Topic 1 of 6

Separable and first-order linear equations

Step 20 · Differential equations topics

Solve first-order equations by separating the variables or with an integrating factor, and use them to model growth, cooling and mixing.

Separable equations

A differential equation links an unknown function with its derivatives, and a solution is a function that makes it true. In a separable equation, $\frac{dy}{dx} = g(x)h(y)$ , move every y to one side and every x to the other, then integrate both sides.

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

Example: Exponential growth

Solve $\frac{dy}{dt} = 0.3y$ with y(0) = 50.

Separate: $\frac{dy}{y} = 0.3\,dt$ .
Integrate: $\ln|y| = 0.3t + C$ , so $y = Ae^{0.3t}$ .
y(0) = 50 gives A = 50.

Answer $y = 50e^{0.3t}$

Example: A product

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

Separate: $y^{-2}\,dy = x\,dx$ .
Integrate: $-\frac{1}{y} = \frac{x^{2}}{2} + C$ , and y(0) = 1 gives C = −1.
So $\frac{1}{y} = 1 - \frac{x^{2}}{2}$ .

Answer $y = \frac{2}{2 - x^{2}}$

Integrating factors

A first-order linear equation has the form y′ + p(x)y = q(x). Multiply through by the integrating factor $\mu = e^{\int p\,dx}$ . The left side becomes the derivative of μy, and one integration finishes the job.

(\mu y)' = \mu q

Example: A constant coefficient

Solve y′ + 2y = 6 with y(0) = 1.

μ = $e^{2x}$ , and the equation becomes $(e^{2x}y)' = 6e^{2x}$ .
Integrate: $e^{2x}y = 3e^{2x} + C$ , so $y = 3 + Ce^{-2x}$ .
y(0) = 1 gives C = −2.

Answer $y = 3 - 2e^{-2x}$

Example: A varying coefficient

Solve $y' + \frac{1}{x}y = x$ for x > 0.

$\mu = e^{\ln x} = x$ , so (xy)′ = x².
Integrate: $xy = \frac{x^{3}}{3} + C$ .

Answer $y = \frac{x^{2}}{3} + \frac{C}{x}$

Modeling with first-order equations

Situation	Equation	Behavior
Growth or decay	$y' = ky$	Exponential
Cooling	$T' = -k(T - T_{s})$	Approaches the surroundings
Logistic growth	$P' = rP\left(1 - \frac{P}{K}\right)$	Levels off at K
Mixing tank	Rate in minus rate out	Approaches a steady level

Example: Cooling coffee

Coffee at 90°C sits in a 20°C room and cools to 60°C in 10 minutes. Find its temperature t minutes in.

T − 20 decays exponentially: $T = 20 + 70e^{-kt}$ .
At t = 10: $70e^{-10k} = 40$ , so $k = \frac{1}{10}\ln\frac{7}{4}$ , about 0.056.

Answer $T \approx 20 + 70e^{-0.056t}$

Example: A mixing tank

A 100 L tank of pure water has brine with 2 g of salt per liter flowing in at 5 L/min, and the mixture drains at 5 L/min. How much salt is there after a long time?

The salt S changes at $S' = 10 - \frac{5S}{100}$ : 10 g a minute in, and 5% of the salt out each minute.
The steady level is where S′ = 0: S = 200.

AnswerClose to 200 g: the tank ends up at 2 g a liter, like the brine

Try it yourself

Work each one on paper first, then open the answer to check.

Solve $\frac{dy}{dt} = -0.5y$ with y(0) = 8.
Show answer
$y = 8e^{-0.5t}$
Solve $\frac{dy}{dx} = \frac{x}{y}$ with y(0) = 2.
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$y = \sqrt{x^{2} + 4}$
Find the integrating factor for y′ + 3y = x.
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$e^{3x}$
Solve y′ + y = 4 with y(0) = 0.
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$y = 4 - 4e^{-x}$
A substance has a half-life of 5 years. What is k in $y = y_{0}e^{-kt}$ ?
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ln 2 ÷ 5, about 0.139
In $P' = 0.2P\left(1 - \frac{P}{500}\right)$ , what level does P approach?
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500