Lesson

A differential equation is an equation that contains an unknown function, say $y(x)$`y`(`x`), and at least one of its derivatives $y'$`y`′,$y''$`y`′′, $y'''$`y`′′′, etc. Here are five examples:

- $y'-2y+6=0$
`y`′−2`y`+6=0 - $\frac{dy}{dx}=2xy$
`d``y``d``x`=2`x``y` - $x^2\frac{dy}{dx}-2xy=0$
`x`2`d``y``d``x`−2`x``y`=0 - $\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0$
`d`2`y``d``x`2−5`d``y``d``x`+6`y`=0 - $y''-(x^2+1)y'=2x+1$
`y`′′−(`x`2+1)`y`′=2`x`+1

The highest derivative in a first order differential equation is the first derivative $y'$`y`′, so that equations 1, 2 and 3 are all first order differential equations. Equations 4 and 5 are examples of second order equations.

Sometimes the coefficients of the $y$`y` variable and its derivatives are constants, and other times they are an assortment of functions of $x$`x`. When the coefficients are all constants, just like equation 4, then we refer to it as a "differential equation with constant coefficients". Equations 3 and 5 have variable coefficients.

You will often see the word linear in the description of some differential equations. The word linear as used here is a tricky concept to describe, and it mainly concerns how combinations of solutions of higher order differential equations can be combined to create other solutions. What we can say however is that equations that have squared derivatives, or squared $y$`y`'s or contain products of $y$`y` and $y'$`y`′ etc are not linear equations.

We can also state that a first order linear differential equation is an equation that can be expressed in the form $y'+P(x)y=Q(x)$`y`′+`P`(`x`)`y`=`Q`(`x`).

When we solve for the function $y$`y` in such an equation we will create an arbitrary constant $C$`C`, and that solution will correspond to a set of possible solution curves for each possible value of $C$`C`. If we could identify a specific point on the solution curve, then by substitution, we could identify the specific value of $C$`C`. In general, if a $y$`y` value corresponding to a specific $x$`x` value is known, then that $y$`y` value is called a boundary value.

Lets show a very simple example. Suppose we wish to solve the differential equation $\frac{dy}{dx}=2x$`d``y``d``x`=2`x`. It's a first order DE and it's linear because we can express it as $y'+P(x)y=Q(x)$`y`′+`P`(`x`)`y`=`Q`(`x`) where $P(x)$`P`(`x`) is trivially zero and $Q(x)=2x$`Q`(`x`)=2`x`.

Solving it is trivial. From $\frac{dy}{dx}=2x$`d``y``d``x`=2`x`, we have $y=x^2+C$`y`=`x`2+`C`. Note that a first order equation produces one constant of integration. The solution curves can be imagined as nested parabolas that all have their axis of symmetry on the $y$`y` axis.

If we knew just one boundary value we could identify $C$`C`. Suppose at $x=3$`x`=3, the boundary value $y=10$`y`=10. Then by substitution, $10=3^2+C$10=32+`C` would show $C=1$`C`=1, and so the specific solution would be $y=x^2+1$`y`=`x`2+1.

As a second example, the equation $y'-5y=0$`y`′−5`y`=0 is another first order linear differential equation and its also an equation with constant coefficients. Note that $P(x)=-5$`P`(`x`)=−5 and $Q(x)=0$`Q`(`x`)=0. Without showing a formal approach to solving this equation, we know that $y=e^{5x}$`y`=`e`5`x` is a solution, cleverly by substitution.

Let's check this assertion. If $y=e^{5x}$`y`=`e`5`x`, then $y'=5e^{5x}$`y`′=5`e`5`x`, and so $y'-5y=5e^{5x}-5e^5x=e^{5x}(5-5)=0$`y`′−5`y`=5`e`5`x`−5`e`5`x`=`e`5`x`(5−5)=0.

In fact, we can even say that $y=Ce^{5x}$`y`=`C``e`5`x` is the general solution because $y'-5y=Ce^{5x}(5-5)=0$`y`′−5`y`=`C``e`5`x`(5−5)=0. Again, if we knew one specific boundary value say $y=2$`y`=2 when $x=0$`x`=0, we could determine that the specific solution is given by $y=2e^{5x}$`y`=2`e`5`x`.

As a final example of a linear differential equation consider the equation $y'+4xy=x$`y`′+4`x``y`=`x`. Here $P(x)=4x$`P`(`x`)=4`x` and $Q(x)=x$`Q`(`x`)=`x`. Robust techniques have been established to solve this and other more complicated first order equations.

In later chapters you will learn one way to solve this specific equation. It starts by recognising that it can be re-expressed in the form $\frac{dy}{dx}=x(1-4y)$`d``y``d``x`=`x`(1−4`y`) where the two factors separate the things in $x$`x` from the things in $y$`y`.

The general solution turns out to be $y=\frac{1}{4}+Ce^{-2x^2}$`y`=14+`C``e`−2`x`2 and we can verify this by direct substitution:

$y'+4xy=-4xCe^{-2x^2}+4x(\frac{1}{4}+Ce^{-2x^2})=x$`y`′+4`x``y`=−4`x``C``e`−2`x`2+4`x`(14+`C``e`−2`x`2)=`x`

Once again, if we knew that, for $x=0$`x`=0, $y=\frac{5}{4}$`y`=54, then by substitution we have $\frac{5}{4}=\frac{5}{4}+C$54=54+`C` and this means $C=1$`C`=1, so that $y=\frac{1}{4}+e^{-2x^2}$`y`=14+`e`−2`x`2.

Note that the solutions of higher order differential equations involve more constants of integration and therefore more boundary values are required to pin-point specific solutions.

In the next chapters we will discuss three types of first order differential equations. Specifically, first order equations having the forms $\frac{dy}{dx}=f(x)$`d``y``d``x`=`f`(`x`), $\frac{dy}{dx}=g(y)$`d``y``d``x`=`g`(`y`) or $\frac{dy}{dx}=f(x)g(y)$`d``y``d``x`=`f`(`x`)`g`(`y`).

Not all of these forms are generally linear. The first form, $\frac{dy}{dx}=f(x)$`d``y``d``x`=`f`(`x`), always is. The second form, $\frac{dy}{dx}=g(y)$`d``y``d``x`=`g`(`y`), is linear when $g(y)$`g`(`y`) is of degree $1$1 or lower. For example when $g(y)=2y+1$`g`(`y`)=2`y`+1 or when $g(y)=6$`g`(`y`)=6 or when $g(y)=-2y$`g`(`y`)=−2`y`, the differential equation is linear, because we can write these three examples as $y'-2y=1$`y`′−2`y`=1, $y'=6$`y`′=6 and $y'+2y=0$`y`′+2`y`=0 which are all in the form $y'+P(x)y=Q(x)$`y`′+`P`(`x`)`y`=`Q`(`x`).

The third form, $\frac{dy}{dx}=f(x)g(y)$`d``y``d``x`=`f`(`x`)`g`(`y`) could be linear or non-linear depending on the nature of $g(y)$`g`(`y`). For example the equation $\frac{dy}{dx}=x(1-4y)$`d``y``d``x`=`x`(1−4`y`) is linear because, as shown above, it can be written as $y'+4xy=x$`y`′+4`x``y`=`x`.

In general if we rearrange $\frac{dy}{dx}=f(x)g(y)$`d``y``d``x`=`f`(`x`)`g`(`y`) to $\frac{dy}{dx}-f(x)g(y)=0$`d``y``d``x`−`f`(`x`)`g`(`y`)=0 it can immediately be seen that the equation has the form $y'+P(x)y=Q(x)$`y`′+`P`(`x`)`y`=`Q`(`x`) when $g(y)=y$`g`(`y`)=`y` with $P(x)=f(x)$`P`(`x`)=`f`(`x`) and $Q(x)=0$`Q`(`x`)=0.

Form differential equations and interpret the solutions

Apply integration methods in solving problems