PDF file of the Elementary Differential Equation. Problem and Solutions ONLY that found in MATHalino.com

1. Elimination of Arbitrary
Constants
→ equation (1)
Problem 01
Divide by dx
Solution 01
Substitute c to equation (1)
Divide by 3x
Multiply by dx
answer
Problem 02
answer
Another Solution
Solution 02
answer
okay
Problem 03
Problem 04
Solution 03

2. Solution 04
→ equation (1)
Multiply by y3
Substitute c to equation (1)
okay
Separation of Variables | Equations of Order
One
Problem 01
, when
,
Solution 01
answer
HideClick here to show or hide the solution
Another Solution
Divide by y2
when
,

3. Problem 03
, when
then,
Solution 03
answer
Problem 02
, when
,
.
Solution 2
when
when
,
then,
then,
answer
,
,
.

4. answer
Problem 04
, when
,
answer
.
Problem 05
, when
Solution 04
,
.
Solution 05
From Solution 04,
when x = -2, y = 1
Thus,
Therefore,
answer
when x = 2, y = 1
Problem 06
, when
Thus,
Solution 06
,
.

5. From Solution 04,
when x = 0, y = 0
when x = 2, y = -1
thus,
Thus,
answer
Problem 07
, when
,
.
Solution 07
answer
Problem 08
, when
,
.
Solution 08

6. answer
Problem 09
, when
For
Let
,
.
,
,
Solution 09
Then,
when x → ∞, y → ½
Thus,
when θ = 0, r = a

7. answer
Problem 11
Thus,
Solution 11
answer
Problem 10
, when
,
.
Solution 10
answer
Problem 12
when x = xo, v = vo
Solution 12
thus,

8. answer
Problem 15
answer
Solution 15
Problem 13
Solution 13
answer
Problem 16
Solution 16
answer
Problem 14
Solution 14

9. answer
Problem 17
answer
Problem 19
Solution 17
Solution 19
answer
Problem 18
Solution 18
answer
Problem 20
Solution 20

10. ans
wer
Problem 22
Solution 22
answer
Problem 21
By long division
Solution 21
Thus,
By long division
Thus,
answer

11. Problem 23
Solution 23
answer
Homogeneous Functions | Equations of Order
One
Problem 01
From
Solution 01
Thus,
Let
answer
Substitute,
Problem 02
Divide by x2,
Solution 02

12. Let
answer
Problem 03
Substitute,
Solution 03
Let
From

13. From
Thus,
answer
From
Problem 04
Thus,
Solution 04
answer
Let
Exact Equations | Equations of Order One
Problem 01
Solution 01

14. Test for exactness
Step 5: Integrate partially the result in Step 4
with respect to y, holding x as constant
;
;
; thus, exact!
Step 1: Let
Step 6: Substitute f(y) to Equation (1)
Equate F to ½c
answer
Step 2: Integrate partially with respect to x,
holding y as constant
Problem 02
Solution 02
→ Equation (1)
Step 3: Differentiate Equation (1) partially with
respect to y, holding x as constant
Test for exactness
Step 4: Equate the result of Step 3 to N and
collect similar terms. Let
Exact!
Let

15. Solution 03
Integrate partially in x, holding y as constant
→ Equation (1)
Test for exactness
Differentiate partially in y, holding x as constant
Exact!
Let
Let
Integrate partially in y, holding x as constant
Integrate partially in x, holding y as constant
→ Equation (1)
Substitute f(y) to Equation (1)
Differentiate partially in y, holding x as constant
Equate F to c
answer
Problem 03
Let

16. Integrate partially in y, holding x as constant
Integrate partially in x, holding y as constant
→ Equation
Substitute f(y) to Equation (1)
(1)
Differentiate partially in y, holding x as constant
Equate F to c
answer
Problem 04
Let
Solution 04
Integrate partially in y, holding x as constant
Test for exactness
Substitute f(y) to Equation (1)
Exact!
Equate F to c
answer
Let
Linear Equations of Order One

17. Problem 01
Solution 01
Multiply by 2x3
answer
Problem 02
Solution 02
→ linear in y
Hence,
Integrating factor:
→ linear in
y
Thus,
Hence,

18. Integrating factor:
Integrating factor:
Thus,
Thus,
Using integration by parts
,
,
Mulitply by (x + 2)-4
answer
Multiply by 4e-2x
answer
Problem 03
Problem 04
Solution 03
Solution 04
→ linear in y
Hence,

19. → linear in x
Multiply 20(y + 1)-4
Hence,
answer
Integrating Factors Found by Inspection
Problem 01
Integrating factor:
Solution 01
Divide by y2
Thus,
Using integration by parts
,
Multiply by y
answer
,
Problem 02

20. Divide by x both sides
Solution 02
Divide by y3
answer
Problem 04
Solution 04
answer
Problem 03
Solution 03

21. - See more at:
Problem 06
Solution 06
Multiply by s2t2
answer
Problem 05
Problem 05
answer
Problem 07
Solution 07 - Another Solution for Problem 06
Divide by xy(y2 + 1)
answer

22. Resolve into partial fraction
Problem 11
Solution 11
Set y = 0, A = -1
Equate coefficients of y2
1=A+B
1 = -1 + B
B=2
Equate coefficients of y
0=0+C
C=0
Hence,
answer
The Determination of Integrating Factor
Problem 01
Thus,
Solution 01
answer - okay

23. answer
→ a function of
x alone
Problem 02
Solution 02
Integrating factor
Thus,

24. →
a function of x alone
Integrating factor
Thus,
→
neither a function of x alone nor y alone
→ a function of y alone
answer
Problem 03
Integrating factor
Solution 03
Thus,

25. →
neither a function of x alone nor y alone
→
a function y alone
answer
Problem 04
Integrating factor
Solution 04
Thus,

26. answer
Substitution Suggested by the Equation |
Bernoulli's Equation
Problem 01
Solution 01
Let
ans
Problem 02
Thus,
Solution 02
Let
→ vari
ables separable
Divide by ½(5z + 11)
Hence,
→ homogeneo
us equation
Let

27. Divide by vx3(3 + v)
From
Consider
Set v = 0, A = 2/3
Set v = -3, B = -2/3
But
Thus,
answer
Problem 03

28. Solution 03
answer
Problem 04
Let
Solution 04
→ Bernoulli's
equation
From which
Integrating factor,
Thus,
But

29. answer
Problem 05
answer
Solution 05
Elementary Applications
Newton's Law of Cooling
Let
Problem 01
A thermometer which has been at the reading
of 70°F inside a house is placed outside where
the air temperature is 10°F. Three minutes later
it is found that the thermometer reading is
25°F. Find the thermometer reading after 6
minutes.
Solution 01
According to Newton’s Law of cooling, the
time rate of change of temperature is
proportional to the temperature difference.
When t = 0, T = 70°F

30. Thus,
Hence,
When x = 0.5xo
When t = 3 min, T = 25°F
answer
Thus,
After 6 minutes, t = 6
answer
Problem 02
A certain radioactive substance has a half-life of
38 hour. Find how long it takes for 90% of the
radioactivity to be dissipated.
Solution 02
Simple Chemical Conversion
When t = 38 hr, x = 0.5xo
Problem 01
Radium decomposes at a rate proportional to
the quantity of radium present. Suppose it is
found that in 25 years approximately 1.1% of
certain quantity of radium has decomposed.
Determine how long (in years) it will take for
one-half of the original amount of radium to
decompose.
Solution 01
Hence,
When 90% are dissipated, x = 0.1xo
answer
When t = 25 yrs., x = (100% - 1.1%)xo = 0.989xo