Solving 2nd Order Differential Equation Symbolically

80 次查看（过去 30 天）

显示更早的评论

Jordan Stanley 2022-4-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1697495-solving-2nd-order-differential-equation-symbolically

编辑： Jordan Stanley 2022-4-16

采纳的回答： Star Strider

在 MATLAB Online 中打开

Hello,

I have the 2nd order differential equation: y'' + 2y' + y = 0 with the initial conditions y(-1) = 0, y'(0) = 0.

I need to solve this equation symbolically and graph the solution.

Here is what I have so far...

syms y(x)
Dy = diff(y);
D2y = diff(y,2);
ode = D2y + 2*Dy + y == 0;
ySol = dsolve(ode,[y(-1)==0,Dy(0)==0])
a = linspace(0,1,20);
b = eval(vectorize(ySol));
plot(a,b)

But I get the following output.

ySol =

Error using eval

Unrecognized function or variable 'C1'.

I'd greatly appreciate any assistance.

采纳的回答

Star Strider 2022-4-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1697495-solving-2nd-order-differential-equation-symbolically#answer_943620

在 MATLAB Online 中打开

The

constants are the initial condition. They must be defined.

syms y(x) y0

Dy = diff(y);

D2y = diff(y,2);

ode = D2y + 2*Dy + y == 0;

ySol(x,y0) = dsolve(ode,[Dy(0)==0,y(-1)==0,y(0)==y0])

ySol(x, y0) =

% a = linspace(0,1,20);

% b = eval(vectorize(ySol));

figure

fsurf(ySol,[0 1 -1 1])

xlabel('x')

ylabel('y_0 (Initial Condition)')

16 个评论
显示 14更早的评论隐藏 14更早的评论

Jordan Stanley 2022-4-15

在 MATLAB Online 中打开

Hello again,

Upon running your initially suggested code

syms y(x) y0 
Dy = diff(y);
D2y = diff(y,2);
ode = D2y + 2*Dy + y == 0;
ySol(x,y0) = dsolve(ode,[Dy(0)==0,y(-1)==0,y(0)==y0])
% a = linspace(0,1,20);
% b = eval(vectorize(ySol));
figure
fsurf(ySol,[0 1  -1 1])
xlabel('x')
ylabel('y_0 (Initial Condition)') 

I get the following error message.

Error using sym/subsindex

Invalid indexing or function definition. Indexing must follow MATLAB indexing. Function arguments must be symbolic variables, and function body must be sym expression.

Related documentation

I have the following code in the section above. I'm not sure if this causes the issue.

% Finds solution to the DE
syms y(x)
Dy = diff(y);
D2y = diff(y,2);
ode = D2y + 2*Dy + y == 0;
ySol = dsolve(ode,[y(-1)==0,Dy(0)==0])
dySol = diff(ySol,x);
ySol = @(x) exp(-x).*(1+x);
dySol = @(x) -exp(-x).*x;
% Sets up directional field
[x,y]=meshgrid(-4:0.5:4,-4:0.5:4);
u = y;              % x1' = y'
v = - 2*y - x;      % x2' = y'' = - 2*y' - y
u1 = u./sqrt(u.^2+v.^2);
v1 = v./sqrt(u.^2+v.^2);
quiver(x, y, u1, v1, 0.6)
xlabel('x-axis')
ylabel('y-axis')
axis on
axis([-4 4 -4 4]);
title('Direction Field')
% Prints the solution curve corresponding to the initial conditions.
hold on
plot(ySol(-4:0.5:4),dySol(-4:0.5:4),"LineWidth",2);  
hold off

Thanks

Torsten 2022-4-16

编辑：Torsten 2022-4-16

How should it be possible to solve 1. without 2. ? If you don't know the solution, you can't trace a solution curve. Or what's your opinion ?

Anyhow - I think your instructors overlooked that the equation together with its initial conditions does not only give one curve, but infinitly many. So "graphing the solution" will become difficult. But Star Strider's answer for this situation looks fine for me.

But you say you get an error. What's your code and what's the error message ?

Jordan Stanley 2022-4-16

编辑：Jordan Stanley 2022-4-16

Well I should mention that we were supposed to choose a 2nd order differential eqaution initial value problem from our textbook and maybe I just happened to choose a more complicated eqaution to use.

Ok, well interestingly enough upon running the code agin that Star Strider suggested I did not get an error message this time.

请先登录，再进行评论。