direction field and solution curve for differential equation.
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given function dy/dt = -ty^3
the solution of function is +-1/sqrt(t^2+C) and y(0) = +-1/sqrt(c).
I cannot deal with this solution with C
How can I find solution curves for the differential equation and combined graph of solution curve with direction field.
For solution curve, I used
syms y(t) c
sol = dsolve(diff(y, t) == -t*y^3, y(0) == [-1 1]/sqrt(c))
fplot (sol, [-3, 3])
For direction field
[T, Y] = meshgrid(-2:0.2:3, -1:0.2:2);
S = -t*y^3;
L = sqrt(1 + S.^2) <- I don't know what that means actually
quiver(T, Y, 1./L, S./L, 0.5) axis equal tight
Please help me!
esat gulhan on 15 Sep 2020
Maybe like this
if it works please accept the answer
syms y(t) c
sol = diff(y, t) == -t*y^3 %your equation
cond=y(0) == 1/sqrt(c); %condition for 1/sqrt(c)
sol2=(dsolve(sol,cond));%solution of differential in cond
cond2=y(0) == -1/sqrt(c);%condition for -1/sqrt(c)
sol3=(dsolve(sol,cond2));%solution of differential in cond2
for c=linspace(0.25,10,8); %c values for condition, i take it free, you can change it
t=linspace(-3,3,100);%t values in -3 3 with 100 divides
plot(t,subs(sol2),'b','LineWidth',1.5); %plot lines with cond 1
plot(t,subs(sol3),'b','LineWidth',1.5);grid on;%plot lines with cond 2
[t, y] = meshgrid(-3:0.2:3, -2:0.2:2);%limits of meshgrid
S = -t.*y.^3;
L = sqrt(1 + S.^2)
quiver(t, y, 1./L, S./L, 0.5)
set(gca, 'XLim', [-3 3], 'YLim', [-2 2]);%limits of graph
esat gulhan on 16 Sep 2020
Did you copy to code in a new blank script. It is not an extension of a code it is a new indepentend code. It must work. Impossıble not to work. plot(t, subs(sol2), 'b', 'LineWidth',1.5); can not be in line 34 in ,it is in line 12.
Plot will seen like this