Advanced Calculus Differential equations: summary

1 First order equations

1.1 Separable equations

These have the form

\frac{dy}{dx} = f(x)g(y).

We “separate the variables” and integrate

\frac{dy}{g(y)} = f(x) dx,
∫ \frac{dy}{g(y)} = ∫ f(x) dx + C.

1.2 Homogeneous equations

These have the form

\frac{dy}{dx} = F (x, y)

where F (tx, ty) = F (x, y). We substitute y = xv to get

v + x\frac{dv}{dx} = F (x, xv) = F (1, v)

or

\frac{dv}{dx} = \frac{F (1, v) − v}{x} .

This is now a separable equation for v. Separate the variables to solve.

1.3 Linear equations

These have the form

\frac{dy}{dx} + p(x)y = q(x).  (∗)

We introduce the integrating factor

ϕ(x) = exp (∫ p(x) dx) .

Then

ϕ′(x) = p(x)ϕ(x)

and so

\frac{d}{dx}(ϕ(x)y) = ϕ(x)\frac{dy}{dx} + ϕ′(x)y = ϕ(x) (\frac{dy}{dx} + p(x)y) .

By (∗) then

\frac{d}{dx}(ϕ(x)y) = ϕ(x)q(x).

We integrate this to determine ϕ(x)y and so also y.

2 Linear equations with constant coefficients

These have the form

a_{n}y^{(n)} + a_{n−1}y^{(n−1)} + ⋯ + a_{1}y′ + a_{0}y = q(x)

where a_{0}, …, a_{n} are constants.

If q(x) = 0 the equation is homogeneous linear.

2.1 The homogeneous case

First consider the second order case

a_{2}y′′ + a_{1}y′ + a_{0}y = 0.  (†)

If y = e^{mx} then

a_{2}y′′ + a_{1}y′ + a_{0}y = (a_{2}m^{2} + a_{1}m + a_{0})e^{mx}

so this is a solution provided that m is a root of the quadratic equation

a_{2}m^{2} + a_{1}m + a_{0} = 0

which is called the auxiliary equation. If the auxiliary equation has distinct real roots m_{1} and m_{2} then (†) has general solution

y = A_{1}e^{m_{1}x} + a_{2}e^{m_{2}x}.

If the roots are the auxiliary equation are non-real: m_{1} = α + iβ and m_{2} = α − iβ then the above solution is valid, but it may be more conveniently reexpressed as

y = e^{αx}(B_{1} cos βx + B_{2} sin βx).

If the auxiliary solution has a repeated root m_{1} = m_{2} = m then the above is not the general solution, rather it is

y = (A + Bx)e^{mx}.

Now consider the n-th order case

a_{n}y^{(n)} + a_{n−1}y^{(n−1)} + ⋯ + a_{1}y′ + a_{0}y = 0.

The auxiliary equation is

a_{n}m^{n} + a_{n−1}m^{n−1} + ⋯ + a_{1}m + a_{0} = 0.

If it has distinct real roots m_{1}, …, m_{n} then the general solution is

y = A_{1}e^{m_{1}x} + ⋯ + A_{n}e^{m_{n}x}.

If the auxiliary equation has a repeated real root m with multiplicity k then the contribution of m to the general solution is

(B_{0} + B_{1}x + ⋯ + B_{k−1}x^{k−1})e^{mx}.

If the auxiliary equation has a pair of non-repeated complex roots α ± iβ then their contribution of m to the general solution is

e^{αx}(A cos βx + B sin βx).

However if each of these complex roots is repeated with multiplicity k then the contribution is

e^{αx}[(A_{0} + A_{1}x + ⋯ + A_{k−1}x^{k−1}) cos βx + (B_{0} + B_{1}x + ⋯ + B_{k−1}x^{k−1}) sin βx].

2.2 The inhomogeneous case

Consider

a_{n}y^{(n)} + a_{n−1}y^{(n−1)} + ⋯ + a_{1}y′ + a_{0}y = q(x).  (∗)

Thee corresponding homogeneous equation is

a_{n}y^{(n)} + a_{n−1}y^{(n−1)} + ⋯ + a_{1}y′ + a_{0}y = 0.  (†)

The general solution of (∗) is y = y_{CF} + y_{P I} where y_{CF}, the complementary function, is the general solution of (†) and y_{P I}, the particular integral, is any particular solution of (†).

To find a particular integral we use the method of undetermined coefficients (= systematic guesswork!). We write down a likely solution with indeterminates for coefficients, plug into (∗) and derive linear equations for these indeterminates.

If q(x) is a polynomial of degree n we try y = c_{0} + c_{1}x + ⋯ + c_{n}x^{n}.

If q(x) = ae^{mx} and m is not a root of the auxiliary equation we try y = ce^{mx}. However, if m is a root of the auxiliary equation, with multiplicity k, we try y = cx^{k}e^{mx}.

If q(x) = r(x)e^{mx} where r(x) is a polynomial of degree n try y = (c_{0} + c_{1}x + ⋯ + c_{n}x^{n})e^{mx} if m is not a root of the auxiliary equation. However, if m is a root of the auxiliary equation, with multiplicity k, we try y = (c_{0} + c_{1}x + ⋯ + c_{n}x^{n})x^{k}e^{mx}.

If q(x) = ae^{αx} cos βx or ae^{αx} sin βx try y = e^{αx}(b cos βx + c sin βx) if m is not a root of the auxiliary equation. The cases where m is a root of the auxiliary equation and the cases where q(x) = r(x)e^{αx} cos βx or r(x)e^{αx} sin βx with r(x) a polynomial, I’ll leave you to figure out.

3 Other types of equation

3.1 Euler’s equation

This is

a_{n}x^{n}y^{(n)} + a_{n−1}x^{n−1}y^{(n−1)} + ⋯ + a_{1}xy′ + a_{0}y = q(x)

where a_{0}, …, a_{n} are constants. This can be reduced to the case of constant coefficients by setting x = e^{t}. For convenience we only consider the secondorder case

Consider

ax^{2}y′′ + bxy′ + c = q(x).  (∗)

Set x = e^{t} and write ẏ for dy/dt and ÿ for d^{2}y/dt^{2}. Note that dx/dt = e^{t} = x so that dt/dx = 1/x. Then

y′ = \frac{dy}{dx} = \frac{dt}{dx} \frac{dx}{dt} = \frac{ẏ}{x}.

Hence

y′′ = \frac{d}{dx} \frac{ẏ}{x} = −\frac{ẏ}{x^{2}} + \frac{1}{x} \frac{dẏ}{dx} = −\frac{ẏ}{x^{2}} + \frac{1}{x} \frac{dt}{dx} \frac{dẏ}{dt} = −\frac{ẏ}{x^{2}} + \frac{ÿ}{x^{2}} .

Thus xy′ = ẏ and x^{2}y′′ = ÿ − ẏ. Now (∗) becomes

aÿ + (b − a)ẏ + cy = q(e^{t}).

This is now a constant coefficient equation and can be solved as such.

3.2 A particular solution is known

Consider the second-order homogeneous linear equation, with non-constant coefficients:

a(x)y′′ + b(x)y′ + c(x)y = 0.  (∗)

If we know a particular solution y(x) = u(x), substitute y(x) = u(x)v(x) in (∗). After some manipulation (∗) becomes

A(x)v′′(x) + B(x)v′(x) = 0

for some functions A(x) and B(x). This is now a first-order linear equation for v′(x). It can be solved for v′(x) and then we integrate to get v(x). Finally we recall that y(x) = u(x)v(x).

3.3 Simultaneous equations

Suppose we have a pair of constant coefficient equations for two dependent variables y and z:

y′ = ay + bz + p(x)
z′ = ry + sz + q(x).

Write D for the operator d/dx. The equations become

(D − a)y − bz = p(x)
−ry + (D − s)z = q(x).

We can manipulate these carefully to eliminate one unknown. We multiply the first by (D − s) and the second by b to get

(D − s)(D − a)y − b(D − s)z = (D − s)p(x)
−bry + b(D − s)z = bq(x).

Adding now gives a second-order linear equation with constant coefficients for y. This can be solved, and then z can be obtained by substituting in the original equations.

RJC 14/2/2001