# Differential Equation ,Finding solution by sketching the graph

"Use direction field and Isocline make a qualitative sketch of the solution , determine equilibrium values and classify them ?"

$y'=y-\sqrt y$ It is clear that $y=0 , y=1$ are equilibrium points but for these types of question which finding solution for DE is hard how to find which equilibrium point is stable or unstable without sketching vector field ?

• I don't understand the question. – Emanuele Paolini Feb 19 '13 at 21:25
• @EmanuelePaolini Is there a way to find which equilibrium point is stable or unstable without sketching vector field ? – Node.JS Feb 19 '13 at 21:27
• @Hooman: yes. Hint: we have to just look at the derivatives sign or if it equals zero to determine if it is a 'sink'. source or node and can draw a phase line from this, that correlates to the nice phase plot you gave. – Amzoti Feb 19 '13 at 21:32

This is a powerful qualitative tool that helps us get a general understanding of the behavior. When the DEQ is not solvable (other than using numerical methods or otherwise), this is the next best thing - we do qualitative analyses. Just look at how much information you were able to garner just from the direction field and plotting typical solutions using numerical methods on top of those.

We have to just look at the derivatives sign or if it equals zero to determine if it is a 'sink', 'source' or 'node' and can draw a phase line from this, that correlates to the nice phase plot you gave.

Theorem (Stability and Instability Conditions)

Let $f$ and $f′$ be continuous.

• The equation $y′ = f(y)$ has a sink at $y = y_0$ provided $f(y_0) = 0$ and $f′(y_0) \lt 0$ (stable).

• An equilibrium $y = y_1$ is a source provided $f(y_1) = 0$ and $f′(y_1) \gt 0$ (unstable).

• There is no test when $f′$ is zero at an equilibrium (indeterminate). The no-test case can sometimes be decided by an additional test:

• (a) Equation $y′ = f(y)$ has a sink at $y = y_0$ provided $f(y)$ changes sign from positive to negative at $y = y_0$.

• (b) Equation $y′ = f(y)$ has a source at $y = y_0$ provided $f(y)$ changes sign from negative to positive at $y = y_0$.

In your example, $y' = f(y) = y - \sqrt{y}$, which is an autonomous system.

You already know the equilibrium points, now, use the theorem and then compare this qualitative method with what you already have in the diagram.

Regards

• Very nice work! $+ 1 \land \color{green}{\bf\Large\checkmark}$ – amWhy May 1 '13 at 2:46
• Awesome, good for you! Have a wonderful slumber! – amWhy May 1 '13 at 4:30

You have an autonomous equation: $$y' = f(y).$$ Every zero of $f$ identifies a constant solution: if $f(y_0)=0$ then $y(x)=y_0$ is a solution. On the other hand when $f$ is positive the solutions are increasing, when $f$ is negative the solutions are decreasing. Hence if $f'(y_0) > 0$ the solution is instable (as $x$ increases) while if $f'(y_0)<0$ the solution is stable.