Order, Linearity and Solutions

In What a Differential Equation Is you met equations whose unknown is a function, such as $y' = 2y$. This course is about solving them. Before choosing a method, we sort equations by two features, order and linearity, because those decide which tools will work.

Key idea

The order is the highest derivative that appears. An equation is linear if it can be written $$a_n(t)y^{(n)} + \dots + a_1(t)y' + a_0(t)y = g(t),$$ where the coefficients depend on $t$ only. A solution on an interval $I$ is a function that makes the equation true for every $t$ in $I$.

Order

Count derivatives, not powers. In $(y'')^{3} + y^{5} = \sin t$ the highest derivative is $y''$, so the order is 2, even though the largest exponent is 5. A first-order equation needs one starting value to pin down a solution; a second-order equation needs two (a position and a velocity), and so on.

Linear or nonlinear

In a linear equation, $y$ and its derivatives appear only to the first power, each multiplied by a function of $t$ alone. The coefficients may be complicated ($e^{t}$, $\ln t$, $\sin t$); what matters is how $y$ itself appears. Terms such as $y^{2}$, $yy'$, $\sin y$, $e^{y}$ or $\sqrt{y}$ make an equation nonlinear.

Why it works

Write the left side as an operator: $L[y]$ $= a_n(t)y^{(n)} + \dots + a_0(t)y$. Because derivatives of sums are sums of derivatives, $$L[c_1y_1 + c_2y_2] = c_1L[y_1] + c_2L[y_2].$$ That single property is what lets us combine solutions, use integrating factors and, later, Laplace transforms. A term like $y^{2}$ destroys it: $(y_1 + y_2)^{2} \ne y_1^{2} + y_2^{2}$.

Sorting a few equations.
Equation	Order	Linear?	Why
$t^{2}y'' + 3ty' - y = \ln t$	2	yes	each term is a function of $t$ times $y$, $y'$ or $y''$
$y' = y^{2}$	1	no	$y$ is squared
$y''' + e^{t}y = 0$	3	yes	$e^{t}$ multiplies $y$; that is allowed
$y'' + \sin y = 0$	2	no	$y$ sits inside a sine
$yy' = t$	1	no	$y$ multiplies its own derivative

Common mistake

Calling $y' = y^{2}$ linear because no derivative is multiplied by another. Linearity is about every appearance of $y$, not only the derivatives. A quick test for an equation with right side 0: if $y$ is a solution of a linear equation, so is $2y$. For $y' = y^{2}$, $y = -\frac{1}{t}$ works, but $2y = -\frac{2}{t}$ gives $y' = \frac{2}{t^{2}}$ and $y^{2} = \frac{4}{t^{2}}$, which do not match.

Checking a solution

To check a proposed solution, differentiate it, substitute, and see whether both sides agree for every $t$ in the interval. You do not need to know how it was found.

Worked example: Substitute and compare

Show that $y = 3e^{2t} - 1$ solves $y' = 2y + 2$.

Differentiate: $y' = 6e^{2t}$.
Right side: $2y + 2 = 2(3e^{2t} - 1) + 2$ $= 6e^{2t}$.
Both sides equal $6e^{2t}$ for every $t$, so it is a solution on $(-\infty, \infty)$.

Answer: It is a solution for all $t$.

Worked example: Which exponentials work?

For which $r$ is $y = e^{rt}$ a solution of $y'' + y' - 6y = 0$?

$y' = re^{rt}$ and $y'' = r^{2}e^{rt}$.
Substitute: $(r^{2} + r - 6)e^{rt} = 0$.
$e^{rt}$ is never 0, so $r^{2} + r - 6 = (r + 3)(r - 2)$ $= 0$.

Answer: $r = 2$ or $r = -3$. (This idea returns as the characteristic equation.)

Worked example: A solution on an interval

Is $y = \frac{1}{t}$ a solution of $t y' + y = 0$?

$y' = -\frac{1}{t^{2}}$, so $ty' + y = -\frac{1}{t} + \frac{1}{t} = 0$.
The formula is undefined at $t = 0$, so it is a solution on $(0, \infty)$ and on $(-\infty, 0)$, not on an interval containing 0.

Answer: Yes, on $(0, \infty)$ (or on $(-\infty, 0)$).

The direction field of $y' = 2y + 2$ (the horizontal axis plays the role of $t$). The blue curve through $(0, 2)$ is $y = 3e^{2t} - 1$; drag the point to see other solutions. The horizontal solution $y = -1$ is also a solution.

Order, Linearity and Solutions

Order

Linear or nonlinear

Checking a solution

Practice

Builds on

Leads to

Sources and further reading