首页 > 代码库 > 微分方程笔记

微分方程笔记

2024-07-04 07:47:38   222人阅读

线性常微分方程解法


一阶线性微分方程

dydx+P(x)y=Q(x)
<script id="MathJax-Element-1" type="math/tex; mode=display">\frac{{dy}}{{dx}} + P(x)y = Q(x)</script>
对应的齐次线性方程
dydx+P(x)y=0
<script id="MathJax-Element-2" type="math/tex; mode=display">\frac{{dy}}{{dx}} + P(x)y = 0</script>
此齐次方程可以用分离变量法求得通解: y=Ce?∫P(x)dx<script id="MathJax-Element-3" type="math/tex">y = C{e^{-\int {P(x)dx} }}</script>

常数变易法求非齐次线性方程的通解:
将齐次方程的通解中的C换成u(x): y=ue?∫P(x)dx<script id="MathJax-Element-4" type="math/tex">y = u{e^{ - \int {P(x)dx} }}</script> 带入非齐次线性方程,可求得其解为:
y=Ce?∫P(x)dx+e?∫P(x)dx∫Q(x)e∫P(x)dxdx
<script id="MathJax-Element-5" type="math/tex; mode=display">y = C{e^{ - \int {P(x)dx} }} + {e^{ - \int {P(x)dx} }}\int {Q(x){e^{\int {P(x)dx} }}dx} </script>
即非齐次线性方程的通解等于对应的齐次方程的通解加非齐次方程的一个特解

二 伯努利方程

dydx+P(x)y=Q(x)yn
<script id="MathJax-Element-6" type="math/tex; mode=display">\frac{{dy}}{{dx}} + P(x)y = Q(x){y^n}</script>
可变换为一阶线性微分方程:
设z=y1?n<script id="MathJax-Element-7" type="math/tex">z=y^{1-n}</script>,可化为:
dzdx+(1?n)P(x)z=(1?n)Q(x)
<script id="MathJax-Element-8" type="math/tex; mode=display">\frac{{dz}}{{dx}} + (1 - n)P(x)z = (1 - n)Q(x)</script>

三 可降阶的高阶微分方程

1)
y(n)=f(x)
<script id="MathJax-Element-9" type="math/tex; mode=display">y^{(n)}=f(x)</script>
两边积分,可降为n-1阶的微分方程
y(n?1)=∫f(x)dx+C1
<script id="MathJax-Element-10" type="math/tex; mode=display">y^{(n-1)}=\int f(x)dx+C_1</script>
连续n次积分,可得方程含有n个任意常数的通解
2)
y′′=f(x,y′)
<script id="MathJax-Element-11" type="math/tex; mode=display">y‘‘ = f(x,y‘)</script>
设 y′=p<script id="MathJax-Element-12" type="math/tex">y‘=p</script>,则y′′=p′<script id="MathJax-Element-13" type="math/tex">y‘‘=p‘</script>, 原方程变换为关于变量x,p的一阶微分方程
p′=f(x,p)
<script id="MathJax-Element-14" type="math/tex; mode=display">p‘=f(x,p)</script>
其通解为
dydx=φ(x,C1
<script id="MathJax-Element-15" type="math/tex; mode=display">\frac{dy}{dx}=\varphi(x,C_1</script>
积分可得原方程通解:
y=∫φ(x,C1)dx+C2
<script id="MathJax-Element-16" type="math/tex; mode=display"> y=\int \varphi(x,C_1)dx+C_2</script>
3)
y′′=f(y,y′)
<script id="MathJax-Element-17" type="math/tex; mode=display">y‘‘=f(y,y‘)</script> (方程中不含自变量x的显式)
令y′=p<script id="MathJax-Element-18" type="math/tex">y‘=p</script>, 有
y′′=dpdx=dpdy dydx=pdpdy
<script id="MathJax-Element-19" type="math/tex; mode=display">y‘‘=\frac{dp}{dx} =\frac{dp}{dy} \frac{dy}{dx} =p \frac{dp}{dy} </script>原方程化为:
pdpdy=f(y,p)
<script id="MathJax-Element-20" type="math/tex; mode=display">p\frac{dp}{dy}=f(y,p)</script>
其通解为
y′=p=φ(y,C1)
<script id="MathJax-Element-21" type="math/tex; mode=display"> y‘=p=\varphi(y,C_1)</script>
分离变量并积分可得原方程的通解:
∫dyφ(y,C1)=x+C2
<script id="MathJax-Element-22" type="math/tex; mode=display"> \int \frac{dy}{\varphi(y,C_1)}=x+C_2</script>




来自为知笔记(Wiz)


联系
我们