Kenneth Kuttler

564 CHAPTER 29. FIRST ORDER SCALAR ODE

This is a homogeneous equation. Let z = vu . Then

uz′ =6− zz−2

− z =1

z−2(−z2 + z+6

)Separating the variables,

(2− z)dzz2− z−6

=duu

This is easily solved,

C−(

ln |z+2|+ 15

ln |z−3|)= ln |u|

The in terms of the original variables,

C =

(45

ln∣∣∣∣y−2x−1

+2∣∣∣∣+ 1

5ln∣∣∣∣y−2x−1

−3∣∣∣∣)+ ln |x−1|

Then to contain the ordered pair, you need

C =45

ln2+15

ln3 =15

ln(48)

PROCEDURE 29.7.4 To solve affine linear equations of the form

(px+qy+ r)dx+(αx+βy+ γ)dy = 0,

do the following:

1. Change the variables u = x− a, v = y− b, plug in and choose a,b to make theresulting equation homogeneous.

2. Solve the resulting homogeneous equation. Then substitute back in x− a for u andy−b for v. Pick the constant to satisfy initial conditions.

29.8 Linear and Nonlinear Differential EquationsRecall initial value problems for linear differential equations are those of the form

y′+ p(t)y = q(t) , y(t0) = y0 (29.9)

where p(t) and q(t) are continuous functions of t. Then if t0 ∈ [a,b] , an interval, thereexists a unique solution to the initial value problem given above which is defined for allt ∈ [a,b]. The following theorem which is really something of a review gives a proof.

Theorem 29.8.1 Let [a,b] be an interval containing t0 and let p(t) and q(t) be continuousfunctions defined on [a,b] . Then there exists a unique solution to 29.9 valid for all t ∈ [a,b] .

Proof: Let P′ (t) = p(t) ,P(t0) = 0. For example, let P(t)≡∫ t

t0 p(s)ds. Then multiplyboth sides of the differential equation by exp(P(t)). This yields

(y(t)exp(P(t)))′ = q(t)exp(P(t))

564 CHAPTER 29. FIRST ORDER SCALAR ODEThis is a homogeneous equation. Let z= 7. Then,_ O-Z | ol 2ua = > == ( z+z+6)Separating the variables,(2—z)dz__ du2—-z-6 ouThis is easily solved,4 1c- (Finle+2] +5 Inle~3)) =Inh5 5The in terms of the original variables,4. |y-2 1. jy—2C= | =Inj|—— +2/+4 =InJ—— —3 Injx—1(3 nfo + [+ gin)? )) +m |Then to contain the ordered pair, you need4 1 1C= ZIn2+ <In3 = <In(48)PROCEDURE 29.7.4 1 solve affine linear equations of the form(px+qy+r)dx+(ax+ Byt+y)dy =0,do the following:I. Change the variables u = x—a, v= y—b, plug in and choose a,b to make theresulting equation homogeneous.2. Solve the resulting homogeneous equation. Then substitute back in x — a for u andy—b for v. Pick the constant to satisfy initial conditions.29.8 Linear and Nonlinear Differential EquationsRecall initial value problems for linear differential equations are those of the formy+p(t)y=q(t), y(t.) =yo (29.9)where p(t) and q(t) are continuous functions of t. Then if to € [a,b], an interval, thereexists a unique solution to the initial value problem given above which is defined for allt € [a,b]. The following theorem which is really something of a review gives a proof.Theorem 29.8.1 Let [a,b] be an interval containing to and let p(t) and q(t) be continuousfunctions defined on |a,b| . Then there exists a unique solution to 29.9 valid for allt € [a,b].Proof: Let P’ (t) = p(t) ,P (to) = 0. For example, let P(t) = Sip p(s) ds. Then multiplyboth sides of the differential equation by exp(P(t)). This yields(y(t)exp (P(t) = a(r)exp(P())