1stODE

Separable Variable Method¶

for any ODE satisfying

\[ \frac{dy}{dx}=g(x)h(y) \]

linear Equation¶

standard form¶

\[y'+P(x)y=f(x)\]

integrating factor¶

\[ M(x)=e^{\int{P(x)dx}} \]

solving method¶

standard form

\[y'+Py=f\]
multiple M(x)

\[y'⋅M+P⋅My=f⋅M\]
notice that

\[M'=P⋅M\]
thus (form 2.)

\[M\cdot y'+M'\cdot y=M\cdot f\]
according to product rule

\[\frac{d}{dx}(M\cdot y)=M\cdot f\]
integrate both sides

\[M\cdot y = \int{M\cdot f dx} + c\]
check initial value and singular points ¹

Exact Equation¶

Any 1^st ODE can be written as

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

If \(\exists f(x, y)\), s.t.

\[df(x,y)=\frac{∂f(x,y)}{∂x}dx+\frac{\partial f(x,y)}{∂y}dy=0\]

which means

\[\frac{∂f(x, y)}{∂x}=M(x, y)\]

\[\text{and}\]

\[\frac{∂f(x, y)}{∂x}=N(x, y)\]

, then

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

is an Exact Equation.

for an exact equation

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

due to

\[df(x,y)=\frac{∂f(x,y)}{∂x}dx+\frac{\partial f(x,y)}{∂y}dy=0\]

we know that

\[ f(x,y) = c \]

solution to exact equation¶

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

check the 1^st ODE is exact equation

\[\frac{∂M}{∂y}=\frac{∂N}{∂x}=\frac{∂f}{∂x∂y}\]
then

\[f(x,y) = \int{M\cdot dx} + g(y)\]
and we can get \(g(y)\) by

\[\frac{\partial}{\partial y}f(x,y)=N(x, y)\]
solve

\[f(x, y)=c\]

modify non-exact equations to exact equations¶

given a 1^st ODE which isn't exact equation

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

meaning

\[\frac{∂M}{∂y}\neq \frac{∂N}{∂x}\]

we can try to find an integrating factor \(\mu\) s.t.

\[ \mu M dx + \mu N dy = 0 \]

is an exact equation.

check if this equation satisfy either

1. > > \(\(h(y)=\frac{N_x-M_y}{M}\)\) > > is independent to \(x\). (dependent to \(y\) alone)

or

2. > > \(\(h(x)=\frac{M_y-N_x}{N}\)\) > > is independent to \(y\). (dependent to \(x\) alone)

if neither of these condition were satisfied, the equation cannot be exact equation.

otherwise, the equation can be multiplied by \(\mu\) to become exact equation, and solve by the method mentioned above.

\[\mu(y)=e^{\int {h(y)dy}}\]

or

\[\mu(x)=e^{\int {h(x)dx}}\]

see proof.

homogeneous Equation¶

There are two different definition of homogeneous equation.

homogeneous Linear Equation¶

The constant term of linear Equation is zero

In 1^st ODE case

\[y'+P(x)y=f(x)\]

The equation is homogeneous if f(x) = 0

see further definition.

homogeneous 1st ODE¶

以下兩種定義等價。

ODE in \(y' = f(x,y)\) form

The equation is homogeneous if

\[y'=f(x,y)=f(tx,ty)\]
ODE in \(Mdx + Ndy = 0\) form

If \(M(x,y)\) and \(N(x,y)\) are homogeneous functions of the same degree, then the 1^st ODE is homogeneous.

Quick Check Guide : sum of powers (指數之和)

Solution to 1st ODE¶

Set \(y=ux\), \(dy = udx +xdu\),

then use separable variable method

Bernoulli's Equations¶

aka 白努力 equation

standard form¶

\[y'+P(x)y=f(x)\cdot y^n\]

Solution¶

Assume

\[u=y^{1-n}\]

according to chain rule, we can write

\[\frac{dy}{dx}=\frac{1}{1-n} u^{\frac{n}{1-n}} \frac{du}{dx}\]

代入 standard form

\[ \begin{align} \frac{1}{1-n} u^{\frac{n}{1-n}} \frac{du}{dx} &+ P(x)\cdot u^{\frac{1}{1-n}} &= f(x)\cdot u^{\frac{n}{1-n}} \\ \frac{du}{dx} &+ P(x)(1-n)\cdot u &= (1-n)f(x) \end{align} \]

then use the method of solving linear 1^st ODE.

Ax + By + C¶

If the 1^st ODE has the form

\[\frac{dy}{dx} = f(Ax+By+C),\quad (B\neq 0)\]

set \(u=Ax+By+C\)

then

\[ \begin{gather} du = Adx + Bdy\\ \downarrow\\ \frac{dy}{dx} = \frac{1}{B}\frac{du}{dx}-\frac{A}{B}\\ \downarrow\\ \frac{1}{B}\frac{du}{dx}-\frac{A}{B}=f(u) \end{gather} \]

and solve it by any method above.

singular points

\(\forall x\) s.t.

\[P(x)\rightarrow \infty \quad \text{or} \quad f(x) \rightarrow \infty\]

singular points cannot be solution.

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