Solving a Second Order Differential Equation
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Steven Lord
il 29 Gen 2021
syms x t y(t)
dx = diff(x, t)
dy = diff(y, t)
x is not a function of t so its derivative with respect to t is 0.
y is a function of t so its derivative with respect to t is not 0. [It could be, if you substituted a constant into dy for y, but it's not always zero.]
dyForConstant = subs(dy, y, 5)
dyForNonconstant = subs(dy, y, sin(t))
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James Tursa
il 29 Gen 2021
Modificato: James Tursa
il 29 Gen 2021
Here is what I get:
>> syms y(t)
>> eqn = 4*diff(y,t,2)+32*diff(y,t)+24*y^3 == 192;
>> Dy = diff(y,t);
>> cond = [y(0)==1.5, Dy(0)==0];
>> ySol(t) = dsolve(eqn,cond)
Warning: Unable to find symbolic solution.
> In dsolve (line 216)
ySol(t) =
[ empty sym ]
Maybe not surprising that it can't find a symbolic solution given the y^3 term. So you could use a numeric solver instead. E.g.,
f = @(t,y) [y(2);(192 - 32*y(2) - 24*y(1)^3)/4];
[t,y] = ode45(f,[0 10],[1.5;0]);
plot(t,y(:,1)); grid on
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