求下列一阶微分方程的解

如题所述

第1个回答  2022-11-10

求下列一阶微分方程的解


(1):y = (√x + C/√x)²
(2):y = - cosx + sinx/x + C/x
(3):4x = 2y² + 2y + 1 + Ce^(2y)

过程如下。
(1):
xy' + y = 2√(xy),为Bernoulli方程
令v = √y --> y=v² --> dy/dx = 2v*dv/dx
x*2v*dv/dx + v² = 2√x*v
dv/dx + v/2x = 1/√x,积分因子e^∫ dx/(2x) = √x
√x * (dv/dx+v/2x) = 1
√x*v = ∫ dx = C + x
√y = (C + x)/√x = √x + C/√x
y = (√x + C/√x)²

(2):
dy/dx + y/x - sinx = 0
dy/dx + y/x = sinx,积分因子e^∫ dx/x = x
x*(dy/dx+y/x) = xsinx
xy = ∫ xsinx dx = - xcosx + sinx + C
y = - cosx + sinx/x + C/x

(3):
dy/dx = 1/(2x - y²)
dx/dy = 2x - y²
dx/dy - 2x = - y²,积分因子e^∫ (- 2) dy = e^(- 2y)
e^(- 2y)*(dx/dy-2x) = - y²e^(- 2y)
e^(- 2y)*x = - ∫ y²e^(- 2y) dy
x = e^(2y)*[(1/4)(2y² + 2y + 1)e^(- 2y) + C]
4x = 2y² + 2y + 1 + Ce^(2y)

一阶微分方程的解法

这是一阶线性非齐次方程,先解相应的齐次方程;
dx/dt=x,
dx/x=dt,
ln|x|=t+C1,
x=Ce^t.
再用常数变易法,设x=ue^t,
dx/dt=(du/dt)e^t+ue^t=x+t=ue^t+t,
(du/dt)e^t=t,
du=te^(-t)dt,
u=C-(t+1)e^(-t),
x=Ce^t-t-1.

matlab一阶微分方程的解法

Examples:

dsolve('Dx = -a*x') returns

ans = exp(-a*t)*C1

x = dsolve('Dx = -a*x','x(0) = 1','s') returns

x = exp(-a*s)

y = dsolve('(Dy)^2 + y^2 = 1','y(0) = 0') returns

y =
[ sin(t)]
[ -sin(t)]

S = dsolve('Df = f + g','Dg = -f + g','f(0) = 1','g(0) = 2')
returns a structure S with fields

S.f = exp(t)*cos(t)+2*exp(t)*sin(t)
S.g = -exp(t)*sin(t)+2*exp(t)*cos(t)

Y = dsolve('Dy = y^2*(1-y)')
Warning: Explicit solution could not be found; implicit solution returned.
Y =
t+1/y-log(y)+log(-1+y)+C1=0

dsolve('Df = f + sin(t)', 'f(pi/2) = 0')
dsolve('D2y = -a^2*y', 'y(0) = 1, Dy(pi/a) = 0')
S = dsolve('Dx = y', 'Dy = -x', 'x(0)=0', 'y(0)=1')
S = dsolve('Du=v, Dv=w, Dw=-u','u(0)=0, v(0)=0, w(0)=1')
w = dsolve('D3w = -w','w(0)=1, Dw(0)=0, D2w(0)=0')
y = dsolve('D2y = sin(y)'); pretty(y)

matlab解一阶微分方程

在Matlab下输入:edit,然后将下面两行百分号之间的内容,复制进去,储存
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function dy=zhidao_kingwolf(t,y)
%y如果在你的区间[-0.0015,0.0015],用你的i的表示式,
%不在区间内的话,我直接将i赋成0
if ((y>=-0.0015)&&(y<=0.0015))
i=(0.0006455*exp(0.4182*(10*y/0.001+21)) + (-3.971e+004)*exp(-0.4854*(10*y/0.001+21)))*0.1/100;
else
i=0;
end
dy=240*sin(2*pi*50*t)-i;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
在Matlab下面输入:
t_end=0.04;
x0=0;
[t,x]=ode45('zhidao_kingwolf',[0,t_end],x0);
plot(t,x);

求一阶微分方程解法

模仿高数上册最后一章中的齐次方程去解,左右除以Z方,令y/z=t,得出dy/dz与t的关系,之后再模仿书中的解法即可

一阶微分方程求解的方法?

一阶微分其实就是一介导数,对于刚学高数的来说,要很快改变高中导数的写发有点…把它写成导数就可以熟悉的解了

解一个一阶微分方程

还是把原题打出来吧,这样做有点儿困难

解一道一阶微分方程

dy/dx=e^2x/e^y
e^ydy=e^2xdx
积分:e^y=0.5e^2x+C
代入y(0)=0, 得:1=0.5+C
得C=0.5
故e^y=0.5e^2x+0.5
y=ln(0.5e^2x+0.5)

数学 一阶微分方程

解:∵y²+y'=1
==>y'=1-y²
==>dy/(1-y²)=dx
==>[1/(1+y)+1/(1-y)]dy=2dx
==>ln│1+y│-ln│1-y│=2x+ln│C│ (C是常数)
==>(1+y)/(1-y)=Ce^(2x)
==>y=[Ce^(2x)-1]/[Ce^(2x)+1]
∴原方程的通解是y=[Ce^(2x)-1]/[Ce^(2x)+1]。

一阶微分方程问题

说明题目的条件给多了,并且出现了矛盾。
如果给出的条件是:在t=10s时,速度为50m/s,
或者 在t=10s时,外力为4N,
就不会出现问题了。