Differential equations: solutions and equilibria

Three quick questions from earlier lessons.

A differential equation links a function with its derivatives. Is $y = e^{2x}$ a solution of $y' = 2y$ ?

A solution is a function, not a number.

A differential equation relates

y

, its derivatives and the variable. A function is a solution if substituting it makes both sides equal for every

x

. A constant solution

y = c

(an equilibrium) has

y' = 0

.

Find the equilibrium of $\dfrac{dy}{dt} = 3(y - 5)$ , and check $y = 5 + 2e^{3t}$ .

3(y - 5) = 0 \Rightarrow y = 5

Constant solution: y' = 0.

y' = 6e^{3t}

Differentiate the proposed solution.

It is a solution

Right side: 3(2e^{3t}) = 6e^{3t}. Equal.

Your turn: the equilibrium of $\dfrac{dy}{dt} = 2(4 - y)$ .

y = 4

Set 2(4 − y) = 0.

Which function is a solution of $\dfrac{dy}{dx} = 5y$ ?
1. $y = 3e^{ 5x}$
2. $y = e^{3x}$
3. $y = 5xy$
4. $y = e^{x} + 5$
Which function is a solution of $\dfrac{dy}{dx} = 6y$ ?
1. $y = 3e^{ 6x}$
2. $y = e^{3x}$
3. $y = 6xy$
4. $y = e^{x} + 6$

$\dfrac{dy}{dt} = 5(y - 3)$ . Find the constant (equilibrium) solution $y = c$ .
$\dfrac{dy}{dt} = 8(y - 2)$ . Find the constant (equilibrium) solution $y = c$ .

Three questions to check you have got it.

$\dfrac{dy}{dt} = 5(y - 3)$ . Find the constant (equilibrium) solution $y = c$ .
$\dfrac{dy}{dt} = 8(y - 2)$ . Find the constant (equilibrium) solution $y = c$ .
$\dfrac{dy}{dt} = 8(y - 1)$ . Find the constant (equilibrium) solution $y = c$ .

Differential equations

Watch out: equilibrium means $y' = 0$ , not $y = 0$ .

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