How can i make the integral curves?

Question

Yoanna Vasileva 2021-1-14

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/717183-how-can-i-make-the-integral-curves

评论： Alan Stevens 2021-1-16

采纳的回答： Alan Stevens

My task is to find the general equation of 2x^2y′ − xy − y3 = 0

and draw in the square {x ∈ [1,6], y ∈ [-3,2]} a vector field

tangent to the integral curves of the equation.

and Draw the integral curves of the equation.

[x,y] = meshgrid(1:0.5:6,-3:0.5:2);
u = ones(size(x));
v = (x.*y+y.^3)/2*x.^2;
r = sqrt(u.^2 + v.^2);
plot([0,0],[-3,2])
hold on
plot([1,6],[0,0])
quiver(x,y,u./r,v./r,'k');
axis ([1 6 -3 2]);
sol = dsolve('2*t^2*Dy=t*y+y^3');
for k = 2.1:0.4:2.6
    sol1 = subs(sol,'C7', k^3);
    xx = 1:.1:6;
    yy = subs(sol1,'t',xx);
    hold on
    plot(xx,yy,'r');
    hold off;
end

This is my code the error which is giving me is "Error using plot

A numeric or double convertible argument is expected"

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Answer 1

Alan Stevens 2021-1-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/717183-how-can-i-make-the-integral-curves#answer_598835

Like so?

[x,y] = meshgrid(1:0.5:6,-3:0.5:2);
u = ones(size(x));
v = (x.*y+y.^3)/2*x.^2;
r = sqrt(u.^2 + v.^2);
plot([0,0],[-3,2])
hold on
plot([1,6],[0,0])
quiver(x,y,u./r,v./r,'k');
axis ([1 6 -3 2]);
sol = dsolve('2*t^2*Dy=t*y+y^3');
xx = 1:.1:6;
for k = 2.1:0.4:2.6
    sol1 = subs(sol,'C1', k^3);   %%% C1 not C7
    yy = double(subs(sol1,'t',xx));
end
plot(xx,yy,'r');

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Yoanna Vasileva 2021-1-15

Thank you for your response. But is giving me this errors now:

Error using symengine (line 59)

DOUBLE cannot convert the input expression into a double array.

If the input expression contains a symbolic variable, use VPA.

Error in sym/double (line 583)

Xstr = mupadmex('symobj::double', S.s, 0);

Error in new (line 14)

yy = double(subs(sol1,'t',xx));

I am not sure if my sol is okay I mean if the given equation is equivalent to the sol solution. Also I am using 2015 MatLab and this may the problem. How would you solve it. I don`t know if my way of solving is right.