math 3113 practice exam 1 answer key

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1. Just plug in to check 2. Just plug in to check 3. f (x, y)= x ln y and ∂f ∂y = x y , and the point in question is (1, 1). Both f and ∂f/∂y are continuous at the point, therefore there exists a unique solution. 4. f (x, y)= x 2 - y 2 and ∂f ∂y = -2y , and the point in question is (0, 1). Both f and ∂f/∂y are continuous at the point, therefore there exists a unique solution. 5. f (x, y)=1+ x 2 + y 2 and ∂f ∂y =2y , and the point in question is (0, 2). Both f and ∂f/∂y are continuous at the point, therefore there exists a unique solution. 6. y = tan C - x - 1 x 7. y = C sin x 8. y = -1+ 1 3 C - 3 tan -1 x 9. y = x 2 sin x - 3x 2 10. y = 1 2 +4 ( x 2 +4 ) - 3 2 11. y =3xe 2x 12. (x + e y ) 2 =2x 2 + C 13. y = 3 Ce x - 3x 4 - 12x 3 - 36x 2 - 72x + 72 14. 2y + p x 2 +4y 2 = Cx 3 2 15. Just plug in to check 1

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Solution to First Practice Exam for Differential Equations

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Page 1: Math 3113 Practice Exam 1 Answer Key

1. Just plug in to check

2. Just plug in to check

3. f(x, y) = x ln y and∂f

∂y=

x

y, and the point in question is (1, 1). Both f and ∂f/∂y are

continuous at the point, therefore there exists a unique solution.

4. f(x, y) = x2 − y2 and∂f

∂y= −2y , and the point in question is (0, 1). Both f and ∂f/∂y

are continuous at the point, therefore there exists a unique solution.

5. f(x, y) = 1+x2+y2 and∂f

∂y= 2y , and the point in question is (0, 2). Both f and ∂f/∂y

are continuous at the point, therefore there exists a unique solution.

6. y = tan

(C − x− 1

x

)7. y = C sinx

8. y = −1 + 13√C − 3 tan−1 x

9. y = x2 sinx− 3x2

10. y =1

2+ 4

(x2 + 4

)− 32

11. y = 3xe2x

12. (x+ ey)2 = 2x2 + C

13. y =3√Cex − 3x4 − 12x3 − 36x2 − 72x+ 72

14. 2y +√x2 + 4y2 = Cx

32

15. Just plug in to check

1