Test Bank For INSTRUCTOR'S SOLUTIONS MANUAL to accompany ADVANCED ENGINEERING MATHEMATICS PETER V. O'NEIL (Complete And Verified Study material) (532pages) LEARNEXAMS

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.

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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

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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