![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/965510/image.png)
Solving 2nd Order Differential Equation Symbolically
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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 = ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/965485/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/965485/image.png)
Error using eval
Unrecognized function or variable 'C1'.
I'd greatly appreciate any assistance.
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採用された回答
Star Strider
2022 年 4 月 15 日
The
constants are the initial condition. They must be defined.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/965510/image.png)
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)')
.
16 件のコメント
Torsten
2022 年 4 月 16 日
編集済み: Torsten
2022 年 4 月 16 日
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 ?
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