Differential Equations

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

Separating variables gives $\frac{dy}{2\sqrt{y}}=dx$, so $\sqrt{y}=x+C$, and the initial condition forces $C=0$. The solution is $y=x^2$.

It is a solution. It is not the only one. The constant function $y\equiv 0$ also satisfies both the equation and the initial condition, and it was destroyed the moment we divided by $\sqrt{y}$.

Worse, there are infinitely many more. For any $a>0$, the function that is $0$ up to $x=a$ and $(x-a{)}^2$ afterwards is differentiable everywhere — both one-sided derivatives at $a$ are zero — and satisfies the equation throughout. So a single initial value problem has an entire family of solutions.

Nothing pathological was needed to arrange this: the right-hand side $2\sqrt{y}$ has an infinite slope in $y$ at the origin, which is exactly what a uniqueness theorem forbids. The practical lesson is the one that costs marks: dividing by an expression is only legal where that expression is non-zero, and every root of the divisor must be checked separately as a possible solution.

1. Order, degree and forming an equation

The order of a differential equation is the highest derivative it contains. Its degree is the power of that highest derivative once the equation has been made a polynomial in all the derivatives — and if it cannot be, the degree is undefined. An equation such as $\sin \frac{dy}{dx}=x$ has order one and no degree at all.

Forming an equation runs the other way. A family of curves with $n$ arbitrary constants is described by an equation of order $n$: differentiate $n$ times and eliminate the constants between the $n+1$ resulting relations.

Illustration 1

Form the differential equation of all circles that touch the $x$-axis at the origin.

Such circles have centre $(0,a)$ and radius $|a|$, so their equation is $x^2+y^2=2ay$ — one arbitrary constant, so a first-order equation is expected. Differentiating,

$2x+2y\frac{dy}{dx}=2a\frac{dy}{dx}\Rightarrow a=\frac{x+y{y}^{\prime }}{y^{\prime }}.$

Substituting this into $a=\frac{x^2+y^2}{2y}$ and clearing denominators gives

$\left(x^2-y^2\right)\frac{dy}{dx}=2xy.$

The count is the check: one constant, one differentiation, first order. If eliminating the constants had required two differentiations, the family would have been described wrongly.

Illustration 2

Form the differential equation of all circles of fixed radius $a$ whose centres lie on the $x$-axis.

Such circles are $(x-h{)}^2+y^2=a^2$, with $h$ the only arbitrary constant. Differentiating,

$2(x-h)+2y\frac{dy}{dx}=0\Rightarrow x-h=-y\frac{dy}{dx}.$

Substituting back into the equation of the family eliminates $h$ and gives

$y^2\left[1+{\left(\frac{dy}{dx}\right)}^2\right]=a^2.$

Note that $a$ survives, because it is not arbitrary — it was given as fixed. Distinguishing a genuinely arbitrary constant from a stated parameter is the step that decides the order, and reading the phrase "of fixed radius" carelessly turns a first-order answer into a second-order one.

The general rule follows the count. A family such as $y=mx+c$ carries two arbitrary constants, so its equation is of order two, and differentiating twice gives $\frac{d^{2}y}{d{x}^2}=0$ with both constants gone. All parabolas $y=a{x}^2+bx+c$ carry three and satisfy $\frac{d^{3}y}{d{x}^3}=0$.

2. Separable equations, and what separation costs

An equation of the form $\frac{dy}{dx}=g(x)h(y)$ separates into $\frac{dy}{h(y)}=g(x)dx$, and both sides are integrated. The division by $h(y)$ is the step to watch: every root of $h$ gives a constant solution, and none of them appears in the general solution obtained afterwards.

Three words are used precisely here and are worth separating. The general solution of an equation of order $n$ contains $n$ arbitrary constants. A particular solution is what remains once initial conditions fix those constants.

A singular solution satisfies the equation but is obtainable from the general solution for no value of the constants at all, and it is exactly what a careless division produces. The opening example had both a particular solution and infinitely many singular ones through the same point, which is why the phrase "the solution" should be treated as a claim to be checked rather than an assumption.

Illustration 3

Solve $\frac{dy}{dx}=y^2-1$ completely.

Separating, $\frac{dy}{y^2-1}=dx$, and partial fractions give $\frac{1}{2}\ln \left|\frac{y-1}{y+1}\right|=x+C$, so

$\frac{y-1}{y+1}=A{e}^{2x}.$

But the division assumed $y\ne \pm 1$, and both $y\equiv 1$ and $y\equiv -1$ satisfy the original equation. The first is recovered by taking $A=0$; the second is not obtainable for any finite $A$ and is a genuinely singular solution that must be stated separately.

3. Homogeneous equations and near misses

If $\frac{dy}{dx}$ can be written as a function of $\frac{y}{x}$ alone, the substitution $y=vx$ makes the equation separable. Differentiating the substitution gives $\frac{dy}{dx}=v+x\frac{dv}{dx}$, and the equation becomes $x\frac{dv}{dx}=F(v)-v$, which separates.

Geometrically the substitution works because such an equation assigns the same slope at every point of a ray through the origin, so the solution curves are scaled copies of one another.

Illustration 4

Solve $\left(x^2+y^2\right)dx=2xydy$.

Dividing by $x^2$ shows the right-hand side depends only on $\frac{y}{x}$, so put $y=vx$:

$v+x\frac{dv}{dx}=\frac{1+v^2}{2v}\Rightarrow x\frac{dv}{dx}=\frac{1-v^2}{2v}.$

Separating and integrating, $-\ln \left|1-v^2\right|=\ln |x|+C$, so $x\left(1-v^2\right)=A$ and, restoring $v=\frac{y}{x}$,

$x^2-y^2=Ax.$

These are circles through the origin only if the sign works out; in general they are hyperbolas, and the check is to differentiate the answer and recover the original equation.

Illustration 5

Solve $\frac{dy}{dx}=\frac{x+y-1}{x+y+1}$.

The right-hand side is a ratio of linear expressions, but the two lines are parallel, so no shift of origin makes it homogeneous. Instead substitute $z=x+y$, giving $\frac{dz}{dx}=1+\frac{dy}{dx}$ and

$\frac{dz}{dx}=1+\frac{z-1}{z+1}=\frac{2z}{z+1}.$

Separating, $\frac{z+1}{2z}dz=dx$, so $\frac{1}{2}\left(z+\ln |z|\right)=x+C$ and, restoring $z=x+y$,

$y-x+\ln |x+y|=C.$

When the two lines are not parallel, the alternative is to shift the origin to their point of intersection, which removes both constants and leaves a homogeneous equation.

When the two lines are not parallel the shift works cleanly. For $\frac{dy}{dx}=\frac{x+2y-3}{2x+y-3}$ the lines $x+2y=3$ and $2x+y=3$ meet at $(1,1)$, so putting $X=x-1$ and $Y=y-1$ leaves the constants behind and gives $\frac{dY}{dX}=\frac{X+2Y}{2X+Y}$, which is homogeneous and yields to $Y=vX$. The shift changes neither derivative, since $dX=dx$ and $dY=dy$, so nothing is lost.

4. Linear equations, with the integrating factor derived

A first-order linear equation has the form $\frac{dy}{dx}+P(x)y=Q(x)$. Multiplying through by a function $\mu (x)$ gives

$\mu \frac{dy}{dx}+\mu Py=\mu Q,$

and the left side is the derivative of $\mu y$ precisely when ${\mu }^{\prime }=\mu P$. That is itself a separable equation, with solution $\mu =e^{\int Pdx}$. So the integrating factor is not a formula to remember but the answer to the question "what makes the left side a single derivative?"

With that in hand, $\frac{d}{dx}\left(\mu y\right)=\mu Q$, and integrating gives

$y{e}^{\int Pdx}=\int Q{e}^{\int Pdx}dx+C.$

Illustration 6

Solve $x\frac{dy}{dx}+y=x^3$.

Divide by $x$ to reach standard form: $\frac{dy}{dx}+\frac{y}{x}=x^2$, so $P=\frac{1}{x}$ and $\mu =e^{\ln x}=x$. The equation becomes $\frac{d}{dx}(xy)=x^3$, giving

$xy=\frac{x^4}{4}+C\Rightarrow y=\frac{x^3}{4}+\frac{C}{x}.$

Here the left side was already a derivative before multiplying, which is worth noticing: whenever an equation reads $x{y}^{\prime }+y=\cdots$, the left side is $(xy{)}^{\prime }$ and no integrating factor need be computed.

Illustration 7

Solve $\frac{dy}{dx}=\frac{1}{x+y^2}$.

As it stands this is neither separable nor linear in $y$. But inverting both sides,

$\frac{dx}{dy}=x+y^2\Rightarrow \frac{dx}{dy}-x=y^2,$

which is linear in $x$ with $P=-1$ and $\mu =e^{-y}$. Then $\frac{d}{dy}\left(x{e}^{-y}\right)=y^2{e}^{-y}$, and integrating by parts twice,

$x{e}^{-y}=-e^{-y}\left(y^2+2y+2\right)+C\Rightarrow x=C{e}^y-\left(y^2+2y+2\right).$

Treating $x$ as the dependent variable is the standard rescue when the equation is linear in one variable and not the other, and Advanced sets it deliberately.

Illustration 8

Solve $\frac{dy}{dx}+\frac{y}{x}=y^2\ln x$.

The right side carries a power of $y$, so this is not linear — but dividing throughout by $y^2$ makes it so. Writing $v=y^{-1}$, so that $\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}$,

$-\frac{dv}{dx}+\frac{v}{x}=\ln x\Rightarrow \frac{dv}{dx}-\frac{v}{x}=-\ln x,$

a linear equation with $\mu =\frac{1}{x}$. The general recipe is that $y^{\prime }+Py=Q{y}^n$ becomes linear under $v=y^{1-n}$.

5. Recognising a differential you already know

Some equations are neither separable nor linear but become trivial once a familiar combination is spotted. The four worth carrying are

$xdy+ydx=d(xy),\frac{xdy-ydx}{x^2}=d\left(\frac{y}{x}\right),$

$\frac{xdy-ydx}{xy}=d\left(\ln \frac{y}{x}\right),\frac{xdy-ydx}{x^2+y^2}=d\left({\tan }^{-1}\frac{y}{x}\right).$

Each is the quotient or product rule read backwards, and the tell is always the combination $xdy\pm ydx$ appearing in the numerator. When it does, the right move is to divide by whatever makes one of these four exact rather than to expand.

Illustration 9

Solve $xdy-ydx=\left(x^2+y^2\right)dx$.

Dividing throughout by $x^2+y^2$ makes the left side exact:

$\frac{xdy-ydx}{x^2+y^2}=dx\Rightarrow d\left({\tan }^{-1}\frac{y}{x}\right)=dx.$

Integrating both sides gives ${\tan }^{-1}\frac{y}{x}=x+C$, so $y=x\tan (x+C)$.

Attacking this as a homogeneous equation also works, but it takes a substitution, a partial-fraction integration and a resubstitution to reach the same place. Recognition is worth several minutes here.

6. Curves defined by a geometric condition

A tangent or normal condition translates into a differential equation through two standard lengths. At a point $(x,y)$ on a curve, the subtangent is $\frac{y}{y^{\prime }}$ and the subnormal is $y{y}^{\prime }$; the tangent meets the axes at points whose coordinates are read off the tangent's equation.

Illustration 10

Find the curves whose subnormal at every point is a constant $k$.

The condition is $y\frac{dy}{dx}=k$, which separates immediately to $\frac{y^2}{2}=kx+C$, that is $y^2=2kx+C^{\prime }$. The curves are parabolas with axis along the $x$-axis, which is the classical characterisation of a parabola by its subnormal.

Illustration 11

Find the curve through $(0,1)$ for which the normal at every point passes through the origin.

The normal at $(x,y)$ has slope $-\frac{1}{y^{\prime }}$, and the line joining $(x,y)$ to the origin has slope $\frac{y}{x}$. Equating them,

$\frac{y}{x}=-\frac{1}{y^{\prime }}\Rightarrow y\frac{dy}{dx}=-x.$

Separating gives $\frac{y^2}{2}=-\frac{x^2}{2}+C$, that is $x^2+y^2=C^{\prime }$, and the point $(0,1)$ fixes $C^{\prime }=1$.

So the curve is the unit circle, and more generally the only curves whose normals all pass through a fixed point are circles centred there — which is the geometric fact the calculation has just proved.

7. Orthogonal trajectories

A family of curves has a differential equation. The family cutting it at right angles everywhere has the same equation with $\frac{dy}{dx}$ replaced by $-\frac{dx}{dy}$, since perpendicular slopes multiply to $-1$.

Illustration 12

Find the orthogonal trajectories of the family $y=c{x}^2$.

First eliminate $c$: differentiating gives $y^{\prime }=2cx$, and $c=\frac{y}{x^2}$, so the family satisfies $y^{\prime }=\frac{2y}{x}$.

Replacing $y^{\prime }$ by $-\frac{1}{y^{\prime }}$ gives $-\frac{dx}{dy}=\frac{2y}{x}$, that is $xdx=-2ydy$, which integrates to

$\frac{x^2}{2}+y^2=C\Rightarrow x^2+2y^2=C^{\prime }.$

The trajectories are ellipses. The elimination of $c$ before the replacement is essential: substituting into an equation that still contains the parameter produces nonsense.

8. Growth, decay and mixing

Any statement that a rate is proportional to a quantity is a first-order equation, and the modelling step is to write down which quantity and with what sign.

Illustration 13

A body cools so that the rate of fall of its temperature is proportional to the excess over the surrounding temperature $T_0$. Find $T(t)$.

Newton's law says $\frac{dT}{dt}=-k\left(T-T_0\right)$ with $k>0$. Separating,

$\ln \left|T-T_0\right|=-kt+C\Rightarrow T=T_0+\left(T_1-T_0\right)e^{-kt},$

where $T_1$ is the initial temperature. The excess decays exponentially and the body approaches, but never reaches, the surrounding temperature — which is the check that the sign of $k$ was chosen correctly.

Illustration 14

A tank holds $100$ litres of pure water. Brine containing $2$ g of salt per litre enters at $5$ litres per minute, and the well-stirred mixture leaves at the same rate. Find the amount of salt after $t$ minutes.

Let $S(t)$ be the salt in grams. Salt enters at $10$ g/min and leaves at $5\cdot \frac{S}{100}$ g/min, so

$\frac{dS}{dt}+\frac{S}{20}=10,$

a linear equation with $\mu =e^{t/20}$. Solving with $S(0)=0$ gives $S=200\left(1-e^{-t/20}\right)$.

The long-run value $200$ g is the concentration of the incoming brine times the tank volume, which is the physical check the algebra must reproduce.

Illustration 15

The population of a colony grows at a rate proportional to its size and doubles in $3$ hours. How long does it take to become eight times its initial size?

From $\frac{dN}{dt}=kN$ we get $N=N_0{e}^{kt}$, and doubling in $3$ hours gives $e^{3k}=2$. Then $N=8N_0$ needs $e^{kt}=8=2^3$, so $kt=3k\cdot 3$ and $t=9$ hours.

Recognising that eight is the cube of two removes the need to compute $k$ at all.

Summary

Dividing by an expression while separating variables destroys every solution on which that expression vanishes, and those solutions must be listed separately. The equation $\frac{dy}{dx}=2\sqrt{y}$ with $y(0)=0$ has infinitely many solutions for exactly this reason, so a general solution with one arbitrary constant need not be the whole story.

The order of an equation is the highest derivative and the degree is its power once the equation is polynomial in the derivatives, with no degree at all otherwise. A family with $n$ arbitrary constants yields an equation of order $n$, obtained by differentiating $n$ times and eliminating.

An equation whose slope depends only on $\frac{y}{x}$ becomes separable under $y=vx$, because it assigns the same direction all along each ray from the origin. A ratio of linear expressions is handled by shifting the origin to the intersection of the two lines, or, when they are parallel, by substituting for the repeated combination.

Some equations are neither separable nor linear and yield instead to recognition: the combination $xdy\pm ydx$ signals one of the four standard exact differentials, and dividing by the right factor turns the whole equation into a single integration.

The integrating factor $e^{\int Pdx}$ is the solution of ${\mu }^{\prime }=\mu P$, which is the condition that the left side collapse into one derivative. When an equation is not linear in $y$, check whether it is linear in $x$, and whether a power $y^n$ on the right can be removed by $v=y^{1-n}$. Geometric conditions enter through the subtangent $\frac{y}{y^{\prime }}$ and the subnormal $y{y}^{\prime }$, and orthogonal trajectories are found by eliminating the parameter first and only then replacing $y^{\prime }$ by $-\frac{1}{y^{\prime }}$.