The Initial Value Problem y' = f(x, y); y(c) = d 65
THEOREM SOLUTION OF THE HOMOGENEOUS LINEAR
FIRST-ORDER DIFFERENTIAL EQUATIONIf a( x) is a continuous function on the interval (a, /3), then the solution
on (a, /3) of the homogeneous linear first-order differential equation(23) y' = a(x)y
is(25) y(x) = K ef a(x)dx
where K is an arbitrary constant.EXAMPLE 4 Solution of a Homogeneou s Linear First-Order IVPa. Find the solution of y' = (tanx)y.
b. Solve the initial value problem(26) y' = (tanx)y; y(2) = 3and specify the interval on which the solution exists and is unique.SOLUTION
a. In this example, the function a(x) = tanx is defined and continu-
ous on each of the intervals In = ((2n - l)7r/2, (2n + l)7r/2) for n =
0, ±1, ±2, .... By the previo us theorem the solution of y' = (tanx)y on
any interval In is(27)
y(x) = Kef tanxdx = Ke-Inlcosxl = __ K = Csecx
I cos x i
where C is an arbitrary constant.b. Setting x = 2 in equation (27) and imposing the initial condition,
y(2) = 3, requires that C satisfy the equation y(2) = 3 = C sec 2.
Thus, C = 3/ sec 2 and the solution of the IVP (26) is y = 3 sec x/ sec 2.
Since x = 2 E Ii = ( 7f /2, 37f /2), this solution exists and is unique on the
interval (7r/2, 37r/2).