questions ode

8/2/2019 Questions Ode

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From the problem statement, we know that:

0 = 1200

t0 = 0

Solution:

Step 1: is the approximate temperature at

Step 2:

For i = 1, t1 = 240, 1 = 106.09

is the approximate temperature at

1 240 106.09 110.322176

2 480 110.32

1

2


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Thus, The exact solution of the ODE is:

The solution of this equation at t= 480 seconds is

The Method of Reduction of Order

As a first assumption we discuss the solution of the homogenous differential equation is

(1.2)On the assumption that we have know one solution, say and only need to find thesecond solution. We will look for a solution of the form . Differentiating using the product rule gives

If we substiute these expressions into (1.2) we obtain

( ) We can now collect terms to get

( )


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Now, since is a solution of (1.2), the term multiplying is zero. We have therefore obtained adifferential equation for , and ,by defining , have

( )

Dividing through by we have

Which can be integrated directly to yield

|| ||

Where is a dummy variable, for some constant . Thus

Where . This can then be integrated to give

For some constant . The solution is therefore

We can recognize as the part of the complementary function that we knew to start with,and

(1.3)


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As the second part of the complementary function. This is called the reduction of order of

formula

EXAMPLE:

Determine the solution of the differential equation below

Given that is a solution.

SOLUTION

Firstly, we write the equation in standard form as

Comparing this with (1.2), we have . After noting that

We can express the integrand in term of its partial fractions as

This gives the second solution of (1.2) as

{ } [ ( )]

( ) And hence the general solution is


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{ ( ) }

Question 2:

A 10-kilogram mass is attached to a spring which is thereby stretched 0.7 meters from its natural

length. The mass is started in motion from the equilibrium position with an initial velocity of 1

meter /sec in the upward direction. Find the resulting motion if the force due to air resistance is

-90 newtons.Solution :

The motion is damped (air resistance) and free (no external force). Our differential equation from

Newtons second law is :

= -90 .(a)Where = the mass (10 kg), and is the spring constant.Since = ,

{Note that a mass of 10 kg exerts a force or weight of (9.8)(10) = 98 newtons). Substituting into

(a) we have

10 + 90 + 140 = 0Or

..(b)The characteristic equation is


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2

+ 9+14 = 0

or ( + 2) ( + 7) = 0

Thus the roots are

= -2 and = -7

Our general solution is, thus

+ ..(c)Where and are arbitrary constants to be determined by the initial conditions which are

0

And t=0 = -1(the initial velocity is in the negative x-direction).

Thus + ,0 = + ..(d)

And

t=0 = -2

+

,

-1 = -2+ 7 = 1Or 2 + 7= 1 (e)Since by (d) , (e) becomes2(-) + 7 = 1

= And - Our final solution is


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- .

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