Solution Of A Linear Ode By Using An Integration Factor (michael De Silva)

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Solution of a Linear ODE by Using an Integration Factor

The purpose of this exercise is to demonstrate how a linear ODE can be solved by the use of an integration factor, but more importantly, to prove that the method devised for formulating the general solution is mathematically sound. The methodology for solving a linear ODE with the aid of an integrating factor, in its most concise form, is presented below. T HEORY: S OLUTION OF A L INEAR ODE BY U SING AN I NTEGRATION FACTOR If a first-order ODE can be written in the linear form, as shown below, y 0 + p(t)y = q(t) the ODE can be solved using an integrating factor µ(t) = e

R

p(t) dt

,

µ(t) [y 0 + p(t)y] = µ(t)q(t) (µ(t)y)0 = µ(t)q(t) Z µ(t)y =

µ(t)q(t) dt + C

Dividing through by µ(t), we have the general solution of the linear ODE.

To prove that this methodology will indeed work, we need to first establish that (µ(t)y)0 = µ(t) [y 0 + p(t)y]. R

Since, µ(t) = e

p(t) dt

(µ(t)y)0 = (e =

R

, consider,

p(t) dt

y)0

R d  R p(t) dt  dy e y + e p(t) dt dt dt

=e

R

p(t) dt

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0

[y + p(t)y]

= µ(t) [y 0 + p(t)y] We will now consider two examples. In the first example, we will opt to apply the Chain Rule first; it should be noted how many more steps are required to arrive at the solution. Ideally, one should be able to 2 “recognize” that (e−t y)0 can be directly solved by applying the Product Rule of differentiation, and this is demonstrated in the final example.

Michael M. Wijetunge de Silva Email: [email protected]

E XAMPLES:

1. Let us now consider the following linear ODE, y 0 − 2ty = t,

(2)

It is in the following form, where y 0 + p(t)y = q(t) such that p(t) = −2t, q(t) = t, and an integrating 2 2 2 factor e−t . It can be seen that (e−t y)0 = e−t (y 0 − 2ty) by applying the Chain Rule, as defined below, T HEORY: T HE C HAIN RULE OF D IFFERENTIATION The Chain Rule is a formula for the derivative of the composite of two functions. In intuitive terms, if a variable, y, depends on a second variable, u, which in turn depends on a third variable, t, then, dy du dy = dt du dt The derivative of a function f (t) = h(g(t)) is given by the Chain Rule as, f 0 (t) = h0 (g(t))g 0 (t)

Hence, 2

f (t) = h(g(t)) = e−t y

(3)

We will use a change of variable here, to aid in applying the Chain Rule. Let g(t) = u = t2 thus, g 0 (t) = du/dt = 2t, and h(u) = e−u y. Evaluate h0 (u), by applying the the Product Rule, as defined below, T HEORY: T HE P RODUCT RULE OF D IFFERENTIATION In calculus, the product rule also called Leibniz’s law, governs the differentiation of products of differentiable functions. It may be stated thus: (u · v)0 = u0 · v + u · v 0 or in Leibniz notation, thus d v u (u · v) = u · + v · dx x x

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Michael M. Wijetunge de Silva Email: [email protected]

Hence, h0 (u) = e−u y


0

= e−u

dy − e−u y du

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From the definition of the Chain Rule of differentiation, it can be seen that, dy dy dt = du dt du

(5)

and since we want to perform a change of variable from u → t, the above is substituted since it is known from g 0 (t) that dt/du = 1/2t, h0 (u) = e−u y


0

= e−u

dy dt e−u 0 − e−u y = y (t) − e−u y dt du 2t

2

e−t 0 2 y − e−t y h0 (g(t)) = 2t

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Hence, "

f 0 (t) = (e

−t2

# " 2 # −t2 −t e e 2 2 2 y)0 = y 0 − e−t y · g 0 (t) = y 0 − e−t y · 2t = e−t (y 0 − 2ty) 2t 2t

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2. It is also possible to solve f 0 (t), intuitively, by directly applying the Product Rule, 2

f 0 (t) = e−t y


0

=

d −t2 2 2 d 2 2 (e )y + e−t y = −2te−t y + e−t y 0 = e−t (y 0 − 2ty) dt dt

It can be seen that above, in solving k 0 (t) = and h(u) = e−u , ( g 0 (t) 0

h (u)

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d −t2 (e ), the Chain Rule is employed as, g(t) = u = t2 , dt

= 2t = −e−u

k 0 (t) = h0 (g(t))g 0 (t) = −2te−t

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Hence, the general solution of y 0 − 2ty = t is, Z 1 2 2 −t2 e y = e−t t dt + C = − e−t + C 2 1 2 y = Cet − 2

(10)

The particular solution may be evaluated if the initial values are specified, making this an Initial Value Problem.

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