We consider the differential equation $$\frac{\dd y}{\dd t}=\frac{4-2t}{3y^2-5}$$

Assignments

1. Determine the general solution of the differential equation in implicit form by the method of separation of variables.
2. Draw a slope field and use the contour function to draw some integral curves in it.

Worked-out solution

1. We write the differential equation in differential form and apply separation of variables: $$(3y^2-5)\,\dd y=(4-2t)\,\dd t$$ When we integrate on both sides we get $$\int (3y^2-5)\,\dd y=\int (4-2t)\,\dd t$$ So: $$y^3 -5y=4t-t^2+C$$ for some constant $C$. In other words, $$F(t,y)=C$$ where $$F(t,y)=y^3-5y+t^2-4t$$
2. You are asked to draw a slope field and some integral curves in one diagram. The last aspect concerns the drawing of contours of $$F(t,y)=y^3-5y+t^2-4t$$ The following code produces the diagram below:

   >> clear
   >> [t,y] = meshgrid(-3:0.5:7, -4:0.5:4);
   >> dy = (4-2*t)./(3*y.^2-5);
   >> dt = ones(size(dy));
   >> L = sqrt(dt.^2 + dy.^2);
   >> dyu = dy./L;
   >> dtu = dt./L;
   >> figure
   >> % drawing the slope field
   >> quiver(t, y, dtu, dyu, 0.3, 'r'), axis tight
   >> xlabel 't', ylabel 'y';
   >> hold on  % continue drawing in the same window
   >> % Draw integral curves in blue
   >> [t,y] = meshgrid(-3:0.1:7, -4:0.1:4);
   >> contour(t,y, y.^3 - 5*y + t.^2 - 4*t, [-8,-4, 0, 4, 8], 'b');

The integral curves describe more than one solution curve. The outer integral curve does this for 3 solution curves that depend on an initial value. We distinguish as solution curves

1. the top part with $y$ values greater than $\frac{1}{3}\sqrt{15}$;
2. the middle part with $y$ values between $-\frac{1}{3}\sqrt{15}$ and $\frac{1}{3}\sqrt{15}$;
3. the lower part smaller than $-\frac{1}{3}\sqrt{15}$