The Initial Value Problem y' = f(x, y); y(c) = d 63where K is any arbitrary real constant. Since the initial condition, y(O) = 3,
in the IVP (18) is specified at x = 0, we set x = 0 in equation (20) and find
that K must satisfyy(0)+2=Ke^0 or 3+2=5=K.
So the expli cit solution of the IVP (18) isy(x) = 5ex2
/^2 - 2.Notice that this solution exists on the interval (-oo, oo ).EXERCISES 2.3.2In exercises 1-10 solve the given initial value problem by finding
the solution of the separable differential equation and then deter-
mining the value of the constant of integration which satisfies the
initial condition. Specify the interval on which each solution exists.
l. y' = 3y; y(O) = -1 2. y' = -y + l; y(O) = 1
3. y' = -y + l; y(O) = 2 4.
2
y' = xev-x ; y(O) = 0- y' = y/x; y(-1) = 2 6. y' = 2x/y; y(O) = 2
7. y' = -2y + y2; y(O) = 1 8. y' = xy + x; y(l) = 2
- xeY dx + dy = O; y(O) = 0 10. y dx - x^2 dy = O; y(l) = 1
- Verify that y^2 - x = 1 is an implicit solution of the differential equation
2yy' = 1 on the interval ( -1, oo).
12. Verify that xy^2 +x = 1 is an implicit solution of the differential equation
2xyy' + y^2 = -1 on the interval (0, 1).
13. Verify that x = exv is an implicit solution of the differential equation
y' = (1 - xy)/x^2 on the interval (0, oo).
- Verify that the relation xy^2 + yx^2 = 1 formally satisfies the differential
equation y' = -y(2x + y)/x(2y + x). - Verify that the relation y = exy formally satisfies the differential equa-