Lesson Overview

Differential equations are frequently found in mathematical models for mechanical systems, biological populations, chemical reactions, and electrical circuits.  Even though the differential equations considered in this lesson are relatively simple, some interesting questions can be formulated -- and answered.  The key to solving a differential equation is manipulating it into a form requiring the evaluation of one or more indefinite integrals.

This lesson has two objectives:

•  begin to develop the ability to write differential equations corresponding to a physical situation
•  present a small number of methods for solving simple differential equations

The fact that solving a differential equation frequently involves finding antiderivatives means that it should be possible to verify that a function does, in fact, satisfy a given differential equation.

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What is a Differential Equation?

An equation  is a mathematical sentence that states that two quantities are equal, i.e., two expressions joined with an equal sign.  A differential equation  is an equation that shows a relationship between a function, its derivative(s), and the independent variable.  Solving a differential equation means finding eliminating the derivative(s) from the differential equation to produce an explicit formula for the function or an implicit relationship between the dependent and independent variables.

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Example 1: Verifying Solutions

Determine which of the following functions are solutions to the differential equation

>	alias( y=y(t) ): ode1 := diff( y, t$2 ) + 4*y = 0: ode1;

>	y1a := y = sin( 2*t ): y1a;

>	y1b := y = sin( t ): y1b;

>	y1c := y = cos( 2*t ): y1c;

>	y1d := y = t*cos( 2*t ): y1d;

>	y1e := y = C[1]*rhs(y1a) + C[2]*rhs(y1c): y1e;

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To check if a given function satisfies a differential equation insert the function into the differential equation, evaluate all derivatives, and simplify.

>	eval( ode1, y1a );

If the result is a true statement -- for all relevant values of the independent variable -- then the function does satisfy the differential equation.  In this case, the equation  is true for all values of  and so  is a solution to this differential equation.

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Repeating these steps for the second function leads to

>	eval( ode1, y1b );

As this equation is true only when t is an integer multiple of ,  is not  a solution to this differential equation.

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The same process can be used for the remaining three functions:

>	y1c; eval( ode1, y1c );

>	y1d; eval( ode1, y1d );

>	y1e; eval( ode1, y1e );

These results show that  and  are also solutions to this differential equation while  is not a solution.

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Example 2: General Solution

To find all functions that satisfy the differential equation

>	alias( y=y(x) ): ode2 := diff( y, x ) = 2/x^2 + 3: ode2;

note that this is really asking for all antiderivatives of

>	f2 := rhs( ode2 ): f(x) = f2;

Example 3: Particular Solution

Consider the same differential equation as in Example 2:

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ode2;

This time, find the function that satisfies the above differential equation and passes through the point ( 1, 5 ).

This extra condition, called an initial condition , is used to select a single solution from the infinite number of solutions in the general solution.  Together, a differential equation and initial condition are called an initial value problem .  The solution to an initial value problem is called a particular solution .

Thus, another way to state the problem for this examle is:  Find the particular solution that satisfies the initial value problem

>	ic3 := [ x=1, y=5 ]: ode2; ic3;

The general solution is known from Example 2:

>	y = gensol2;

When the initial condition is substituted into this equation, the result is the following equation for :

>	q3 := eval( y=gensol2, ic3 ): q3;

This equation is easily solved:

>	c3 := isolate( q3, C ): c3;

The particular solution solving this initial value problem is

>	partsol3 := eval( gensol2, c3 ): y = partsol3;

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To verify this solution does satisfy the initial value problem, check that the solution satisfies the differentiatl equation

>	eval( ode3, y=partsol3 );

and check that the initial condition is satisfied:

>	eval( y=partsol3, ic3 );

Because both of these tests returned an equation that is true, provided , we have found the solution to the initial value problem. (Technically, this solution makes sense on the largest interval where the differential equation is satisifed that contains the initial condition.  Here, that means for  > 0.)

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Example 4: Graphical Solution

Consider, the same initial value problem as in Example 3:

>	ode2; ic3;

The general solution was found in Example 2 and a particular solution was found in Example 3.  For many differential equations it is not feasible -- or necessary -- to obtain a symbolic representation of the solution; frequently a graphical solution can provide the necessary information.  While it is beyond the scope of the current course to go into great deatil about the graphical analysis of differential equations, you should be aware of some of the basic tools for preparing a graphical solution to a differential equation or initial value problem.

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Separable Equations

Only differential equations that can be written in the form

that can be solved by directly finding an antiderivative.  If the differential equation involves both y(x) and diff(y(x),x), then another approach must be used.  If the differential equation can be written in the form

then dividing the equation by G(y) (assuming G(y)<>0) it is possible to ``separate variables'':

     .  

If antiderivatives of both sides of this equation can be found, this will yield an implicit solution to the differential equation:

   .  

In general, the indefinite integral on the left-hand side will need to be evaluated with the Generalized Power Rule (with ).  Finally, if an initial condition is supplied, it should be used now to obtain an equation for .  If the antiderivatives work out particularly nice, it might be possible to solve for  to obtain an explicit solution.

Enough of the general discussion.  Let's look at a few specific examples.

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Example 5: A First Separable Example

Find the general solution to the differential equation

>	alias( y=y(x) ): ode5 := diff( y, x ) = -sqrt(x)/y: ode5;

This differential equation cannot be solved by direct antidifferentiation, but is separable. Multiply the equation by  yields

>	q1 := ode5*y: q1;

The indefinite integral, with respect to , of both sides of the separated equation leads to the equation

>	q2 := Int( lhs(q1), x ) = Int( rhs(q1), x ): q2;

These integrals are fairly easy to evaluate by the Generalized Power Rule (with  and ) and the Power Rule (with ).  The result, after manually inserting a single constant of integration, is

>	q3 := value( lhs(q2) ) = value( rhs(q2) ) + C: q3;

Other than multiplying this equation by 2

>	q4 := 2*q3: q4;

and replacing  with a new constant, say ,

>	q5 := eval( q4, C=K/2 ): q5;

there is about as simple as the general solution can be written.  Taking the square root requires the consideration that  could be positive or negative.

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Example 6: Projectile Motion

An object moving in a straight line has acceleration (in centimeters per second per second) given by

>	A := a = 10*(1+t)^(-5) + 3*sin(Pi*t): A;

Assume the object begins from rest with  initial position .  Find the velocity and position of the object for any time  > 0.

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Example 7: Wolf Population

The wolf population, W, in a certain state has been growing at a rate proportional to the cube root of the current population size.  The population was estimated to be 1000 in 1980 and 1700 in 1990.

Example 8: Escape Velocity

The gravitational attraction  exerted by Earth on an object with mass  located at a distance  from Earth's center is

>	alias( v=v, s=s ): Fgrav := F = -m*g*R^2/s^2: Fgrav;

where  is Earth's radius and  is the acceleration of gravaity at Earth's surface.  Assume the initial velocity is  and neglect air resistance.  For small values of  an object launched from Earth's surface will return to Earth; for large values of  the object will not fall back to Earth.  The escape velocity is the minimum value of  for which the object does not return to Earth.  Find the escape velocity as a function of , , and .

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Lesson Summary

This lesson is an introduction to differential equations and their use as models to physical problems.  The differential equations considered in this lesson could be solved by direct integration once the independent and dependent variables are separated.  The general solution to a differential equation corresponds to a family of antiderivatives; a particular solution to an initial value problem is found by using additional information to identify a single function in the general solution that satisfies the differential equation and the initial condition.  All solutions to differential equation can be checked by sustituting the proposed solution back into the differential equation.

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What's Next?