Differential Equations
Equations whose unknown is a whole function — reading off order and degree, the standard recipes that unwind them (separation, substitution, the integrating factor), and the slope fields and growth laws that reveal what a differential equation really describes
- The order and degree of a differential equation, and what a general versus particular solution means.
- How to form a differential equation by eliminating the arbitrary constants of a family of curves.
- The variables-separable method and homogeneous equations solved by \(y=vx\).
- First-order linear equations and the integrating factor.
- Direction fields — the geometric meaning of a first-order equation.
- Applications: exponential growth and decay, and orthogonal trajectories.
Order, Degree & Solutions
A differential equation relates a function to its derivatives; solving it means finding the function. Its order is the highest derivative that appears, and its degree is the power of that highest derivative once the equation is a polynomial in the derivatives. A solution containing the full quota of arbitrary constants is the general solution; fixing those constants with initial conditions gives a particular one.
A first-order equation has one constant; a second-order equation has two. Each independent initial condition removes one constant, so \(n\) conditions pin down a unique particular solution.
Degree is meaningful only when the equation can be written as a polynomial in the derivatives — free of radicals and transcendental functions of \(y',y'',\dots\). So \(\big(\tfrac{d^2y}{dx^2}\big)^3+\big(\tfrac{dy}{dx}\big)^2=0\) has order \(2\), degree \(3\), but \(\sin\!\big(\tfrac{dy}{dx}\big)=x\) has no degree at all.
Forming a Differential Equation
Run the previous idea backwards. Given a family of curves carrying arbitrary constants — all circles centred at the origin, say, or all lines through a point — you can produce the single differential equation they all obey by differentiating enough times to eliminate every constant.
The number of differentiations equals the number of constants, and the result is an equation of order \(n\). For \(x^2+y^2=r^2\), one differentiation gives \(2x+2y\,y'=0\), i.e. \(y'=-x/y\) — the constant \(r\) has vanished.
Variables Separable
The simplest equations to solve are those where the \(x\)-stuff and the \(y\)-stuff can be herded onto opposite sides of the equation. Once separated, you integrate each side on its own.
Move every \(y\) (with \(dy\)) to one side and every \(x\) (with \(dx\)) to the other, then integrate. A single arbitrary constant appears — combine the two constants of integration into one. The result is usually left as an implicit relation between \(x\) and \(y\).
Homogeneous Equations
Some equations are not separable as they stand, yet the right-hand side depends only on the ratio \(y/x\). A single substitution then converts them into a separable equation in a new variable.
Writing \(y=vx\) gives \(\tfrac{dy}{dx}=v+x\tfrac{dv}{dx}\), and the equation becomes separable in \(v\) and \(x\). Solve for \(v\), then substitute back \(v=y/x\). An equation qualifies whenever every term has the same total degree in \(x\) and \(y\).
Linear Equations & the Integrating Factor
A first-order equation is linear when \(y\) and \(\tfrac{dy}{dx}\) appear only to the first power. Multiplying by a cleverly chosen integrating factor turns the whole left-hand side into the derivative of a single product — after which one integration finishes the job.
Multiplying by \(\mu=e^{\int P\,dx}\) makes the left side exactly \(\tfrac{d}{dx}(y\,\mu)\), so integrating both sides gives the solution. The one essential first step is to put the equation in standard form — coefficient of \(\tfrac{dy}{dx}\) equal to \(1\) — before reading off \(P\) and \(Q\).
Direction Fields
A first-order equation \(\tfrac{dy}{dx}=f(x,y)\) has a vivid geometric meaning: at every point of the plane it prescribes a slope. Drawing a short segment of that slope at many points produces a direction field, and a solution curve is any curve that stays tangent to it throughout.
Solution curves never cross where \(f\) is well behaved, because each point carries exactly one slope. Picking a starting point selects the single curve through it — the geometric face of an initial condition. The field also shows behaviour, such as where solutions level off, even when no formula is available.
Growth & Decay
The most famous differential equation in science says that a quantity changes at a rate proportional to its current size. Populations, radioactive atoms, cooling bodies and compound interest all obey it — and its solution is the exponential.
The constant \(y_0\) is the starting amount and \(k\) the relative rate. With \(k>0\) the quantity grows and doubles every \(\tfrac{\ln 2}{k}\); with \(k<0\) it decays with a half-life of \(\tfrac{\ln 2}{|k|}\).
Orthogonal Trajectories
Two families of curves are orthogonal trajectories of one another if every member of one cuts every member of the other at a right angle — think of the equipotential lines and field lines of physics. Finding them is a neat application of everything in this chapter.
First form the differential equation of the given family (Section 28-2). Then replace \(\tfrac{dy}{dx}\) by \(-\tfrac{dx}{dy}\) — since perpendicular slopes multiply to \(-1\) — and solve the new equation. Its solution family is the orthogonal trajectories.
Putting It to Work
Problem. State the order and degree of \(\Big(\dfrac{d^{2}y}{dx^{2}}\Big)^{3}+\Big(\dfrac{dy}{dx}\Big)^{2}+y=0\).
Solution. The highest derivative is the second; it is raised to the power three, and the equation is polynomial in the derivatives:
Problem. Form the differential equation of the family \(y=mx\) (all lines through the origin).
Solution. Differentiate to get \(\tfrac{dy}{dx}=m\), then eliminate \(m=y/x\):
Problem. Solve \(y\,\dfrac{dy}{dx}+x=0\).
Solution. Separate and integrate \(y\,dy=-x\,dx\):
Problem. Solve \(\dfrac{dy}{dx}=\dfrac{x+y}{x}\).
Solution. The right side is \(1+\tfrac yx\); put \(y=vx\), so \(v+x v'=1+v\), giving \(x v'=1\):
Problem. Solve \(\dfrac{dy}{dx}+y=e^{x}\).
Solution. Here \(P=1\), so the integrating factor is \(\mu=e^{\int 1\,dx}=e^{x}\):
Problem. A substance decays at a rate proportional to the amount present. If half of it decays in \(1600\) years, find the decay constant \(k\).
Solution. From \(\tfrac{dN}{dt}=-kN\) we get \(N=N_0e^{-kt}\); impose \(N=\tfrac{N_0}{2}\) at \(t=1600\):
Chapter Summary
Order = highest derivative; degree = its power (polynomial form only).
Differentiate \(n\) times to eliminate \(n\) arbitrary constants.
\(\int\tfrac{dy}{h(y)}=\int g(x)\,dx\) after separating.
Put \(y=vx\) to make it separable in \(v\).
\(\mu=e^{\int P\,dx}\); then \(y\,\mu=\int Q\,\mu\,dx+C\).
\(y=y_0e^{kt}\) for growth/decay; \(\tfrac{dy}{dx}\to-\tfrac{dx}{dy}\) for orthogonals.
Problems
Classify the equation first — separable, homogeneous, or linear — then apply the matching recipe. Difficulty rises down the list.
- State the order and degree of \(\dfrac{d^{2}y}{dx^{2}}+3\dfrac{dy}{dx}+2y=0\).
- State the order and degree of \(\Big(\dfrac{dy}{dx}\Big)^{3}+2y=x\).
- Form the differential equation of the family \(y=Cx^{2}\).
- Solve \(\dfrac{dy}{dx}=x\,e^{-y}\).
- Solve \(\dfrac{dy}{dx}=\dfrac{y}{x}\).
- Solve the homogeneous equation \(\dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}\).
- Solve the linear equation \(\dfrac{dy}{dx}+2y=6\).
- Solve \(\dfrac{dy}{dx}+\dfrac{1}{x}\,y=x^{2}\).
- Solve \(\dfrac{dy}{dx}=e^{\,x+y}\).
- Find the particular solution of \(\dfrac{dy}{dx}=2x\) satisfying \(y(0)=3\).
- A population grows at a rate proportional to its size and doubles in \(20\) years. Find the growth constant \(k\).
- Solve the homogeneous equation \(\dfrac{dy}{dx}=\dfrac{x+y}{x-y}\).
- Find the orthogonal trajectories of the family of lines \(y=Cx\).