Skip to content

初等积分法 | Elementary Integration Method

This chapter gives primary method of solving special differential functions, which plays a great role in future study.

Outline

  • 一阶线性微分方程, 齐次与非齐次

  • 变量分离方程

  • 齐次方程, 化简齐次的方法

  • 全微分方程、积分因子、分组求积分因子

  • Bernouli 方程、Riccati 方程

  • 隐式微分方程, 可解出x, 可解出y, 双曲函数参数法

恰当方程 | Exact Equation

We focus on the symmetrical form

\[ \begin{equation} M(x,y)dx + N(x,y)dy = 0 \label{eq-exact} \end{equation} \]

This can bring us great convenience for digging into one-order ODE because it can gives us both the relation \(y=f(x)\) or \(x=g(y)\).

全微分方程、恰当方程的定义 | Definition of Exact Equation

If there exists a \(\mathit{\varphi}(x, y) \in C^{1}(D)\) such that

\[ d\mathit{\varphi}(x, y) = M(x,y)dx + N(x,y)dy \]

then equation \(\ref{eq-exact}\) is called Exact Equation.

There are some questions to answer:

  • How to judge an equation to be exact Equation?
  • If so, how to find original function \(\varphi(x, y)\)?
  • If not, how to transform it into one exact Equation?

In this pattern, we answer the first two equation and leave the third one after learning LFODE.

方程是恰当的充要条件 | Necessary and Sufficient Condition for exact Equation

Assume \(D\) is a simply connected region, and \(M(x, y)\), \(N(x, y) \in C(D)\) with \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x} \in C^{1}(D)\). Then equation \(\ref{eq-exact}\) is exact Equation if and only if

\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]

Prove it.

\(\Rightarrow\) is easy, by using second-order mixed partial derivatives of \(\mathit{\varphi}\).

\(\Leftarrow\). Using Green Formula/Theorem.

\[ \begin{align*} &\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \\ \Leftrightarrow \ &\int_{\gamma}P(x, y)dx+Q(x,y)dy = 0 \quad \forall\text{closed loop } \gamma\\ \Leftrightarrow \ &\int_{\gamma}P(x, y)dx+Q(x,y)dy = C \quad \forall\text{curve } \gamma \text{ connecting } (x_0, y_0), (x, y) \\ \Leftrightarrow \ &\exists \mathit{\varphi}(x,y) \in C^{1}(D) \text{ s.t } d\mathit{\varphi}(x, y) = P(x, y)dx+Q(x,y)dy \end{align*} \]

积分因子 | Integral Factor

This part we hope to find \(\mu(x, y)\) so when we multiply it to both sides of equation \(\ref{eq-exact}\) and we get

\[ \mu(x,y)M(x,y)dx + \mu(x,y)N(x,y)dy = 0 \]

such that there exists \(\mathit{\varphi}(x, y)\) which satisfies

\[ d\mathit{\varphi}(x,y) = \mu(x,y)M(x,y)dx + \mu(x,y)N(x,y)dy \]

Naively, if \(\mathit{\varphi}(x,y)\in C^2\), then

\[ \frac{\partial (\mu M)}{\partial y} = \frac{\partial^2 \mathit{\varphi}}{\partial x\partial y}=\frac{\partial (\mu N)}{\partial x} \]

Theoretically speaking, we have to solve a PDE

\[ \begin{equation} M(x,y)\frac{\partial \mu}{\partial y} - N(x,y)\frac{\partial \mu}{\partial x} = \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\mu(x,y) \label{eq-pde} \end{equation} \]

However, actually, it is very hard to solve the above PDE. So we focus on some special case like \(\mu(x,y)=\mu(x)\), \(\mu(y)\), \(\mu(x+y)\), \(\mu(xy)\).

Now we have the following theorem to judge whether we can get the above form of integral factors.

方程有特殊类型的积分因子的充要条件 | Necessary and Sufficient Condition of special integral factor of ODE

(i) Equation \(\ref{eq-pde}\) has solution \(\mu(x)\) depending only on \(x\), if and only if

\[ \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M} \overset{\Delta}{=} G(x) \]

only depends only on \(x\). Then

\[ \mu(x) = e^{\int_{x_0}^{x}G(t)dt} \]

(ii) Similarly, Equation \(\ref{eq-pde}\) has solution \(\mu(y)\) depending only on \(y\), if and only if

\[ \frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{N} \overset{\Delta}{=} G(y) \]

only depends only on \(y\). Then

\[ \mu(y) = e^{\int_{y_0}^{y}G(t)dt} \]

(iii) More generally, equation \(\ref{eq-pde}\) has solution \(\mu(\varphi(x,y))\), if and only if

\[ \frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{N\frac{\partial \varphi}{\partial x}-M\frac{\partial \varphi}{\partial y}} \overset{\Delta}{=} f(\varphi(x,y)) \]

变量分离方程 | Variable Separation Equation

This chapter we discuss how to solve equation when it is not Exact Equation. The basic idea is, through transformation, we can convert an equation into an exact Equation.

变量分离方程的定义 | Definition of Variable Separation Equation

If there exists \(M_1(x), M_2(y), N_1(x), N_2(y) \in C^1(D)\) such that

\[ M(x, y) =M_1(x)M_2(y), N(x, y) = N_1(x), N_2(y) \]

then we call equation \(\ref{eq-exact}\) Variable Separation Equation.

For this type of equation, we can multiply both sides

\[ \begin{equation} \frac{1}{M_2(y)N_1(x)} \label{eq-sep-factor} \end{equation} \]

equation \(\ref{eq-exact}\) becomes

\[ \frac{M_1(x)}{N_1(x)}dx + \frac{M_2(y)}{N_2(y)}dy = 0 \]

This is an exact equation, and \(\ref{eq-sep-factor}\) is called an Integral Factor of the equation.

we can get its integral

\[ \int_{x_0}^{x}\frac{M_1(t)}{N_1(t)}dt + \int_{y_0}^{y}\frac{M_2(s)}{N_2(s)}ds = c \]

which is easily seen a solution of the original equation.

And don't forget that if there exists \(a_i (i= 1,2,\cdots, m)\) such that \(N_1(a_i) = 0\), or exists \(b_j (j=1,2,\cdots, n)\) such that \(M_2(b_j) = 0\), then of course \(x = a_i, y = b_j\) are also solutions of the original solution.

  • 齐次方程 | Homogeneous Equation

The following equation can also be transferred into Variable Separation Equation.

齐次方程的定义 | Definition of Homogeneous Equation

We call \(f(x, y)\) Homogeneous Function of degree \(n\) if

\[ f(tx, ty) = t^n f(x, y) \]

and call equation \(\ref{eq-exact}\) Homogeneous equation if \(M(x, y), N(x, y)\) are Homogeneous Function.

When we let \(y = u x\), then \(dy = xdu + udx\), substitute in the equation and get

\[ \begin{align*} M(x, ux)dx+N(x,ux)(xdu+udx)&=0 \\ \Leftrightarrow [M(x, ux)+N(x,ux)u]dx+N(x,ux)xdu&=0 \end{align*} \]

extract \(x\) out by definition of Homogeneous Equation:

\[ x^n[M(1, u)+N(1,u)u]dx+x^{n+1}N(1,u)du=0 \]

If \(x^{n+1}[M(1, u)+N(1,u)u]\neq 0\), then we divide both sides by this and get

\[ \frac{1}{x}dx + \frac{N(1,u)}{M(1,u)+uN(1,u)}du=0 \]

which is also Variable Separation Equation.

一阶线性微分方程 | Linear First-Order Differential Equation

Now we focus on a really important expression of ODE: Linear First-Order Differential Equation(LFODE):

\[ \begin{equation} \frac{dy}{dx} + p(x)y = q(x) \label{eq-LFODE} \end{equation} \]
  • Homogeneous LFODE(H-LFODE)

We let \(q(x)\equiv 0\) in \(\ref{eq-LFODE}\), we get

\[ \begin{equation} \frac{dy}{dx} + p(x)y = 0 \label{eq-H-LFODE} \end{equation} \]

rewrite it into symmetrical form:

\[ p(x)ydx + dy = 0 \]

which is Variable Separation Equation.

So when \(y\neq 0\), multiply both sides \(1/y\) and integrate

\[ \ln{|y|} + \int_{x_0}^{x}p(t)dt = C \]

get \(y\) out of form \(x\):

\[ \begin{align} y &= \pm e^{C}\cdot e^{-\int_{x_0}^{x}p(t)dt} \nonumber \\ &= C_1\cdot e^{-\int_{x_0}^{x}p(t)dt} \end{align} \]

where \(C_1=\pm e^{C} \neq 0\), but we can include trivial solution \(y \equiv 0\) by letting \(C_1 = 0\).

  • Non-Homogeneous linear First-Order Differential Equation

we have two ways to get the answer.

Making use of Integral Factors.

To begin with, we convert equation \(\ref{eq-LFODE}\) into symmetrical form:

\[ \begin{equation} (p(x)y-q(x))dx+dy=0 \label{eq-SYM-LFODE} \end{equation} \]

Multiply \(e^{\int_{x_0}^{x}p(t)dt}\) to both sides of equation \(\ref{eq-SYM-LFODE}\):

\[ d\left(\ e^{\int_{x_0}^{x}p(t)dt} y \right) - e^{\int_{x_0}^{x}p(t)dt} q(x)dx = 0 \]

That is,

\[ d\left(\ e^{\int_{x_0}^{x}p(t)dt} y- \int_{x_0}^{x}e^{\int_{x_0}^{s}p(t)dt} q(x)ds \right) = 0 \]

integrate and get

\[ e^{\int_{x_0}^{x}p(t)dt} y - \int_{x_0}^{s}e^{\int_{x_0}^{x}p(t)dt} q(s)ds + C = 0 \]

extract \(y\) out and get:

\[ y = e^{-\int_{x_0}^{x}p(t)dt} \left( C + \int_{x_0}^{s}e^{\int_{x_0}^{x}p(t)dt} q(s)ds \right) \]

Through Variation of Constants.

We make a brave treatment: assume one special solution to equation \(\ref{eq-LFODE}\) is

\[ y = u\cdot e^{-\int_{x_0}^{x}p(t)dt} \]

where \(u\) is a new variable.

Subsititute in equation \(\ref{eq-LFODE}\) and get

\[ \left(p(x) u e^{-\int_{x_0}^{x}p(t)dt} -q(x) \right) dx - u p(x) e^{-\int_{x_0}^{x}p(t)dt} dx + e^{-\int_{x_0}^{x}p(t)dt} du = 0 \]

That is

\[ -q(x) dx + e^{-\int_{x_0}^{x}p(t)dt} du = 0 \]

multiply \(e^{\int_{x_0}^{x}p(t)dt}\) to both sides and integrate

\[ u = \int_{x_0}^{x}e^{\int_{x_0}^{s}p(t)dt}q(s)ds \]

So the special solution is

\[ y = e^{-\int_{x_0}^{x}p(t)dt}\cdot \int_{x_0}^{x}e^{\int_{x_0}^{s}p(t)dt}q(s)ds \]

一阶隐式微分方程 | First-order Implicit Differential Equation

Now we focus on equation

\[ \begin{equation} F(x,y, y') = 0 \label{eq-para} \end{equation} \]

where \(y'\) cannot be explicitly solved out.

  • 参数法 | parametric method

If we let \(p = y'\), then equation

\[ \begin{equation} F(x,y,p) = 0 \label{eq-para-p} \end{equation} \]

represents a curved surface in 3-D space.

And we can juggle equation \(\ref{eq-para-p}\) and \(dy=pdx\) to get a curve in the space.

奇解 | Singular Solution

Definition of Singular Solution

Assume \(y=\phi(x)\) defined on region \(J\) is a special solution of ODE

\[ F(x,y,y')=0 \]

its integral curve is

\[ \varGamma=\{(x,y): y=\phi(x),x\in J\} \]

If \(\forall M\in \varGamma\), there exists \(\delta>0\), such that a solution \(\psi(x)\) different from \(\phi(x)\) which is tangent to \(\phi(x)\) at \(M\) in \(O_\delta(M)\). Then we call \(y=\phi(x)\) is a singular solution of the above ODE.

In every point of \(y-\phi(x)\), there exists another solution. That is, uniqueness is broken.

Here we gives the necessary condition for the existence of singular solution.

necessary condition for the existence of singular solution

Assume \(F(x,y,p)\in C(G)\), where \(G\subset \mathbb{R}^3\), and has continuous partial derivatives with respect to \(y\) and \(p\), i.e. \(F_y'(x,y,p)\) and \(F_p'(x,y,p)\). If \(y=\phi(x)(x\in J)\) is a singular solution of the ODE, and

\[ (x,\phi(x),\phi'(x))\in G,\quad x\in J \]

then \(y=\phi(x)\) satisfies

\[ F(x,y,p)=0,\quad F'_p(x,y,p)=0. \]

The first equation is easy because of singular solution is also a solution.

We have to prove the second equation by contradiction.

Then there exists \((x_0,y_0,p_0)\in G\), such that we can use Implicite function theorem.

Using the two equations are called p-test equation, we can get a formula of singular solution, but we still have to test whether it is the solution of ODE and whether it is singular solution. Here we give another theorem.

Theorem for singular solution

Assume \(F(x,y,p)\in C^2(G)\), and \(y=\phi(x)\) obtained from p-test equation is a solution of ODE. If

\[ \begin{align} &F_y(x,\phi(x),\phi'(x))\neq 0 \label{F-y}\\ &F_{pp}(x,\phi(x),\phi'(x))\neq 0\label{F-pp}\\ \nonumber &F_p(x,\phi(x),\phi'(x))=0 \end{align} \]

then \(y=\phi(x)\) is the singular solution of ODE.

Note: if equation \(\ref{F-y}\) and \(\ref{F-pp}\) does not satisfies, we still could not ensure whether \(\y=phi(x)\) is a singular solution.