linear Equations

First-order linear equation

Form:
$$x^{\prime }=a(t)x+f(t)$$
Homogeneous if $f(t)=0$, form:
$$x^{\prime }=a(t)x$$
$a(t),f(t)$ are called coefficients of the equation.

Solution of the homogeneous equation

$$\begin{gathered} \frac{dx}{x}=a(t)dt?ln|x|=\int a(t)dt+C \\ |x|=e^{\int a(t)dt+C}=e^C{e}^{\int a(t)dt} \end{gathered}$$

Let $A$ be a constant, which can be either >0, <0 or =0
$$x(t)=A{e}^{\int a(t)dt}$$

Solution of the inhomogeneous equation

The equation
$$x^{\prime }=ax+f$$

Rewrite the equation as
$$x^{\prime }?ax=f$$

Multiply by the integrating factor
$$u(t)=e^{?\int a(t)dt}$$
So that the equation becomes
$$(ux{)}^{\prime }=u(x^{\prime }?ax)=uf$$

Integrate this equation
$$u(t)x(t)=\int u(t)f(t)dt+C$$

Solve for x(t)

An alternate solution – Variation of parameters

The equation
$$y^{\prime }=ay+f$$

Get a partialicular solution to the homogeneous equation $y_h^{\prime }=a{y}_h$
$$y_h(t)=e^{\int a(t)dt}$$

Substitute $y=v{y}_h$ into the inhomogeneous equation to find v, or remember that
$$v^{\prime }=\frac{f}{y_h}$$
And solve $v$

Write the v into the general solution $y=v{y}_h$

Theorem – Structure of the Solution

Every solution to the inhomogeneous equation is of the form
$$y(t)=y_p(t)+A{y}_h(t)$$
Where $A$ is an arbitrary constant,

$y_p$ is a partialicular solution to the inhomogeneous equation $y^{\prime }=a(t)y+f(t)$,

and $y_h$ is a partialicular solution to the associated homogeneous equation $y^{\prime }=a(t)y$.

3. Exact Differential Equations

Definition

The differential of a continuously differentiable function F is the differential form
$$dF=\frac{?F}{?x}dx+\frac{?F}{?y}dy$$
A differential form is said to be exact if it is the differential of a continuously differentiable function.

Theorem

Let $\omega =P(x,y)dx+Q(x,y)dy$ be a differential form where both P and Q are continuous and differentiable

(a) if $\omega$ is exact, then
$$\frac{?P}{?y}=\frac{?Q}{?x}$$
(b) if (a) is true in a rectangle R, then $\omega$ is exact in R

Solving Exact Differential Equation

If the equation $P(x,y)dx+Q(x,y)dy=0$ is exact, the solution is given by $F(x,y)=C$, where F is found by

Solve $\frac{?F}{?x}=P$ by integration:
$$F(x,y)=\int P(x,y)dx+?(y)$$

Solve $\frac{?F}{?y}=Q$
$$\frac{?F}{?y}=\frac{?}{?y}\int P(x,y)dx+{?}^{\prime }(y)=Q(x,y)$$

Integrating Factors

The form $Pdx+Qdy$ has an integrating factor depending on one of the variables under the following conditions.

• If
  $$h=\frac{1}{Q}\left(\frac{?P}{?y}?\frac{?Q}{?x}\right)$$
  is a function of x only, then $\mu (x)=e^{\int h(x)dx}$ is an integrating factor.

• If
  $$g=\frac{1}{P}(\frac{?P}{?y}?\frac{?Q}{?x})$$
  is a function of y only, then $\mu (y)=e^{?\int g(y)dy}$ is an integrating factor.

Separable Equations

If the equation has the form
$$P(x)dx+Q(y)dy=0$$
The solution is given by
$$F(x,y)=\int P(X)dx+\int Q(y)dy=C$$

Homogenous Equations

A function G(x, y) is homogenous of degree n if
$$G(tx,ty)=t^{n}G(x,y)$$
A differential equation $Pdx+Qdy=0$ is said to be homogenous if both of the coefficients P and Q are homogeneous of the same degree

Solve Homogenous Equations

Substitute y = xv

Then $P(x,y)dx+Q(x,y)dy=0$ turns into $P(x,xv)dx+Q(x,xv)(vdx+xdv)=0$
$$\begin{gathered} P(x,xv)=x^{n}P(1,v) \\ Q(x,xv)=x^{n}Q(1,v) \end{gathered}$$
Dividing $x^n$, and collecting terms
$$(P(1,v)+vQ(1,v))dx+xQ(1,v)dv=0$$
The integrating factor is
$$\frac{1}{x(P(1,v)+vQ(1,v))}$$
Separate variables, and solve the equation
$$\frac{dx}{x}+\frac{Q(1,v)dv}{P(1,v)+vQ(1,v)}=0$$

總結

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