1550078481-Ordinary_Differential_Equations__Roberts_

62 Ordinary Differential Equations

EXAMPLE 3 Solving an IVP Using Separation of Variables

Solve the initial value problem

(18) y' = xy + 2x; y(O) = 3.

SOLUTION

Writing the derivative y' as the ratio of differentials, dy / dx, and factoring.

the right-hand side of the differential equation in (18), we obtain the equivalent

equation

dy

dx = x(y + 2).

Multiplying by dx and dividing by (y + 2) , we get

Integration yields

dy

--= xdx, provided y =f. -2.

y+2

J y; 2 = J xdx

and then the implicit solution

x2

ln IY + 21 =

2

+ C , provided y =f. -2

and where C is an arbitrary constant. Exponentiating, we find

Both the left-hand side and the right-hand side of (19) are positive. Since

IY + 21 = ±(y + 2) and e^0 is an arbitrary positive constant, we may remove

the absolute value appearing in equation (19), if we replace ec by a new

arbitrary constant K which may be positive or negative. Doing so, we obtain

the following solution of the differential equation y' = xy + 2x :

y + 2 = Kex2/2

where K =f. 0. Noting that y = -2 is a particular solution of the differential
equation y' = xy + 2x, we can remove the restriction K =f. 0 and obtain the
explicit solution of y' = xy + 2x

(20)