在Octave中，有没有办法解决1个变量的2个变量的方程

Question

我有一个MATLAB脚本，使用拉普拉斯变换解决了不均匀的一阶线性IVP。 （对于此示例，脚本设置为解决IVP ， 。）

syms x(t) s X;

a0 = -3;
x0 = 4;
rhs = t^2;

lhs = diff(x,t) + a0*x;
ode = lhs - rhs

Lx = X;
LDx = s*X - x0;
LHS = LDx + a0*Lx;
RHS = laplace(rhs,t,s);
IVP = LHS - RHS;

IVP = collect(IVP,X);

X = solve(IVP, X);
X = partfrac(X);

sol = ilaplace(X, s, t)
check1 = diff(sol,t) - 3*sol
check2 = vpa(subs(sol, t, 0))

如果我替换＆＃34;因素＆＃34; for＆＃34; collect＆＃34;，脚本几乎在Octave上工作，符号包链接到SymPy，除了＆＃34;解决＆＃34;命令https://www.mathworks.com/help/symbolic/solve.html。

是否有任何Octave（或SymPy，如果它可以作为一种变通方法）命令，它将作为MATLAB符号工具箱使用＆＃34;解决＆＃34;命令所以我可以用拉普拉斯变换用脚本解决IVP，所以我不必手动解决X，然后使用＆＃34; ilaplace＆＃34;？

提前感谢您提供的任何帮助。

Answer 1

以下是一些Octave脚本（大约至少）重现上述MATLAB脚本。您必须根据X将IVP = 0的解决方案手动输入到每个脚本的第2部分，但它们确实起作用。如果有人有任何方法让Octave在X方面解决IVP = 0，就像MATLAB求解函数一样，我很乐意听到它。

这对解决了$ \ dot {x} - 3x = t ^ 2 $，$ x（0）= 4 $。

第1部分：

syms x(t) s X;

a0 = -3;
x0 = 4;
rhs = t^2;

lhs = diff(x,t) + a0*x;
ode = lhs - rhs

Lx = X;
LDx = s*X - x0;
LHS = LDx + a0*Lx;
RHS = laplace(rhs,t,s); % The t and s in laplace aren't necessary, as they are default
IVP = LHS - RHS;

coeff = coeffs(IVP,X);
IVP = coeff*[1;X]

第2部分：

syms x(t) s X;

X = -1*((-4*s^3-2)/s^3)/(s-3)

X = partfrac(X);

sol = ilaplace(X, s, t)
check1 = diff(sol,t) - 3*sol
check2 = vpa(subs(sol, t, 0))

此对解决$ \ ddot {x} - 2 \ dot {x} - 3x = t ^ 2 $，$ x（0）= 4 $，$ \ dot {x}（0）= 5 $。< / p>

第1部分：

syms x(t) s X;

a1 =-2;
a0 = -3;
x0 = 4;
xdot0 = 5;
rhs = t^2;

Dx = diff(x,t);
D2x = diff(x,t,2);
lhs = D2x + a1*Dx + a0*x;
ode = lhs - rhs

Lx = X ;
LDx = s*X - x0;
LD2x = s^2*X - x0*s - xdot0;
LHS = LD2x + a1*LDx + a0*Lx;
RHS = laplace(rhs,t,s); % The t and s in laplace aren't necessary, as they are default
IVP = LHS - RHS;

coeff = coeffs(IVP,X);
IVP = coeff*[1;X]

第2部分：

syms x(t) s X;

a1 = -2;
a0 = -3;

X = -1*((-4*s^4 + 3*s^3 - 2)/s^3)/(s^2 - 2*s - 3)

X = partfrac(X);

sol = ilaplace(X, s, t)

Dsol = diff(sol,t);
D2sol = diff(sol,t,2);
check1 = D2sol + a1*Dsol + a0*sol
check2 = vpa(subs(sol, t, 0))
check3 = vpa(subs(Dsol, t, 0))

非常感谢所有的帮助和建议！真的很感激！

Answer 2

好的，所以，我的一个学生解决了这个问题（我将在本周晚些时候与他联系，看他是否希望公开承认他的解决方案）。

您只需要将coeff*[1;X]的结果定义为等于0的等式集，比如IVPEQ = coeff*[1;X] == 0，然后在此等式solve上使用符号包命令X = solve(IVPEQ, X) }。

以下是我之前的一阶IVP求解器版本与我学生的修改

syms x(t) s X;

a0 = -3;
x0 = 4;
rhs = t^2;

lhs = diff(x,t) + a0*x;
ode = lhs - rhs

Lx = X;
LDx = s*X - x0;
LHS = LDx + a0*Lx;
RHS = laplace(rhs,t,s); % The t and s in laplace aren't necessary, as they are default
IVP = LHS - RHS;

coeff = coeffs(IVP,X);
IVPEQ = coeff*[1;X] == 0;

X = solve(IVPEQ,X);

X = partfrac(X);

sol = ilaplace(X, s, t)

Dsol = diff(sol,t);
check1 = Dsol + a0*sol
check2 = vpa(subs(sol, t, 0))

这里是学生修改的二阶IVP求解器

syms x(t) s X;

a1 =-2;
a0 = -3;
x0 = 4;
xdot0 = 5;
rhs = t^2;

Dx = diff(x,t);
D2x = diff(x,t,2);
lhs = D2x + a1*Dx + a0*x;
ode = lhs - rhs

Lx = X ;
LDx = s*X - x0;
LD2x = s^2*X - x0*s - xdot0;
LHS = LD2x + a1*LDx + a0*Lx;
RHS = laplace(rhs,t,s); % The t and s in laplace aren't necessary, as they are default
IVP = LHS - RHS;

coeff = coeffs(IVP,X);
IVPEQ = coeff*[1;X] == 0;

X = solve(IVPEQ,X);

X = partfrac(X);

sol = ilaplace(X, s, t)

Dsol = diff(sol,t);
D2sol = diff(sol,t,2);
check1 = D2sol + a1*Dsol + a0*sol
check2 = vpa(subs(sol, t, 0))
check3 = vpa(subs(Dsol, t, 0))

再次感谢@Tasos_Papastylianou，感谢您的大力帮助！