1550078481-Ordinary_Differential_Equations__Roberts_

56 Ordinary Differential Equations

b. Solve the initial value problem

I 1

Y = (x + 2); y(l) =^4

and specify the interva l on which the solution exists and is unique.

c. Solve the initial value problem

I 1

y = (x + 2); y(-7) =^3

and sp ecify the interva l on which the solution exists and is unique.

SOLUTION

a. Integrating equation (3), we find the general solution to b e

(4)

i

x 1

y(x ) = (t+ 2 )dt+C=lnlx+2l+C.

Since the function g ( x ) = 1 / ( x + 2) is defined and continuous on the

interva ls (-oo, -2) and (-2, oo ) , equation ( 4) is the general solution of

differentia l equation (3) on the interva ls (-oo, -2) and (- 2, oo ).

b. To solve the initia l value problem

(5)

I 1

y = (x + 2); y(l) =^4

a ll we n eed to do is find the value of C in the general solution ( 4) which

will yield y(l) = 4. Setting x = 1 in (4), we see that C must satisfy

y(l) =lnl1+2l+C=4.

Solving for C, we get C = 4 - ln3. Substituting this value into (4)

produces the following general solution to the IVP (5)

(6)

Ix+ 21

y( x ) = ln Ix + 21+4 - ln3 = ln -

3

Since t he initia l condition, y(l) = 4, is sp ecified a t x = 1 E (-2, oo) ,

equation (6) is the unique solution of the IVP (5) on the interval (-2, oo ).

c. To solve the initia l value problem

(7) y'

1

y(-7) = 3

(x + 2 )'

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