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UNIT III. Solutions of First Order Differential Equations
Overview
This unit gives you an insight regarding the reverse process operation of
differential equation. Further, it gives you the analytical skill regarding the application of
each solution, and finally, it helps you recognized the derivation of particular and general
solution from the given differential equation.
Learning Objectives:
At the end of the unit, I am able to:
- Identify whether the differential equation is homogeneous, exact or linear;
- Apply separation of variables, solution of homogenous, exact and linear
equations to find the general and particular equation of the given differential
equations.
Setting Up
Directions. Answer each question clearly and completely. Write your answer on the
spaces provided.
- Transform the equation ( 1 + ๐ฆ
2
2
)๐๐ฆ = 0 in the form ๐ด(๐ฅ)๐๐ฅ +
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- What is homogeneous function? Give example.
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- What is exact differentia equation?
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- The equation
๐๐ฆ
๐๐ฅ
๐ฆ = ๐(๐ฅ) is a linear equation in ๐ฅ. Write the equivalent
linear equation in ๐ฆ.
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โ 2 ๐ก
2
โ๐๐
๐
0
0
๐
โ๐๐
๐
Example 2. Obtain the general and particular solution of
2
2
Example 3. ๐๐๐๐ฃ๐ ๐กโ๐ ๐๐๐๐ก๐๐๐ข๐๐๐ ๐ ๐๐๐ข๐ก๐๐๐ ๐๐ ๐ฆโฒ = ๐ฅ๐๐ฅ๐(๐ฆ โ ๐ฅ
2
โ๐
๐
๐
โ๐๐
๐
๐
Answer
Ans. ๐๐๐๐ = ๐๐๐๐๐
Assessing Learning
Name ___________________________________ Score ____________
Year and Sec. _____________________________ Date _____________
Activity No. 3.
Obtain the general and particular solution of the following:
2
Ans. ๐ฆ
2
5
2
๐ฅ
2
Ans. ๐ฆ
3
๐ฅ
2
4
2
Ans. ๐ฆ๐๐๐( 1 โ ๐ฅ) = 1
II. HOMOGENEOUS EQUATIONS
The differential equation
is homogeneous if
In general,
๐
example of Homogeneous differential equation ( 2 x
2
2
)dx + xydy = 0
let, ๐
= ( 2 x
2
2
)dx + xydy
2
2
)dx + (๐๐ฅ )( ๐๐ฆ)๐๐ฆ
2
[( 2 x
2
2
)dx + xydy]
2
Thus, the degree is 2.
to solve for ๐ธ๐๐ข๐๐ก๐๐๐ ( 1 )) let
or
after substitution, resulted differential equation will lead to ๐ฃ๐๐๐๐๐๐๐ ๐ ๐๐๐๐๐๐๐๐.
Example 1. Solve the solution of
3(3x
2
2
)dx โ 2xydy = 0
Solution:
3(3x
2
2
)dx โ 2xydy = 0
Let
Substitute,
3(3x
2
2
x
2
)dx โ 2vx
2
(vdx + xdv) = 0
3(3 + v
2
)x
2
dx โ 2vx
2
(vdx + xdv) = 0
Divide by x
2
3(3 + v
2
)dx โ 2v(vdx + xdv) = 0
9dx + 3v
2
dx โ 2v
2
dxโ2vxdv = 0
9dx + v
2
dx โ 2vxdv = 0
(9 + v
2
)dx โ 2vxdv = 0
( 9 +๐ฃ
2
)๐๐ฅ
๐ฅ( 9 +๐ฃ
2
)
2 ๐ฃ๐ฅ๐๐ฃ
๐ฅ( 9 +๐ฃ
2
)
2
2
lnx โ ln(9+v
2
) = lnc
ln x/(9+v
2
) = lnc
x/(9+v
2
) = c
x = c(9+v
2
From
๐ฆ
๐ฅ
Thus, ๐ฅ = ๐ ( 9 +
๐ฆ
2
๐ฅ
2
x
3
= c(9x
2
+y
2
) answer
Assessing Learning
Name ______________________________________ Score __________________
Year and Section _____________________ Date ___________________
Activity No. 3.
Determine the general solution of the following:
- xydx + (x
2
2
)dy = 0
- x
2
yโ = 4x
2
2
- 3xydx + (x
2
2
- (x โ y)dx + (3x + y) = 0 ; when x = 3, y = - 2 ; find the particular solution.
- xydx โ (x + 2y)
2
dy = 0
- [x csc(y/x) โ y]dx + xdy = 0
Step 1: Let
Step 2: Integrate partially with respect to x, holding y as constant
2
Step 3: Differentiate ๐ธ๐๐ข๐๐ก๐๐๐
partially with respect to ๐ฆ, holding ๐ฅ as constant
Step 4: Equate the result of Step 3 to N and collect similar terms. Let
Step 5: Integrate partially the result in Step 4 with respect to y, holding x as constant
2
Step 6: Substitute f(y) to Equation (1)
2
2
2
2
๐
๐
Example 2
2
Ans. 3x
2
+ xy
2
- y
3
= c
Assessing Learning
Name ________________________________ Score _________________
Year and Sec. __________________________ Date __________________
Activity No. 3.
Test the exactness of each of the following equations and solve. If not exact solve using
the previous method.
- (2xy + y)dx + (x
2
- (x โ 2y)dx + 2(y โ x)dy = 0
- (1 + y
2
)dx โ (x
2
y + y)dy = 0
- (cosx cosy โ cotx)dx โ sinxsiny dy = 0
- [2x + y cos(xy)]dx + x cos(xy)dy = 0
- (1 โ xy)
dx + [y
2
2
(1 โ xy)
]dy = 0 ; when x = 2, y = 1
โซ ๐๐๐ฅ
โซ ๐๐๐ฅ
โซ ๐๐๐ฅ
๐ข
๐ข
๐ข
๐ข
๐ข
๐ข
๐ข
โซ ๐๐๐ฅ
โซ ๐๐๐ฅ
Example 1. Solve the solution of (๐ฅ
5
5
5
4
4
4
โซ ๐๐๐ฅ
โซ (โ 3 /๐ฅ)๐๐ฅ
โซ ๐๐๐ฅ
โ 3 โซ ๐๐ฅ/๐ฅ
โซ ๐๐๐ฅ
โ 3 ๐๐๐ฅ
โซ ๐๐๐ฅ
โ 3
โซ ๐๐๐ฅ
โซ ๐๐๐ฅ
โ 3
4
โ 3
โ 3
3
2
3
๐
๐
Example 2. 2 ( 2 ๐ฅ๐ฆ + 4 ๐ฆ โ 3 )๐๐ฅ + (๐ฅ + 2 )
2
Ans. y = 2 (x+ 2 )
โ 1
โ 4
