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1.4 Initial Value Problems
As we have seen, most differential equations have more than one solution. For a first-order equation, the general solution usually involves an arbitrary constant C, with one particular solution corresponding to each value of C. What this means is that knowing a differential equation that a function y(x) satisfies is not enough information to determine y(x). To find the formula for y(x) precisely, we need one more piece of information, usually called an initial condition. For example, suppose we know that a function y(x) satisfies the differential equation
y′^ y.
It follows that
y(x) Cex
for some constant C. If we want to determine C, we need at least one more piece of information about the function y(x). For example, if we also know that
y( 0 ) 3 ,
the the value of C must be 3 , and hence y(x) 3 ex^.
Initial Value Problems An initial value problem consists of
- A first-order differential equation y′^ f (x, y), and
- An initial condition of the form y(a) b.
For example, y′^ y, y( 0 ) 3
is an initial value problem, whose solution is
y 3 ex^.
In general, we expect that every initial value problem has exactly one solution. We can find this solution using the following procedure.
Solving Initial Value Problems Given an initial value problem
y′^ f (x, y), y(a) b,
we can solve it using the following procedure:
- Find the general solution to the given differential equation, involving an arbitrary constant C.
- Substitute x a and y b to get an equation for C.
- Solve for C and then substitute the answer back into the formula for y.
EXAMPLE 1 Find the solution to the following initial value problem:
y′^ −y^2 , y( 0 ) 5.
SOLUTION We previously found the general solution to this differential equation:
y 1 x + C
,
Plugging in x 0 and y 5 gives the equation
5 (^0) +^1 C.
Solving for C gives C 1 / 5 , so y (^) x + 1 ( 1 / 5 ).
This simplifies to In this last step we multiplied the numerator and denominator by 5 to simplify the fraction of fractions.
y (^5) x 5 + 1
EXAMPLE 2 Find the solution to the following initial value problem:
y′^ 2 y, y( 0 ) 5.
SOLUTION The given differential equation isn’t very different from the equation
y′^ y.
In that case, the general solution was y Cex^. How can we modify this solution to account for the extra 2? A few moments of thought reveals the answer: More generally, the solution to any equation of the form y′^ k y (where k is a constant) is y Cekx^.
y Ce^2 x
So this is the general solution to the given equation. Plugging in x 0 and y 5 gives the equation 5 Ce^0 , so C 5 and the solution is y 5 e^2 x
The Fundamental Theorem of ODE’s (Optional) As a general rule, we expect any initial value problem of the form
y′^ f (x, y), y(a) b
to have a unique solution. The following theorem gives specific conditions which guarantee that this holds.
Fundamental Theorem of ODE’s (Geometric Version) Consider a first-order differential equation of the form
y′^ f (x, y),
where the function f (x, y) is continuously differentiable. Then:
- The solution curves for this differential equation completely fill the plane, and
- Solution curves for different solutions do not intersect.
Here statement (1) is the same as saying that every point (a, b) lies on at least one solution curve, i.e. every initial condition gives at least one solution. Statement (2) is the same as saying that no point (a, b) lies on more than one solution curve, i.e. every initial condition has at most one solution.
EXERCISES
1–2 Solve the given initial value problem.
1. y′^ xex^ , y( 0 ) 3 2. y′^ 3 y, y( 2 ) 4
