Chapter One First Order Differential Equations

Section 1.1 Preliminary Concepts

1. For z > 1,2 = 2√2-1 1 = 1, so is a solution.

22-1

2. With p(x)=ce we have +-ce+ce = 0, so is a solution.

1

3. For 0, we rewrite the equation as 2x + 2y <2 xss=removed>

-2xc

27

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4. For z ≠±√2, we have C -2 = (x-2)2 = 2-2 2 so is a solution. 2-

5. On any interval not containing z = 0 we have, z 112 3 272 = 2+ -(-)- 2

() = 2-4, so is a solution.

6. For all z, += -ce+(1+ce) = 1, thus p(x)=1+ce is a solution for all 2.

OF OF

In 7 11, recall that for y defined implicitly by F(x, y) = C we have (x, y) for which the partials Fr and Fy exist.

+0 for all Эх ду

7. With F(x,y) = y²+xy-2x23x-2yC, we have y-4x-3+(2y+x-2)=0.

8. With F(x,y)=zyyC, we have y³ (3xy-1) = 0.

9. With F(x,y) = x²-4x²+C, we have &ye(2+xe)y = 0.

10. With F(x,y)=8inz-2+4-2x+6yC, we have 0. Solving for y' gives y' = x-2y 3x6y+4 8 2-2y+4 +( (24+4+6)= 2+ -16

(+) 11. With F(x, y) = tan-1 (1)+2=C, we have -y +y 2+2+2x- -0.

12. Direct integration gives y=2²+ C. The initial condition y(2) 1 gives 1 = 4 + C во

2. The нelque solution in

2