Answer to Question #213425 in Differential Equations for johny

Question

sketch the gradient field for the differential equation dy/dt=t-y² for t=-1......,1

y=-1,.......,1

Answer 1

Let's calculate the values of the function $f\left(t,y\right)=t-y^2$ in the area $A=\left\{\left(t,y\right)\left|-1\le t\le1,\,\,\,-1\le y\le 1\right.\right\}$ .To be more detailed, let's take a step of $\Delta t=\Delta y=0.5$ . Then,

$$t=-1 : \left\{\begin{array}{l} f(-1,-1)=-1-\left(-1\right)^2=-2\\[0.3cm] f(-1,-0.5)=-1-\left(-0.5\right)^2=-1.25\\[0.3cm] f(-1,-0)=-1-0^2=-1\\[0.3cm] f(-1,0.5)=-1-0.5^2=-1.25\\[0.3cm] f(-1,1)=-1-1^2=-2 \end{array}\right.\\[0.3cm] t=-0.5 : \left\{\begin{array}{l} f(-0.5,-1)=-0.5-\left(-1\right)^2=-1.5\\[0.3cm] f(-0.5,-0.5)=-0.5-\left(-0.5\right)^2=-0.75\\[0.3cm] f(-0.5,-0)=-0.5-0^2=-0.5\\[0.3cm] f(-0.5,0.5)=-0.5-0.5^2=-0.75\\[0.3cm] f(-0.5,1)=-0.5-1^2=-1.5 \end{array}\right.\\[0.3cm] t=0 : \left\{\begin{array}{l} f(0,-1)=0-\left(-1\right)^2=-1\\[0.3cm] f(0,-0.5)=0-\left(-0.5\right)^2=-0.25\\[0.3cm] f(0,-0)=0-0^2=-0\\[0.3cm] f(0,0.5)=0-0.5^2=-0.25\\[0.3cm] f(0,1)=0-1^2=-1 \end{array}\right.\\[0.3cm] t=0.5 : \left\{\begin{array}{l} f(0.5,-1)=0.5-\left(-1\right)^2=-0.5\\[0.3cm] f(0.5,-0.5)=0.5-\left(-0.5\right)^2=0.25\\[0.3cm] f(0.5,-0)=0.5-0^2=0.5\\[0.3cm] f(0.5,0.5)=0.5-0.5^2=0.25\\[0.3cm] f(0.5,1)=0.5-1^2=-0.5 \end{array}\right.\\[0.3cm] t=1 : \left\{\begin{array}{l} f(1,-1)=1-\left(-1\right)^2=0\\[0.3cm] f(1,-0.5)=1-\left(-0.5\right)^2=0.75\\[0.3cm] f(1,-0)=1-0^2=1\\[0.3cm] f(1,0.5)=1-0.5^2=0.75\\[0.3cm] f(1,1)=1-1^2=0 \end{array}\right.$$

In the figure it looks like this