Differential Equations

Lessons

Senior 6 · Mathematics

Differential Equations

5 lessons

Lesson 2 of 5

Homogeneous and linear first order equations

35 min

When the variables will not separate

Try to separate $\frac{dy}{dx} = \frac{x + y}{x}$ or $\frac{dy}{dx} + y = x$ : the right-hand side cannot be split into a factor in $x$ times a factor in $y$ . This lesson gives two more classes of first order equation, each with its own method. The first is solved by a substitution that makes it separable; the second by multiplying by a well-chosen factor that turns the left side into one derivative.

Definition

Homogeneous function of degree $n$ : a function with $f (t x, t y) = t^{n} f (x, y)$ for all $t \gt 0$ . For example, $x^{2} + 3 x y$ is homogeneous of degree 2.

Homogeneous first order equation: an equation that can be written as

\frac{dy}{dx} = F\left(\frac{y}{x}\right),

which happens when $\frac{dy}{dx} = \frac{M(x, y)}{N(x, y)}$ with $M$ and $N$ homogeneous of the same degree (divide top and bottom by $x^{n}$ ).

Example: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy} = \frac{1 + \left(\frac{y}{x}\right)^2}{2\frac{y}{x}}$ is homogeneous.

Important

Method for homogeneous equations. Put $y = v x$ , where $v$ is a new unknown function of $x$ . By the product rule, $\frac{dy}{dx} = v + x\frac{dv}{dx}$ , so the equation becomes

v + x\frac{dv}{dx} = F(v) \quad\Longrightarrow\quad \int \frac{dv}{F(v) - v} = \int \frac{dx}{x} ,

which is separable. Solve for $v$ , then put back $v = \frac{y}{x}$ .

Worked example 1: the substitution $y = v x$

Solve (a) $\frac{dy}{dx} = \frac{x + y}{x}$ , $x \gt 0$ , with $y (1) = 2$ (b) $(x^2 + y^2)\,dx - 2xy\,dy = 0$ , $x \gt 0$ , with $y (1) = 2$ .

Solution

$F (v) = 1 + v$ . Then $v + x v^{'} = 1 + v$ , so $x\frac{dv}{dx} = 1$ , $v = \ln x + C$ , and $y = x\ln x + Cx$ . From $y (1) = 2$ : $C = 2$ , so $y = x\ln x + 2x$ .
$\frac{dy}{dx} = \frac{1 + v^2}{2v}$ , so $x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 - v^2}{2v}$ . Separate: $\int \frac{2v}{1 - v^2}\,dv = \int \frac{dx}{x} \quad\Longrightarrow\quad -\ln|1 - v^2| = \ln x + C \quad\Longrightarrow\quad x(1 - v^2) = A .$ With $v = \frac{y}{x}$ : $x^{2} - y^{2} = A x$ . From $y (1) = 2$ : $1 - 4 = A$ , so $A = - 3$ and $y^{2} = x^{2} + 3 x$ , that is $y = \sqrt{x^2 + 3x}$ (positive root, since $y(1) \gt 0$ ).

Warning

One word, two meanings. "Homogeneous" here means $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ . In the next lesson a linear equation is called homogeneous when its right-hand side is 0, as in $y^{''} + 4 y = 0$ . The two ideas are unrelated; read which one a question means.

Definition

Linear first order equation: an equation that can be written in the standard form

\frac{dy}{dx} + P(x)\,y = Q(x) .

Integrating factor: the function $\mu(x) = e^{\int P(x)\,dx}$ (any one antiderivative will do).

Example: $x\frac{dy}{dx} + 2y = x^2$ is linear; its standard form is $\frac{dy}{dx} + \frac{2}{x}y = x$ , so $\mu = e^{2\ln x} = x^2$ for $x \gt 0$ .

Important

Method for linear equations. Multiply the standard form by $μ \mu$ . Since $\mu' = P\mu$ , the left side is exactly a product-rule derivative:

\mu y' + \mu P y = \mu y' + \mu' y = \frac{d}{dx}(\mu y) .

So $\frac{d}{dx}(\mu y) = \mu Q$ , and integrating once,

y = \frac{1}{\mu(x)}\left(\int \mu(x)\,Q(x)\,dx + C\right).

Worked example 2: an integrating factor

Solve (a) $\frac{dy}{dx} + y = x$ with $y (0) = 1$ (b) $x\frac{dy}{dx} + 2y = x^2$ , $x \gt 0$ , with $y (1) = 1$ .

Solution

$P = 1$ , so $\mu = e^x$ , and $\frac{d}{dx}(e^x y) = xe^x$ . By parts, $e^{x} y = x e^{x} - e^{x} + C$ , so $y = x - 1 + Ce^{-x} .$ From $y (0) = 1$ : $- 1 + C = 1$ , so $C = 2$ and $y = x - 1 + 2e^{-x}$ .
Standard form $y' + \frac{2}{x}y = x$ , $\mu = x^2$ : $\frac{d}{dx}(x^2 y) = x^3$ , so $x^2y = \frac{x^4}{4} + C$ and $y = \frac{x^2}{4} + \frac{C}{x^2}$ . From $y (1) = 1$ : $C = \frac{3}{4}$ , so $y = \frac{x^2}{4} + \frac{3}{4x^2}$ .

Check (a): $y' = 1 - 2e^{-x}$ , and $y' + y = 1 - 2e^{-x} + x - 1 + 2e^{-x} = x$ ✓

The general solution of dy/dx + y = x is y = x − 1 + Ce^(−x). Whatever the starting value, the term Ce^(−x) dies away and every solution approaches the line y = x − 1 (dashed).

Warning

Use the standard form before finding $μ \mu$ . In $x y^{'} + 2 y = x^{2}$ , $P (x)$ is $\frac{2}{x}$ , not 2. Taking $\mu = e^{2x}$ from the unreduced equation gives a left side that is not a derivative of anything simple, and the method fails.

Tip

Which method? Ask in this order. (1) Can the right side be factored as $g (x) h (y)$ ? Separate. (2) Is it a function of $\frac{y}{x}$ only? Put $y = v x$ . (3) Do $y$ and $y^{'}$ appear only to the first power? Use an integrating factor. Some equations, like $y^{'} + 2 y = 4$ , belong to two classes, and either method works.

MIT OpenCourseWare: Integrating Factor for Constant Rate, with Gilbert Strang (Learn Differential Equations: Up Close). Watch how multiplying by an exponential turns the left side into a single derivative, the step used in Worked example 2.

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Practice

Questions 3 and 8 are harder.

Which of these equations are homogeneous? (a) $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$ (b) $\frac{dy}{dx} = \frac{x + y + 1}{x}$ (c) $\frac{dy}{dx} = \frac{y}{x} + \sin\frac{y}{x}$
Solve $\frac{dy}{dx} = \frac{y - x}{x}$ , $x \gt 0$ , with $y (1) = 3$ .
Solve $x\frac{dy}{dx} = y + \sqrt{x^2 + y^2}$ , $x \gt 0$ , with $y (1) = 0$ .
Solve $\frac{dy}{dx} + 2y = 4$ with $y (0) = 0$ , and say what happens as $x \to \infty$ .
Find the general solution of $\frac{dy}{dx} - y = e^{2x}$ .
Solve $x\frac{dy}{dx} - y = x^3$ , $x \gt 0$ , with $y (1) = 2$ .
Find the general solution of $\frac{dy}{dx} + y\cot x = \cos x$ , for $0 \lt x \lt \pi$ .
A 100-litre tank holds pure water. Brine with $0.2\,\mathrm{kg}$ of salt per litre flows in at 4 litres per minute, and the well-mixed liquid flows out at the same rate. Show that the mass $S$ kg of salt satisfies $\frac{dS}{dt} = 0.8 - 0.04S$ , solve it, and find $S$ after 25 minutes.

Show answersHide answers

(a) and (c); in (b) the "+ 1" spoils it.
$x\frac{dv}{dx} = -1$ : $y = -x\ln x + 3x$
$\int \frac{dv}{\sqrt{1 + v^2}} = \ln x$ gives $v + \sqrt{1 + v^2} = x$ , so $v = \frac{x^2 - 1}{2x}$ and $y = \frac{x^2 - 1}{2}$ .
$\mu = e^{2x}$ : $y = 2 - 2e^{-2x} \to 2$
$\mu = e^{-x}$ : $y = e^{2x} + Ce^x$
$y' - \frac{y}{x} = x^2$ , $\mu = \frac{1}{x}$ : $y = \frac{x^3}{2} + \frac{3x}{2}$
$\mu = \sin x$ : $y\sin x = \frac{\sin^2 x}{2} + C$ , so $y = \frac{\sin x}{2} + \frac{C}{\sin x}$ .
Salt enters at $0.2 \times 4 = 0.8$ kg per minute and leaves at $\frac{S}{100} \times 4 = 0.04S$ . $S = 20\left(1 - e^{-0.04t}\right)$ ; $S(25) = 20(1 - e^{-1}) \approx 12.6\,\mathrm{kg}$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from computed values), the worked examples, the practice items and the quiz questions. Every solution was checked by substituting it back into its equation numerically.
Video: Integrating Factor for Constant Rate, Gilbert Strang, MIT OpenCourseWare, embedded from YouTube.

1When the variables will not separate

When the variables will not separate