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FACULTY OF MATHEMATICAL STUDIES
MATHEMATICS FOR PART I ENGINEERING
Lectures
MODULE 12 DIFFERENTIAL EQUATIONS II
Revision of material in Differential Equations I
Separable first order equations
dx dt = f
( (^) x t
- Exact equations
- Linear equations
- Concluding comments
- Revision of material in Differential Equations I
Notation and classification:
e.g. dx dt
- t^2 x = 0 ordinary differential equation (ODE) (only ordinary derivatives in the equation) first order (highest derivative is first order) linear (no powers, or products, of x and derivatives of x ) homogeneous dependent variable x independent variable t
∂^2 y ∂x^2
c^2
∂^2 y ∂t^2 = y^2 partial differential equation (PDE) (partial derivatives appear) second order (highest derivative is second order) nonlinear (because of y^2 term) dependent variable y two independent variables x and t
In this module we consider FIRST ORDER ODEs of the form dx dt = f (x, t).
Four main types of this equation will be solved analytically.
- Separable first order equations
These equations were considered in Differential Equations I. For this type of equation the function f (x, t) is the product of a function of x only and a function of t only, i.e. f (x, t) = g(x) h(t). The ODE then becomes dx dt = g(x) h(t),
and on cross-multiplying (or more accurately by dividing by g(x), integrating both sides of the equation with respect to t and changing the variable on the left-hand side (LHS) to x) we obtain ∫ dx g(x) =
h(t) dt.
Clearly the LHS of the above depends only on x, whereas the right-hand side (RHS) depends only on t. Therefore, it is straightforward in principle to integrate both sides with respect to the named variables and hence find the solution to the ODE. Let us illustrate the method with an example.
Ex 1. Solve dx dt
Writing this equation in the general form by moving t^2 x to the RHS clearly shows that f (x, t) = −t^2 x , and so the function is a product of terms which depend only on t and x respectively. The given ODE is therefore of separable type. Rearranging leads to (^) ∫ dx x
t^2 dt ,
ln |x| = − t^3 3
+ C ,
and on taking exponentials this can be written
|x| = exp
t^3 3
+ C
= e−t (^3) / 3 eC^ = A e−t (^3) / 3 .
Since A is an arbitrary constant it is possible to remove the modulus sign from the LHS (without loss of generality) and so the general solution can be written
x = A e−t
(^3) / 3 .
- dx/dt = f (x/t)
This is the second main class of equation. Here the RHS depends on the variables x and t only through the ratio x/t.
Examples are x t ,^
x^2 t^2 +^
t^2 x^2 ,^
x t sin
t x
The solution procedure for this type of equation is to introduce a new variable y , where y = x t , and then rewrite the equation in terms of the variables y and t (instead of x and t as in the original equation).
General case: dx dt = f
( (^) x t
Introduce y = x t ,^ i.e.^ x^ =^ yt. Differentiating the latter equation with respect to t (using the product rule) gives
dx dt
dy dt t + y. 1 = dy dt t + y.
Substituting this into the original equation leads to
t dy dt
t dy dt = f (y) − y, which is separable so ∫ dy f (y) − y
dt t
In principle, therefore, a solution has been found. It should be emphasised that the substitution y = x/t ALWAYS produces a separable equation (although the integrals may be difficult to evaluate, of course).
This result is considered in more detail in Further Calculus II (module 19). Note the types of derivative appearing in the above identity. The variable x depends only on t , so differentiating it with respect to t gives the ordinary derivative dx/dt. On the other hand h depends on the two variables x and t so differentiation of h leads to the partial derivatives ∂h/∂x and ∂h/∂t. Finally, after use of x(t) , h can be expressed as a function of t only and hence we can use the derivative dh/dt on the LHS of the equation.
After that digression let us return to exact first order ODEs. The comments in (a) and (b) allow us to state the following result:-
If a first order ODE has the form p(x, t) dx dt
and if there exists a function h such that
∂h ∂x = p, ∂h ∂t = q, (2)
then (1) can be written dh dt = 0 , with solution h = constant.
Important question: does h exist?
It can be shown that a necessary condition for the existence of h is
∂p ∂t
∂q ∂x
which is equivalent to ∂
(^2) h ∂t∂x
(^2) h ∂x∂t
Solution procedure for exact equations: (a) check that the given ODE has the general form (1) and hence determine the functions p and q ; (b) check that equation (3) is satisfied, so that the equation is exact; (c) solve equations (2) to determine h(x, t) ; (d) state the solution, h = constant.
Ex 3. Solve (x − xt^2 ) dx dt + 8t^ −^ x
(^2) t = 0.
For this equation
p = x − xt^2 , ∂p ∂t = 0 − 2 xt = − 2 xt ,
q = 8t − x^2 t , ∂q ∂x = 0 − 2 xt = − 2 xt.
Since the two calculated partial derivatives are equal the given ODE is exact, and we must now find h. In order for the LHS of the given ODE to be expressed in the form dh/dt it is necessary to satisfy the equations ∂h ∂x = p = x − xt^2 and ∂h ∂t = q = 8t − x^2 t ,
which on integrating give
h = x
2 2 − x
(^2) t 2 2
(^2) t 2 2
respectively. Note that the equations being integrated both involved partial derivatives and so the usual constant of integration has to be a general function of the other variable. The equations stated above give
two expressions for the same function h , and so it is necessary to spot the form of h which is consistent with both equations. The simplest form, omitting the constant which could be added, is
h(x, t) = x^2 2
x^2 t^2 2
[Observe that this equation is consistent with the two expressions for h given earlier.] The solution of the original ODE is h = constant , which can therefore be written
x^2 2
x^2 t^2 2
- Linear equations
The most general first order linear ODE has the form
dx dt
where p(t) and r(t) are any functions of t. The solution procedure involves multiplying the equation by a function g(t) chosen so that the resulting equation is exact.
Let us first derive the general formula for g(t). On multiplication by g(t) the original ODE becomes
g(t) dx dt
- g(t) p(t) x = g(t) r(t) ,
i.e. g(t) dx dt
Using the main result from section 4, the above equation is exact provided ∂p ∂t = ∂q ∂x , where p(t) = g(t)
and q(t) = g(t)p(t)x − g(t)r(t). Hence the equation is exact if
∂ ∂t (g(t)) =
∂x (g(t)p(t)x − g(t)r(t)) , i.e. dg dt = g(t)p(t).
The latter equation is separable and so an expression for g(t) can easily be found
∫ dg g =
p(t) dt ,
ln |g| =
p(t) dt + C ,
|g| = e
p(t) dt+C (^) = e
p(t) dteC (^) = A e
p(t) dt (^) ,
or g(t) = A e
p(t) dt (^).
It is usual to choose A = 1 for simplicity. Having determined above the appropriate form for the multiplying function g(t) , the procedure to be used in practice for solving linear equations can now be stated.
- Concluding comments
Some first order equations fall into more than one category. It is best, therefore, to check the type of ODE in the following order
separable, linear, dx dt =^ f
( (^) x t
, exact,
and then stop and solve by the first category in which it appears.
[To illustrate the above consider the equation t dx dt
separable? YES since
dx x
dt t
linear? YES dx dt
+^1
t x = 0 , p(t) = 1/t, r(t) = 0 ; dx dt = f
( (^) x t
? YES
dx dt
x t
exact? YES ∂ ∂t (t) = 1 , ∂ ∂x (x) = 1 , equal.
Thus given equation falls into all four categories, but is most easily solved as a separable equation.]
rec/00lode
