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Ordinary differential equations

A differential equation relates a function to its derivatives: it describes how something changes (a temperature that cools, the charge of a capacitor, a mass that oscillates). This topic explains first-order separable and linear equations, second-order equations with constant coefficients and initial conditions.

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What a solution is

A solution of $y^{'} = f (x, y)$ is a function $y (x)$ that satisfies it. The order is that of the highest derivative. The general solution has as many arbitrary constants as the order ( $C$ , or $C_{1}$ and $C_{2}$ …): it is a family of curves. Initial conditions ( $y (0) = 1$ , $y^{'} (0) = 0$ …) pick one curve of the family: the particular solution.

Geometrically, $y^{'} = f (x, y)$ gives the slope at every point of the plane (the slope field); a solution is a curve that follows that slope at every point.

First order

Separable: $y^{'} = g (x) h (y)$ . Separate the variables and integrate each side:

\int \frac{d y}{h ( y )} = \int g (x) d x + C

Linear: $y^{'} + P (x) y = Q (x)$ . Multiply by the integrating factor $μ = e^{\int P d x}$ , which turns the left side into the derivative of a product: $(μ y)^{'} = μ Q$ , and it is enough to integrate.

y = \frac{1}{μ} (\int μ Q d x + C)

Second order with constant coefficients

For $a y^{''} + b y^{'} + cy = 0$ , try $y = e^{r x}$ : this leaves the characteristic equation $a r^{2} + b r + c = 0$ , and its roots give the solution:

Roots	General solution	Behavior
distinct real $r_{1}$ , $r_{2}$	$C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x}$	exponential, no oscillation (overdamped)
double $r$	$(C_{1} + C_{2} x) e^{r x}$	the borderline case (critical damping)
complex $α \pm β i$	$e^{αx} (C_{1} cos β x + C_{2} sin β x)$	oscillates with frequency $β$ (underdamped if $α < 0$ )

The three cases with

y (0) = 1

and

y^{'} (0) = 0

: this is what happens to a mass on a spring with a lot of, just enough or little damping.

If the right side is not 0 (a nonhomogeneous equation), the solution is $y = y_{h} + y_{p}$ : that of the homogeneous one plus a particular one. By undetermined coefficients, $y_{p}$ is tried with the shape of the right side: a polynomial for a polynomial, $A e^{k x}$ for $e^{k x}$ , $A cos k x + B sin k x$ for a sine or cosine; if that shape is already in $y_{h}$ , multiply by $x$ .

The multi-domain simulator solves these same equations numerically when they come from a circuit or a mechanical or thermal system; here they are obtained in exact form. Every solution the calculator gives is substituted into the equation before it is shown.

Common mistakes

Forgetting the constant when integrating, or putting it on one side only and then losing it while solving for $y$ .
Using the integrating factor without first writing the equation as $y^{'} + P (x) y = Q (x)$ (the coefficient of $y^{'}$ must be 1).
Applying the initial conditions to $y_{h}$ instead of to the full solution $y_{h} + y_{p}$ .
With a double root, writing $C_{1} e^{r x} + C_{2} e^{r x}$ (that is a single constant): the $x$ is missing.

Worked examples

Example 1A separable equation

Solve

y^{'} = x y

Solution

Separate: $\frac{d y}{y} = x d x$ . Integrating: $ln ∣ y ∣ = \frac{x ^{2}}{2} + K$ .
$∣ y ∣ = e^{K} e^{x^{2} /2}$ ; with $C = \pm e^{K}$ (and $C = 0$ , also a solution): $y = C e^{x^{2} /2}$ .

y = C e^{x^{2} /2}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y'=x y; press Compute. The Answer is $y = C e^{\frac{x ^{2}}{2}}$ .
The calculator sees it as linear ( $y^{'} - x y = 0$ ) and uses the integrating factor $e^{- x^{2} /2}$ : the result is the same.

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Example 2First-order linear

Solve

y^{'} + 2 y = e^{x}

Solution

$P = 2$ , so $μ = e^{2 x}$ : $(e^{2 x} y)^{'} = e^{3 x}$ .
$e^{2 x} y = \frac{e ^{3 x}}{3} + C$ → $y = \frac{e ^{x}}{3} + C e^{- 2 x}$ .
Check: $y^{'} + 2 y = \frac{e ^{x}}{3} - 2 C e^{- 2 x} + \frac{2 e ^{x}}{3} + 2 C e^{- 2 x} = e^{x}$ ✓.

y = C e^{- 2 x} + \frac{e ^{x}}{3}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y'+2y=e^x; press Compute. The Answer is $y = C e^{- 2 x} + \frac{e ^{x}}{3}$ .
The calculator uses $y_{h} + y_{p}$ (characteristic root $- 2$ and $y_{p} = A e^{x}$ , with $A = \frac{1}{3}$ ): the same result by another route.

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Example 3A cooling coffee

A coffee at 90 °C is left in a room at 20 °C and cools according to

T^{'} = - 0.1 (T - 20)

, with

t

in minutes. Find

T (t)

, the temperature after 10 min and when it reaches 40 °C.

Solution

It is separable (and linear): $\frac{d T}{T - 20} = - 0.1 d t$ → $ln ∣ T - 20∣ = - 0.1 t + K$ → $T = 20 + C e^{- t /10}$ .
$T (0) = 90$ → $C = 70$ : $T (t) = 20 + 70 e^{- t /10}$ .
$T (10) = 20 + 70/ e = 45.75$ °C. For $T = 40$ : $70 e^{- t /10} = 20$ → $t = 10 ln 3.5 = 12.53$ min.

T (t) = 20 + 70 e^{- t /10}

; 45.75 °C at 10 min; 40 °C at 12.53 min.

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y'=-0.1(y-20) and, in Initial conditions (optional), y(0)=90; press Compute. The Answer is $y = 70 e^{- \frac{x}{10}} + 20$ (the calculator calls the time $x$ and the temperature $y$ ).

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Example 4Distinct real roots

Solve

y^{''} + 3 y^{'} + 2 y = 0

Solution

Characteristic: $r^{2} + 3 r + 2 = (r + 1) (r + 2) = 0$ → $r = - 1$ , $r = - 2$ .
$y = C_{1} e^{- x} + C_{2} e^{- 2 x}$ : both exponentials decay, with no oscillation.

y = C_{1} e^{- x} + C_{2} e^{- 2 x}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y''+3y'+2y=0; press Compute. The Answer is $y = C_{1} e^{- x} + C_{2} e^{- 2 x}$ .

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Example 5With initial conditions

Solve

y^{''} + 3 y^{'} + 2 y = 0

with

y (0) = 1

and

y^{'} (0) = 0

Solution

From the general one: $y (0) = C_{1} + C_{2} = 1$ and $y^{'} (0) = - C_{1} - 2 C_{2} = 0$ .
Subtracting: $C_{2} = - 1$ , $C_{1} = 2$ . It is the first curve of the three-cases figure.

y = 2 e^{- x} - e^{- 2 x}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y''+3y'+2y=0 and, in Initial conditions (optional), y(0)=1, y'(0)=0; press Compute. The Answer is $y = 2 e^{- x} - e^{- 2 x}$ .

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Example 6Double root

Solve

y^{''} - 2 y^{'} + y = 0

Solution

$r^{2} - 2 r + 1 = (r - 1)^{2} = 0$ : double root $r = 1$ .
One exponential alone is missing an independent solution: the second is $x e^{x}$ . $y = (C_{1} + C_{2} x) e^{x}$ .

y = (C_{1} + C_{2} x) e^{x}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y''-2y'+y=0; press Compute. The Answer is $y = (C_{2} x + C_{1}) e^{x}$ .

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Example 7Complex roots: oscillation

Solve

y^{''} + 4 y = 0

(a mass on a spring without friction).

Solution

$r^{2} + 4 = 0$ → $r = \pm 2 i$ : $α = 0$ , $β = 2$ .
$y = C_{1} cos 2 x + C_{2} sin 2 x$ : it oscillates without damping, with angular frequency 2 (period $π$ ).

y = C_{1} cos 2 x + C_{2} sin 2 x

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y''+4y=0; press Compute. The Answer is $y = C_{1} cos (2 x) + C_{2} sin (2 x)$ .

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Example 8Nonhomogeneous: undetermined coefficients

Solve

y^{''} + y = sin 2 x

Solution

Homogeneous: $r^{2} + 1 = 0$ → $y_{h} = C_{1} cos x + C_{2} sin x$ .
Particular: try $y_{p} = A cos 2 x + B sin 2 x$ ( $sin 2 x$ is not in $y_{h}$ ). $y_{p}^{''} + y_{p} = - 3 A cos 2 x - 3 B sin 2 x = sin 2 x$ → $A = 0$ , $B = - \frac{1}{3}$ .
$y = C_{1} cos x + C_{2} sin x - \frac{sin 2 x}{3}$ .

y = C_{1} cos x + C_{2} sin x - \frac{sin 2 x}{3}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Differential equations: type y''+y=sin(2x); press Compute. The Answer is $y = C_{1} cos (x) + C_{2} sin (x) - \frac{s i n ( 2 x )}{3}$ .

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Ordinary differential equations

What a solution is

First order

Second order with constant coefficients

Common mistakes

Worked examples

Example 1A separable equation

Solution

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Example 2First-order linear

Solution

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Example 3A cooling coffee

Solution

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Example 4Distinct real roots

Solution

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Example 5With initial conditions

Solution

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Example 6Double root

Solution

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Example 7Complex roots: oscillation

Solution

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Example 8Nonhomogeneous: undetermined coefficients

Solution

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