Module 12 - Differential Equations 1

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 12

DIFFERENTIAL EQUATIONS II

1. Revision of material in Differential Equations I 2. Separable first order equations x dx 3. =f dt t 4. Exact equations 5. Linear equations 6. Concluding comments

1. Revision of material in Differential Equations I Notation and classification: dx + t2 x = 0 e.g. dt 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 ∂2y 1 ∂2y − = y2 ∂x2 c2 ∂t2 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 = f (x, t) . dt

Four main types of this equation will be solved analytically. 2. 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 = g(x) h(t), dt 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 Z Z dx = h(t) dt . g(x) 1

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. dx + t2 x = 0 . dt Writing this equation in the general form by moving t2 x to the RHS clearly shows that f (x, t) = −t2 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 Z Z dx =− t2 dt , x t3 ln |x| = − + C , 3 and on taking exponentials this can be written  3  3 3 t |x| = exp − + C = e−t /3 eC = A e−t /3 . 3 Ex 1.

Solve

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 3

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 .   x2 x t2 x t , sin , ... Examples are + 2, t t2 x t x x The solution procedure for this type of equation is to introduce a new variable y , where y = , and then t rewrite the equation in terms of the variables y and t (instead of x and t as in the original equation). x dx =f . General case: dt t x Introduce y = , i.e. x = yt. t Differentiating the latter equation with respect to t (using the product rule) gives dy dy dx = t+ y.1 = t+y . dt dt dt Substituting this into the original equation leads to dy + y = f (y) dt dy t = f (y) − y, dt Z Z dy dt = . f (y) − y t t

which is separable so

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). 2

x x2 dx = + 2 . dt t t [N.B. You should not remember the general formula for the solution, simply work from first principles each time.] dx dy dy x = t+ y.1 = t+y . Put y = , i.e. x = yt, so t dt dt dt Substituting into the original equation leads to Ex 2.

Solve

t

i.e. −

dy +y dt dy t Z dt dy y2 1 − y

=

x x2 + 2 = y + y2 , t t

= y2 , Z dt , = t = ln |t| + C ,

t = ln |t| + C , x

in original variables .

4. Exact equations Before considering the theory of exact equations two topics need to be looked at. (a) Partial differentiation First order partial derivatives were defined in module 4, and second and higher order partial derivatives can be obtained with an obvious generalisation of the procedures for ordinary derivatives:∂f ∂f = 2xy 2 , = x2 (2y) = 2x2 y e.g. f (x, y) = x2 y 2 , ∂x ∂y and higher derivatives are then obtained, using an obvious notation, by  
∂ ∂2f ∂ ∂f = 2xy 2 = 2y 2 , = ∂x2 ∂x ∂x ∂x  
∂ ∂f ∂ ∂2f = = 2x2 y = 4xy , ∂x∂y ∂x ∂y ∂x  
∂ ∂f ∂ ∂2f = = 2xy 2 = 2x(2y) = 4xy . ∂y∂x ∂y ∂x ∂y For the particular choice f (x, y) = x2 y 2 we have shown above that

∂ 2f ∂2f = , ∂x∂y ∂y∂x

i.e. the mixed second order partial derivatives are equal. One can show that this result holds for ALL “reasonable” functions f (i.e. all functions that are sufficiently smooth – precise mathematical conditions can be written down). (b) Chain rule

You have used the chain rule for functions of one variable: dy dx df dg dy = = . e.g. if y = f (x) and x = g(u) then du dx du dx du [The picture y − x − u can be helpful.]

The chain rule can also be applied to functions of more than one variable. Introduce the function h(x(t), t), then ∂h dx ∂h dh = + . dt ∂x dt ∂t 3

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 + q(x, t) = 0 dt

(1)

∂h = q, ∂t

(2)

and if there exists a function h such that ∂h = p, ∂x then (1) can be written Important question:

dh = 0 , with solution h = constant . dt

does h exist?

It can be shown that a necessary condition for the existence of h is ∂p ∂q = , ∂t ∂x which is equivalent to

(3)

∂2h ∂2h = . ∂t∂x ∂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.

dx + 8t − x2 t = 0 . dt

Solve (x − xt2 )

For this equation p = x − xt2 , q = 8t − x2 t ,

∂p = 0 − 2xt = −2xt , ∂t ∂q = 0 − 2xt = −2xt . ∂x

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 ∂h = p = x − xt2 = q = 8t − x2 t , and ∂x ∂t which on integrating give h=

x2 x2 t2 − + f (t) 2 2

and

h = 4t2 −

x2 t2 + g(x) , 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 4

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) =

x2 t2 x2 − + 4t2 . 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 x2 t2 x2 − + 4t2 = constant . 2 2

5. Linear equations The most general first order linear ODE has the form dx + p(t) x = r(t) , 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 dx + g(t) p(t) x = g(t) r(t) , dt dx g(t) + (gpx − gr) = 0 . dt g(t)

i.e.

Using the main result from section 4, the above equation is exact provided and q(t) = g(t)p(t)x − g(t)r(t) .

∂q ∂p = , where p(t) = g(t) ∂t ∂x

Hence the equation is exact if ∂ ∂ (g(t)) = (g(t)p(t)x − g(t)r(t)) , ∂t ∂x

i.e.

dg = g(t)p(t) . dt

The latter equation is separable and so an expression for g(t) can easily be found Z

dg = g

Z p(t) dt , Z

ln |g| = |g| = e

p(t) dt + C , R

p(t) dt+C

R =e

e =Ae

R or

R

p(t) dt C

g(t) = A e

p(t) dt

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.

5

Solution procedure for linear equations: (a) make sure that the given ODE does have the required general form of a linear equation; dx equal to 1, dividing if necessary, and observe the form of (b) write the ODE with the coefficient of dt p(t) ; R (c) calculate g(t) = e p(t) dt , which is usually called the integrating factor; (d) multiply the ODE (with leading coefficient unity) by g(t), and from the general theory we then know that the resulting equation is always exact and can be written d (g(t) x(t)) = g(t)r(t); dt (e) integrate the latter equation with respect to t to obtain x(t) . Let us now illustrate the above method with two examples. dx + 3x = t3 . dt First make the leading coefficient unity by dividing the ODE by t to give

Ex 4.

Solve t

dx 3 + x = t2 . dt t

(1)

The integrating factor is calculated by R g(t) = e

3t−1 dt

= e3 ln |t| = eln(|t|

3

)

= |t3 | = t3 ,

since the modulus sign can be neglected without any loss in generality. Multiplying equation (1) by t3 gives t3

dx + 3t2 x = t5 , dt d 3 (t x) = t5 . dt

Integrating with respect to t then yields the general solution t3 x =

t6 +C, 6

or

x=

t3 C + 3. 6 t

x dx − 2 = t3 . dt t Here the coefficient of the derivative term is already unity and it is observed that p(t) = −2/t . Hence the integrating factor is obtained as  Z
1 1 2 g(t) = exp − dt = exp (−2 ln |t|) = exp ln(|t|−2 ) = 2 = 2 , t |t| t

Ex 5.

Solve

since the modulus sign can be neglected in the final step without loss of generality. On multiplying the original ODE by the integrating factor 1 dx 2x − 3 = t, t2 dt t d x = t, i.e. dt t2 t2 x +C, = t2 2 6

→

x=

t4 + Ct2 . 2

6. 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 x dx separable, linear, =f , exact, dt t and then stop and solve by the first category in which it appears. dx +x = 0. dt Z Z dt dx =− ; x t

[To illustrate the above consider the equation t separable?

YES

linear?

YES  dx x =f ? YES dt t

exact?

YES

since

dx 1 + x = 0, p(t) = 1/t, r(t) = 0 ; dt t dx x =− ; dt t ∂ ∂ (t) = 1 , (x) = 1 , equal. ∂t ∂x

Thus given equation falls into all four categories, but is most easily solved as a separable equation.] rec/00lode2

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