---
title: 工程數學 03/15+17
---
###### tags: `工程數學 2021`

### 03/15
Ex. $(\sin(x)\cosh(y))dx=(cos(x)sinh(y))dy=0$
→ $M=\sin(x)\cosh(y)$
→ $N=cos(x)sinh(y)$
$\frac{\partial M}{\partial y}$ = $\sin(x) \sinh (y)$
$\frac{\partial N}{\partial x}$ = $\sin(x) \sinh (y)$
$\frac{\partial u}{\partial x}=\sin(x)\cosh(y)$ → $u=-\cos (x) \cosh (y) + k(y)$
$\frac{\partial u}{\partial y} = -cos (x) sinh (y) + \frac{dk(y)}{dy} = N = -\cos (x) \sinh (y)$
→ $\frac{dk(y)}{dy} = 0$ → $k(y) = C^*$




### 03/17

#### First-Order Linear Differential Equations
A 1st order DE is said to be linear if it can be writer as
$y' + p(x)y = r(x)$
1. r(x) = 0 → DE is homogeneous
2. r(x) ≠ 9 → DE is non-homogeneous

1. $y' + p(x)y = 0$
→ $\frac{dy}{dx} = -p(x)y$
→ $\frac{dy}{y} = -p(x)dx$
→ $\ln|y| = -\int{ p dx + c^* }$
→ $y = ce^{- s p dx}$
→ $y = ce^{-h}$, $h = \int{ p dx }$
* A homogeneous always has a trvial solution y(x) = 0
2. $y' = p(x) y = r(x)$
→ $\frac{dy}{dx} + py = r$
→ $(py - r)dx + dy = 0$
____M____ _N_
$\frac{\partial M}{\partial y} = p ≠ \frac{\partial N}{\partial x} = 0$
$\frac{1}{N} ( \frac{\partial M}{\partial y} - \frac{\partial }{\partial X}) = p(x)$ a function of r
$F(x) = e^{ \int{ \frac{1}{n} (\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) }dx }$
$= e^{\int{ pdx }}$
$e^{\int pdx}(y' + py) = e^{\int pdx}r$
→ $( e^{\int pdx} y )' = e^{\int pdx}r$
→ $e^{\int pdx}y = \int{( e^{\int pdx} r )dx + c}$
→ $y = ce^{-\int p dx} + e^{-\int{ pdx }} \int{ e^{\int{pdx}} r dx }$
→ $ce^{-h} + e^{-h} \int{ e^h dx },\ h=\int{pdx}$
* The solution of the homogeneous DE is a speied case of the solution of the non-homogeneous
另解： :+1:

Ex. $y'-y = e^{2x}$
$y' + py = r$
$p=-1, = e^{2x}$
$y= ce^{-h} + e^{-h} \int e^h r dx$
$h = \int{ pdx } = -x$
$e^{-h} = e^x, e^h = e^{0x}$
$y = ce^x + e^x \int{ e^{-x} e^{2x} dx }$
$= ce^x + e^{2x}$
另解： F(x) = e^{-x}

Ex. $xy' - 4y + 2x^2 + 4 = 0$
→ $y' - \frac{4}{x}y = -2x-\frac{4}{x}$
$p(x) = -\frac{4}{x}, (x) = -2x-\frac{4}{x}$
$y = ce^{-h} + e^{-h} \int {e^h r dx}$
$h = \int{pdx} = -4\ln|x|$
$e^{-h} = e^{4\ln|x|} = x^4$
$e^h = x^{-4}$
$y = cx^4 + x^4 \int{ x^{-4} (2x-\frac{4}{x}) dx}$
$= cx^4 + x^4 \int{ -2x^{-3} - 4x^{-5} dx }$
$= cx^4 + x^2 + 1$
另解： $F(x) = x^{-5}$

Certain non-linear DE can be transfered to linear
1. Bernoulli equation
$y' + p(x)y = r(x) y^a$
($a=0$ or $1$. DE is linear)
Let $u = y^{1-a}$
$u' = (1-a)y^{-a} y'$
$= (1-a)y^{-a}(ry^{a}-py)$
$= (1-a)r - (1-a)pu$
→ $u' + (1-a) pu = (1-a)r$
$(y' + py = r)$ linear!

Ex. $xy' + y = x^2 y^2$
→ $y' + x^{-1} y = x y^2$
(y' + py = ry^a)
Let $u = y^{1-2} = y^{-1}$
$u' = - y^{-2}(xy^{2} - x^{-1}y)$
$= -x+x^{-1}u$
→ $u' - x^{-1}u = -x$ linear
$u = ce^{-h} + e^{-h} \int{e^h r dx}$
$h = \int{p dx} = \int{-x^{-1}dx} = -\ln x$
$e^{-h} = e^{\ln x} = x$, $e^h = x^{-1}$
$u = cx+x\int{x^{-1}(-x)dx}$
$= cx-x^2$
$y^{-1} = cx - x^2$ → $y = {1}\over{cx - x^{2}}$

Ex. $y' + {1}\over{x} y = 3x^2y^3$
$(y'+py = ry^a)$
Let $u = y^{1-3} = y^{-2}$
$u' = -y^{-3}(3x^2y^3- {y}\over{x} )$
$ = -x^6 + {2}\over{x}u$
→ $u' - {2}\over{x}u = -6x^2$
$u = ce^{-h} + e^{-h} \int e^h r dx$
$e^{-h} = x^2$
$y = {1}\over{ \sqrt{cx^2-6x^3} }$