Separation of variables and initial conditions

Warm-up

Three quick questions from earlier lessons.

Discover

To solve $\dfrac{dy}{dx} = 2xy$ , Ana integrates both sides with respect to $x$ and writes $y = x^2 y + C$ . Why is that not a solution method?

$y$ is not a constant: the variables must be separated first
It is fine, it just needs a value for $C$

Key idea

Gather each variable with its own differential.

Separable equation

\dfrac{dy}{dx} = g(x)h(y)

: write

\dfrac{dy}{h(y)} = g(x)\,dx

, integrate both sides (one constant

C

), solve for

y

, then use the initial condition to find

C

.

Worked example

Solve $\dfrac{dy}{dx} = 2xy$ with $y(0) = 3$ .

\dfrac{dy}{y} = 2x\,dx

Separate.

\ln|y| = x^2 + C

Integrate.

y = 3e^{x^2}

Exponentiate: y = Ae^{x²}; y(0) = 3 gives A = 3.

Your turn, guided

Your turn: $\dfrac{dy}{dx} = 3y$ , $y(0) = 2$ .

y = 2e^{3x}

ln|y| = 3x + C, so y = Ae^{3x}.

Practice ★

Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 9$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 2$ .

Watch out!

Practice ★★

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 8$ (so $y > 0$ ). Give $y$ in terms of $x$ .
Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 6$ (so $y > 0$ ). Give $y$ in terms of $x$ .

Solve $\dfrac{dy}{dx} = y^2$ with $y(0) = \dfrac{1}{ 3 }$ .

Boss problem. Solve $\dfrac{dy}{dx} = 1x^2 y$ with $y(0) = 4$ .

Exit ticket

Three questions to check you have got it.

Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 9$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 2$ .

Summary card

Separation of variables

$\dfrac{dy}{h(y)} = g(x)\,dx$ .
Integrate both sides, one $+C$ .
Solve for $y$ ; use the initial value.

Watch out: $e^{x + C} = Ae^x$ , not $e^x + C$ .

About this lesson