Midterm 1 Solutions - differential equations PDF

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differential equations...

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UNIVERSITY OF REGINA Department of Mathematics and Statistics Differential Equations I Mathematics 217 (Section 001) Fall 2019 Instructor: Liviu Mare Midterm Test # 1 - Solutions)

Time: 50 min.

Full Name:

Pages: 6

Student Number:

INSTRUCTIONS 1. The left side pages are to be used as scrap paper. They are provided for rough work and checking only and will not be graded unless you expressly indicate there is work to be found there. 2. The marks allocated for each question are found to the left of the question.

Good Luck!

For instructor use only: Page:

3

4

5

Total

Marks:

2

5

3

10

Score:

MATH 217

Page 3

Midterm Test # 1

Marks (5)

1. It is known that the general solution of the differential equation 2 y x

1 2 y x

y is 2

2

. 1 Find a solution of the IVP consisting of this differential equation and the condition 4. y 2 y

cx2

Give the largest interval I over which the solution is defined.

Solution. Substitute into the given solution and get: y

2

2

2 c 2 1

2 . 2c 1

2

Solve for c: 2

2

4

2c 1 2 2 2c 1 2c 1 1

2c 2 c 1. Get y 2

2 x2

1

.

The denominator is x2 1, not defined for x 1 and x 1. One is looking for the largest interval I which doesn’t contain neither 1 nor 1, and contains 2. This is 1, .

/2

MATH 217 (5)

Page 4

Midterm Test # 1

2. Find the general solution of the differential equation dy dx

3xex y .

dy dx

3xex e y .

Solution. Rewrite the equation:

This is a separable DE. We solve it in the usual way: ey dy

3xex dx 3xex dx

ey dy

Take the last integral separately and compute it by parts: 3xex dx

3

3 xex

xex dx

3

d x ex dx dx

x

d x e dx dx

3 xex

ex dx

3 xex

c

ex

c.

Back to the DE: ey y (5)

3 xex ex c ln 3 xex ex c .

3. Find the general solution of the differential equation 2xy3 2 dx

3x2 y 2

8e4y dy

0.

Hint. The equation is exact and you do not need to check this.

Solution. Search for f x, y such that ∂f ∂x ∂f 2 ∂y 1

2xy3

2

3x2 y 2 8e4y

From the first condition, f x, y

2xy3 2 dx g y

x2 y 3

2x g y .

/5

MATH 217

Page 5

Midterm Test # 1

By substituting this into (2), one obtains: 3x2 y 2 g y g y 8e4y

8e4y

8e4y dy

g y

1 4y e 4

8

2e4y .

One obtains: x2 y 3

f x, y

2x 2e4y .

The general solution is: x2 y 3 2x 2e4y (5)

c.

4. Find the general solution of the differential equation dy dx

xy2 .

2xy

Solution. This is a Bernoulli equation. Divide by y 2 and obtain: 2 dy

y Substitute u

1

2xy

dx

x.

y 1 , which gives: du dx

d y 1 dx dy du . y 2 dx dx

y

2 dy

dx

The DE becomes

du 2xu x. dx This is a linear equation, which is solved in the usual way. First rewrite it: du dx

x.

2xu

Then compute x2 .

2x dx The integrating factor is µ x Multiply both sides of the DE by e 2

ex

2

x

du dx

d e dx ex u

x2

2

e u

e

x2

.

and get x2

xe

2xu

xe xe

x2

x2 x2

dx.

/3

MATH 217

Page 6

Midterm Test # 1

The integral in the right hand side can be computed by substitution: v 2xdx and the integral is dv xe

x2

1 v e dv 2

dx

1 v e 2

c

1 e 2

x2

x2 , so

c.

One obtains: e

x2

1 e 2

u 1 2

u

x2

c

2

cex .

Resubstitute y y

1 2

1

1 2

2

cex

1 . cex2

/0...