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Chapter 5
Differential equations
5.1 Ordinary and partial differential equations
A differential equation is a relation between an unknown function and its derivatives. Such equations are extremely important in all branches of science; mathematics, physics, chemistry, biochemistry, economics,... Typical example are
- Newton’s law of cooling which states that
the rate of change of temperature is proportional to the temperature difference between it and that of its surroundings.
This is formulated in mathematical terms as the differential equation dT dt =^ k(T^ −^ T^0 ), where T (t) is the temperature of the body at time t, T 0 the temperature of the surroundings (a constant) and k a constant of proportionality,
- the wave equation, ∂^2 u ∂t^2 = c^2 ∂^2 u ∂x^2
where u(x, t) is the displacement (from a rest position) of the point x at time t and c is the wave speed.
The first example has unknown function T depending on one variable t and the relation involves the first
order (ordinary) derivative dT dt
. This is a ordinary differential equation, abbreviated to ODE. The second example has unknown function u depending on two variables x and t and the relation involves
the second order partial derivatives ∂
(^2) u ∂x^2 and ∂
(^2) u ∂t^2
. This is a partial differential equation, abbreviated to PDE. The order of a differential equation is the order of the highest derivative that appears in the relation. The unknown function is called the dependent variable and the variable or variables on which it depend are the independent variables. A solution of a differential equation is an expression for the dependent variable in terms of the independent one(s) which satisfies the relation. The general solution includes all possible solutions and typically includes arbitrary constants (in the case of an ODE) or arbitrary functions (in the case of a PDE.) A solution without arbitrary constants/functions is called a particular solution. Often we find a particular solution to a differential equation by giving extra conditions in the form of initial or boundary conditions.
Example 5.1 Show that cos ct and sin ct are solutions of the second order ODE
u¨ + c^2 u = 0,
where c is a constant. Deduce that A cos ct + B sin ct is also a solution for arbitrary constants A, B.
Remark It is conventional to use ˙u to denote the derivative (of u) with respect to t and ¨u the second derivative with respect to t. In a similar way we will use u′^ and u′′^ to denotes derivatives with respect to x.
Solution :
Remarks
- A differential equation which contains no products of terms involving the dependent variable is said to be linear. For example, d^2 y dx + x^2 = 0, ut + ux = 0,
are linear but d^2 y dx^2 +^ y
(^2) = 0, ut + uux = 0,
are nonlinear. The ODE in the above example is also linear. As illustrated, any linear combination (A cos ct+B sin ct) of given solutions (cos ct, sin ct) is also a solution. This is true for all linear differential equations and makes them much easier to solve. It is not true of nonlinear differential equations.
- The solution u = A cos ct + B sin ct contains two arbitrary constants and is the general solution of the second order ODE ¨u + c^2 u = 0. This illustrates the fact that the general solution of an nth order ODE contains n arbitrary constants.
Answer: The general solution, y = − 14 log(−4(ex^ + C)) = − 14 log(C − 4 ex),
where we have relabelled the arbitrary constant C (= − 4 × old value of C). The particular solution is y = − 14 log(5 − 4 ex). §
Remark The same techniques may also be used to solve PDEs which are separable in the same sense. For example, the PDE
y ∂u ∂x = u^2 ,
only has a derivative with respect to x and so we regard y as fixed and rewrite it as ∫
y fixed
y du u^2 =
y fixed
dx.
Hence −y
u = x + A(y), i.e., u = −y x + A(y)
where the “constant of integration” is the arbitrary function A(y). This must, in general, depend on y since this variable was fixed during the integration.
Exact type
An ODE of the form d dx φ(x, y) = 0,
is said to be exact and obviously has the general solution φ(x, y) = C, where C is an arbitrary constant. Being able to recognise that an ODE can be expressed in this form is more difficult however. Using the chain rule for functions of two variables, we have d dx φ(x, y) = ∂φ ∂x
dy dx
∂φ ∂y
(see Example ??). Hence an ODE of the form
P (x, y) + Q(x, y) dy dx = 0,
is exact provided there exists some φ(x, y) such that
P = ∂φ ∂x and Q = ∂φ ∂y
For example, the ODE y + x dy dx
is exact since we can take φ = xy giving
∂φ ∂x =^ y^ and^
∂φ ∂y =^ x. In general, exact ODEs are characterised by the following theorem.
Theorem The ODE
P (x, y) + Q(x, y) dy dx
is exact if and only if
∂P
∂y
∂Q
∂x
Proof If the ODE is exact then there exists φ such that
P =
∂φ ∂x and^ Q^ =^
∂φ ∂y ,
and hence it is necessary that
∂P
∂y =^
∂^2 φ ∂y∂x =^
∂^2 φ ∂x∂y =^
∂Q
∂x.
The proof that this condition is enough to guarantee the existence of φ is omitted. §
Example 5.3 Show that the ODE
y + cos(x + y) +
x − y + cos(x + y)
) (^) dy dx
is exact an find its general solution.
Solution :
Answer: The general solution is
y = x^2
2 x^2 dx = x^2 ( 23 x^3 + C) = 23 x^5 + Cx^2.
Example 5.5 Show that the ODE
2 y + 2x^2 y^2 + (x + x^3 y) dy dx
is not exact.
Find an integrating factor xn^ (for some n, to be determined) and hence find its general solution.
Solution :
Answer: The general solution is φ = x^2 y + 12 x^4 y^2 = C. §
Example 5.6 Solve the ODE dy dx
4 xy + 5y^3 4 x^2 − 2 y^3
by finding an integrating factor depending only on y.
Solution :
Answer: The general solution is φ = 2 x^2 y^2 + 5x^ + 2y^ =^ C. §
5.3 Second order ODEs
We will only consider linear second order ODEs, those of the form
a(x) d
(^2) y dx^2
- b(x) dy dx
- c(x)y = f (x). (1)
Euler equation
Euler equation are of the form ax^2 y′′^ + bxy′^ + cy = s(x), (4)
where a, b, c are constants and a 6 = 0. One way of solving Euler equations is to substitute y = xk, where k is a real number, into the corresponding homogeneous equation,
ax^2 y′′^ + bxy′^ + cy = 0,
and solve to find k. You can now apply the Reduction of order method to the original equation (4) by setting y = uv = xkv.
Example 5.7 Find the general solution of
x^2 y′′^ + 3xy′^ + y = x. (1)
Solution :
Answer: y = (^1) x v = 14 x + C log x^ | x|+ Dx , which is the general solution of (1). §
Constant coefficients
The ODE ay′′^ + by′^ + cy = 0,
where a, b, c are constants (with a 6 = 0) always has a solution of the form y = eλt, where λ is a constant (which may be real or complex), to be determined. Making this substitution in the ODE, it is found that λ must satisfy the auxiliary equation aλ^2 + bλ + c = 0,
a quadratic equation which may be solved to find λ. There are three case
Distinct real solutions λ, μ The general solution of the homogeneous equation is y = Aeλx^ + Beμx.
One real (repeated) solutions λ One solution eλt^ is known. Reduction of order is used to find the general solution y = (A + Bx)eλx.
Complex roots α ± iβ The general solution of the homogeneous equation is
y = A′e(α+iβ)t^ + B′e(α−iβ)t^ = eαt(A′eiβt^ + B′e−iβt) = eαt(A cos βt + B sin βt), where A = A′^ + B′^ and B = iA′^ − iB′.
In all case, for an inhomogeneous equation, reduction of order may be used to find the general solution.
Example 5.8 Find the general solutions of
(a) y′′^ + 2y′^ − 3 y = 0, (b) y′′^ − 3 y′^ + 2y = e−x, (c) y′′^ + 2y′^ + 3y = 0, (d) y′′^ + 2y′^ + y = x.
Solution :
Answer:
Solution Answer: z = − xy cos(xy) + A
x y
, where A is an arbitrary function. §
Example 5.10 Solve the PDEs
(a) fx + fy = f, (b) fx + fy = f + y,
by considering the change of variables (x, y) → (u, v) where u = x − y, v = x.
Solution :
Answer: (a) log |f | = v + A(u), and so f = ±eA(u)ev^ = B(u)ev^. In terms of x and y this is f = B(x − y)ex,
where B is an arbitrary function. (b)f = ev^ (−ve−v^ − e−v^ + ue−v^ + A(u)) = u − v − 1 + A(u)ev^ = −y − 1 + A(x − y)ex, where A is an arbitrary function. §
Second order PDEs
The application of the chain rule to calculating second derivatives is a bit more complicated. We consider the second x-derivative as an example. Now, we use the product rule to obtain ∂^2 z ∂x^2
∂x
∂z ∂x
∂^2 u ∂x^2
∂z ∂u
∂u ∂x
∂x
∂z ∂u
∂^2 v ∂x^2
∂z ∂v
∂v ∂x
∂x
∂z ∂v
In this there are two expressions which may be calculated using the general form of the chain rule (1); first,
replacing ∗ by ∂z ∂u , ∂ ∂x
∂z ∂u
∂u ∂x
∂u
∂z ∂u
∂v ∂x
∂v
∂z ∂u
∂u ∂x
∂^2 z ∂u^2 +^
∂v ∂x
∂^2 z ∂v∂u ,
and replacing ∗ by ∂z ∂v
∂x
∂z ∂v
∂u ∂x
∂^2 z ∂u∂v
∂v ∂x
∂^2 z ∂v^2
Solution :
Answer: z = xA
( (^) y x
+ B
( (^) y x
Example 5.13 Solve the PDE
x^2 zxx − y^2 zyy = yzy − xzx,
by using the change of variables
u = xy, v = y x
Solution :
Answer: z(x, y) = B (xy) + C
( (^) y x
End of 2X-Course
