Ordinary Differential Equations Introduction A differential equation is an equation involving an unknown function and its derivatives. A differential equation is called an ordinary differential equation (ODE) if the unknown function depends on only a single independent variable. If the unknown function depends on more than one independent variables, the differential equation is called a partial differential equation (PDE). Here are some examples of differential equations: 1) 2) 3)
đ?‘‘đ?‘Ś
3 đ?‘‘đ?‘Ľ = 2đ?‘Ś − 3 đ?‘‘2 đ?‘Ś đ?‘‘đ?‘Ľ 2 đ?‘‘2 đ?‘Ś đ?‘‘đ?‘Ľ 2
; ODE
đ?‘‘đ?‘Ś
+ đ?‘‘đ?‘Ľ + 5đ?‘Ś = đ?‘Ľ 2 ; ODE đ?‘‘đ?‘Ś
− 2 đ?‘‘đ?‘Ą = 0
; PDE
where đ?‘Ľ is known as an independent variable and đ?‘Ś is a dependent variable or an unknown function. A differential equation is called an ordinary differential equation (ODE) if the unknown function depends on only a single independent variable. If the unknown function depends on more than one independent variables, the differential equation is called a partial differential equation (PDE). The order of a differential equation is the order of the highest derivative in the equation. The degree of a differential equation is the degree of the highest order derivative which appears in the equation. The following table shows examples of ordinary differential equations with different order and degree: Ordinary differential equation dy  e x2 y dx d2y dy ď€5 0 2 dx dx
Order 1
Degree 1
2
1
2
3
1
ďƒŚ d 2 y ďƒś dy ďƒ§ďƒ§ 2 ďƒˇďƒˇ  ď€ 3y  x ďƒ¨ dx ďƒ¸ dx
2
3
3 x  1ď€ŠďƒŚďƒ§ dy ďƒśďƒˇ  d y3  sin x dx ďƒ¨ dx ďƒ¸ 3
1
A solution to a differential equation is any differentiable function đ?‘Ś = đ?‘“(đ?‘Ľ) which satisfies the differential equation. There are two types of solutions: the particular solution and the general solution or primitive solution. A particular solution to be a differential equation is one obtained by assigning definite values to the arbitrary constants. A general solution or primitive solution to a differential equation is a function with arbitrary constants that represents the family of all particular solutions. It is also known as a primitive solution.
Solving The First Order Ordinary Differential Equations In this section, we will deals with finding solution to differential equation of the first đ?‘‘đ?‘Ś
order, đ?‘‘đ?‘Ľ = đ?‘“(đ?‘Ľ, đ?‘Ś) where f is the function of two variables. Methods to be discussed here are only for: 1) Separable equations 2) Homogeneous Equations 3) Linear Equations
Solving Separable Differential Equations A differential equation of first order is said to have separable variables if the equation can be written in the form đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
đ?‘Œ(đ?‘Ś)
= đ?‘‹(đ?‘Ľ) đ?‘œđ?‘&#x;
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
= đ?‘Œ(đ?‘Ś). đ?‘‹(đ?‘Ľ)
The equation is said to be separable since the variable x and y could be separated by equal sign. đ?‘‘đ?‘Ś
For
đ?‘‘đ?‘Ľ
=
đ?‘‹(đ?‘Ś) đ?‘Œ(đ?‘Ľ)
, we rewrite the equation as đ?‘Œ(đ?‘Ś)đ?‘‘đ?‘Ś = đ?‘‹(đ?‘Ľ)đ?‘‘đ?‘Ľ (separating the
variable x and y). Then we get the solution by integrating the terms on both sides: âˆŤ đ?‘Œ(đ?‘Ś)đ?‘‘đ?‘Ś = âˆŤ đ?‘‹(đ?‘Ľ)đ?‘‘đ?‘Ľ . In short, the steps to solve separable equation are: 1) 2) 3)
separate the variables Integrate both sides Apply the initial value problem (if any)
2
Example 1 Solve the following differential equation: (1) (2) (3)
đ?‘‘đ?‘Ś
=
đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
2đ?‘Ľ đ?‘Ś+3
= (1 + đ?‘Ľ)(1 + đ?‘Ś) = đ?‘’3đ?‘Ľ+đ?‘Ś đ?‘‘đ?‘Ś
(4) (đ?‘Ľ 2 + đ?‘Ľ) đ?‘‘đ?‘Ľ = 2đ?‘Ś + 1; đ?‘Ś(1) = 0
Solution (1)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
=
2đ?‘Ľ
The equation is in the form of
đ?‘Ś+3
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
=
đ?‘‹(đ?‘Ľ) . đ?‘Œ(đ?‘Ś)
It is a separable equation
(đ?‘Ś + 3)đ?‘‘đ?‘Ś = 2đ?‘Ľđ?‘‘đ?‘Ľ
Separate the variables
âˆŤ(đ?‘Ś + 3) = âˆŤ 2đ?‘Ľđ?‘‘đ?‘Ľ
Integrate both sides of the equation
đ?‘Ś2 2
+ 3đ?‘Ś + đ?‘?1 = đ?‘Ľ2 + đ?‘?2
đ?‘Ś 2 + 6đ?‘Ś = 2đ?‘Ľ 2 + đ??ś
(2)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
This is a general equation with arbitrary constant, c
= (1 + đ?‘Ľ)(1 + đ?‘Ś)
đ?‘‘đ?‘Ś
The equation is in the form of đ?‘‘đ?‘Ľ = đ?‘‹(đ?‘Ľ)đ?‘Œ(đ?‘Ś) It is a separable equation
1 1+đ?‘Ś
đ?‘‘đ?‘Ś = (1 + đ?‘Ľ)đ?‘‘đ?‘Ľ
Separate the variables
1
âˆŤ 1+đ?‘Ś đ?‘‘đ?‘Ś = (1 + đ?‘Ľ)đ?‘‘đ?‘Ľ
Integrate both sides of the equations
đ?‘Ľ2
ln|1 + đ?‘Ś| + đ?‘?1 = đ?‘Ľ + 2 + đ?‘?2 2 ln|1 + đ?‘Ś| = 2đ?‘Ľ + đ?‘Ľ 2 + đ??ś (1 + đ?‘Ś)2 = đ?‘’ 2đ?‘Ľ+đ?‘Ľ
(3)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
2 +đ??ś
or This is the general solution
= đ?‘’3đ?‘Ľ+đ?‘Ś = đ?‘’3đ?‘Ľ . đ?‘’đ?‘Ś
đ?‘‘đ?‘Ś
The equation is in the form of đ?‘‘đ?‘Ľ = đ?‘‹(đ?‘Ľ)đ?‘Œ(đ?‘Ś) 3
It is a separable equation 1 đ?‘’đ?‘Ś
đ?‘‘đ?‘Ś = đ?‘’3đ?‘Ľ đ?‘‘đ?‘Ľ
Separate the variables
âˆŤ đ?‘’ −đ?‘Ś đ?‘‘đ?‘Ś = âˆŤ đ?‘’ 3đ?‘Ľ đ?‘‘đ?‘Ľ
Integrate both sides of the equation
1
−đ?‘’ −đ?‘Ś + đ?‘?1 = 3 đ?‘’ 3đ?‘Ľ + đ?‘?2 1 đ?‘’đ?‘Ś
(d)
1 3đ?‘Ľ
=− đ?‘’ 3
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
(đ?‘Ľ 2 + đ?‘Ľ) đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
=
+đ??ś
This is the general solution
= 2đ?‘Ś + 1; đ?‘Ś(1) = 0
This is an initial value problem
2đ?‘Ś+1
đ?‘‘đ?‘Ś
It is a separable equation 1 2đ?‘Śâˆ’1
đ?‘‘đ?‘Ś =
1 đ?‘Ľ 2 +đ?‘Ľ
1
đ?‘‘đ?‘Ľ
Separate the variables
1
âˆŤ 2đ?‘Śâˆ’1 đ?‘‘đ?‘Ś = âˆŤ đ?‘Ľ2 +đ?‘Ľ đ?‘‘đ?‘Ľ 1 2
2 1 2
Integrate both sides of the equation 1
1
đ?‘Ľ
đ?‘Ľ+1
ln|2đ?‘Ś + 1| + đ?‘?1 = âˆŤ ( −
1
2 1 2
) đ?‘‘đ?‘Ľ
ln|2đ?‘Ś + 1| + đ?‘?1 = ln|đ?‘Ľ| − ln|đ?‘Ľ + 1| + đ?‘?2 ;
ln|2đ?‘Ś + 1| = ln |
đ?‘Ľ đ?‘Ľ+1
| + đ??ś;
Substitute x = 1 and đ?‘Ś = 0, 1
đ?‘‹(đ?‘Ľ)
The equation is in the form of đ?‘‘đ?‘Ľ = đ?‘Œ(đ?‘Ś).
đ?‘Ľ 2 +đ?‘Ľ
ln|2(0) + 1| = ln |
1 1+1
Applying the initial value y(1)=0
| + đ??ś;
1
ln(1) = ln + đ??ś; 2
1
đ??ś = − ln |2|; 1
đ?‘Ľ
2
đ?‘Ľ+1
ď œ ln|2đ?‘Ś + 1| = ln |
1
| − ln ( ); 2
2đ?‘Ľ
ln|2đ?‘Ś + 1|1â „2 = ln |đ?‘Ľ+1|; 2đ?‘Ľ 2
2đ?‘Ś + 1 = (đ?‘Ľ+1)
This is the particular solution
4
Exercise 1 Solve the following differential equations. 1. (𝑦 + 1)(2𝑥 + 1 )𝑑𝑥 − 𝑑𝑦 = 0 2.
𝑑𝑦 𝑑𝑥
=
1 4𝑦−𝑥 2 𝑦
3. (𝑒 𝑥 + 1)𝑑𝑦 + 𝑒 𝑥 𝑑𝑥 = 0 4.
𝑑𝑦 𝑑𝑥
=√
1+𝑦 1−𝑥2
;
𝑦(0) = 3
Answers 1. 𝑙𝑛|𝑦 + 1| = 𝑥 2 + 𝑥 + 𝐶 1
1
𝑥+2
2. 2 y = 4 ln |𝑥−2| + 𝐶 3. 𝑦 = 𝑙𝑛|𝑒 𝑥 + 1| + 𝑐 4. 2√𝑦 + 1 = 𝑠𝑖𝑛−1 𝑥 + 4
5
Solving Homogeneous Differential Equations
A differential equation
đ?‘‘đ?‘Ś
= đ?‘“(đ?‘Ľ, đ?‘Ś) is said to be a homogeneous first order differential
đ?‘‘đ?‘Ľ
equations if the function đ?‘“(đ?‘Ľ, đ?‘Ś) can be expressed in the terms of one variable. đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
đ?‘Ś
đ?‘Ś
where đ?‘Ł = đ?‘Ľ.
= đ??š(đ?‘Ľ )=F(v)
Some examples of homogeneous first order differential equations are đ?‘‘đ?‘Ś
(1)
=
đ?‘‘đ?‘Ľ
đ?‘Ś đ?‘Ľ
=đ?‘Ł
đ?‘Ľ
(2) (3)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
=
đ?‘Śđ?‘’ đ?‘Ś đ?‘Ľ 2
1
= đ?‘Łđ?‘’ đ?‘Ł 2
+đ?‘Ś = đ?‘Ľ đ?‘Ľđ?‘Ś =
1+đ?‘Ł 2 đ?‘Ł
These types of equations can be transformed into separable equation by changing the dependent variable. The following are steps to solve first order homogeneous differential equation: 1) Write the differential equation đ?‘Ś
2) From đ?‘Ł = đ?‘Ľ or
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
đ?‘Ś đ?‘Ľ
đ?‘Ś = đ?‘Łđ?‘Ľ đ?‘‘đ?‘Ś
Differentiate y with respect to x , we will have 3) Substitute đ?‘Ś = đ?‘Łđ?‘Ľ
đ?‘‘đ?‘Ś
and
đ?‘‘đ?‘Ľ
=đ?‘Ł+đ?‘Ľ
đ?‘‘đ?‘Ł đ?‘‘đ?‘Ľ
đ?‘‘đ?‘Ł
đ?‘Ł + đ?‘Ľ đ?‘‘đ?‘Ľ = đ??š(đ?‘Ł) 4) Separate the variable x and v đ?‘‘đ?‘Ł đ?‘‘đ?‘Ľ
=
đ??ş(đ?‘Ł)−đ?‘Ł
1 đ??ş(đ?‘Ł)−đ?‘Ł
đ?‘Ľ
đ?‘‘đ?‘Ł =
1 đ?‘Ľ
đ?‘‘đ?‘Ľ
5) Integrate both sides of the equation 1
đ?‘Ś
= đ?‘“(đ?‘Ľ, đ?‘Ś) = đ??š ( ) = đ??š(đ?‘Ł) where đ?‘Ł = đ?‘Ľ
1
âˆŤ đ??ş(đ?‘Ł)−đ?‘Ł đ?‘‘đ?‘Ł = âˆŤ đ?‘Ľ đ?‘‘đ?‘Ľ đ?‘Ś
6) Replace v by đ?‘Ľ 7) Apply initial value problem(if given).
6
in n
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
đ?‘‘đ?‘Ľ
=đ?‘Ł+đ?‘Ľ
= đ??š(đ?‘Ł)
đ?‘‘đ?‘Ł đ?‘‘đ?‘Ľ
Example 2 Solve (1)
đ?‘‘đ?‘Ś
3đ?‘Śâˆ’2đ?‘Ľ
=
đ?‘‘đ?‘Ľ
đ?‘Ľ
(2) đ?‘Ľ 3 đ?‘Ś đ?‘‘đ?‘Ś − (đ?‘Ľ 4 + 3đ?‘Ľ 2 đ?‘Ś 2 + đ?‘Ś 4 )đ?‘‘đ?‘Ľ = 0 (3) đ?‘Ľ 2 đ?‘‘đ?‘Ś + (đ?‘Ś 2 + đ?‘Ľđ?‘Ś)đ?‘‘đ?‘Ľ = 0;
đ?‘Ś(0) = 2
Solution (1)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
= =
3đ?‘Śâˆ’2đ?‘Ľ đ?‘Ľ 3đ?‘Łđ?‘Ľâˆ’2đ?‘Ľ đ?‘Ľ
=
đ?‘Ľ(3đ?‘Łâˆ’2) đ?‘Ľ
đ?‘‘đ?‘Ś
= 3đ?‘Ł − 2
đ?‘Ł+đ?‘Ľ
đ?‘‘đ?‘Ł đ?‘‘đ?‘Ľ
đ?‘‘đ?‘Ľ
= 3đ?‘Ł − 2
đ?‘‘đ?‘Ł
đ?‘Ľ đ?‘‘đ?‘Ľ = 2đ?‘Ł − 2 1
đ?‘‘đ?‘Ł = âˆŤ đ?‘Ľ đ?‘‘đ?‘Ľ âˆŤ 2 (đ?‘Łâˆ’1) 1 2
1 2
= đ??š(đ?‘Ł). The equation is homogeneous
Substitution of đ?‘Ś = đ?‘Łđ?‘Ľ and
đ?‘Ś
đ?‘™đ?‘› | − 1| = ln|đ?‘Ľ| + đ?‘? đ?‘Ľ
đ?‘Śâˆ’đ?‘Ľ đ?‘Ľ
− đ?‘™đ?‘›đ?‘Ľ = đ?‘?
Substitute by đ?‘Ł =
Simplify the answer
đ?‘Śâˆ’đ?‘Ľ
đ?‘™đ?‘›âˆš đ?‘Ľ 3 = đ?‘? đ?‘Śâˆ’đ?‘Ľ
√
đ?‘Ľ3
đ?‘‘đ?‘Ľ
=đ?‘Ł+đ?‘Ľ
Integrate both sides of the equation
đ?‘™đ?‘›|đ?‘Ł − 1| = ln|đ?‘Ľ| + đ?‘?
ln √
đ?‘‘đ?‘Ś
Separate the variable x and v 1
1
Substitute y=vx on RHS of the equation and simplify
= đ?‘’đ?‘?
The general solution is
đ?‘Ś = đ??´đ?‘Ľ 3 + đ?‘Ľ
7
đ?‘Ś đ?‘Ľ
đ?‘‘đ?‘Ł đ?‘‘đ?‘Ľ
,
(2)
𝑥 3 𝑦 𝑑𝑦 − (𝑥 4 + 3𝑥 2 𝑦 2 + 𝑦 4 )𝑑𝑥 = 0 𝑑𝑦 𝑑𝑥 𝑑𝑦
𝑑𝑦
𝑑𝑥
𝑥4 +3𝑥2 (𝑣𝑥)2 +(𝑣𝑥)4
=
= 𝑓(𝑥, 𝑦)
𝑣𝑥 4 [1+3𝑣 2 +𝑣4 ]
𝑑𝑦
𝑣
𝑑𝑥
𝑑𝑦
1
𝑑𝑣
= 𝐹(𝑣). The equation is homogeneous
Substitution
1
𝑥 𝑑𝑥 = 𝑣 + 2𝑣 + 𝑣 3 𝑑𝑣
𝑑𝑥
Substitute y= vx on RHS and simplify
𝑥3 (𝑣𝑥)
𝑣 + 𝑥 𝑑𝑥 = 𝑣 + 3𝑣 + 𝑣 3
𝑥 𝑑𝑥 =
𝑑𝑦
𝑥 4 [1+3𝑣 2 +𝑣 4 ]
=
𝑑𝑥
Write the equation in form of
𝑥 3𝑦
=
𝑑𝑥
𝑑𝑦
𝑥 4 +3𝑥 2 𝑦 2 +𝑦 4
=
𝑑𝑦 𝑑𝑥
=𝑣+𝑥
𝑑𝑣 𝑑𝑥
,
Separate the variable x and v
1+2𝑣2 +𝑣4 𝑣 𝑣
1
∫ 1+2𝑣2 +𝑣4 𝑑𝑣 = ∫ 𝑥 𝑑𝑥 𝑣
Integrate both sides of the equation
1
∫ (𝑣2 +1)2 𝑑𝑣 = ∫ 𝑥 𝑑𝑥 1
1
𝑑𝑢 = ln|𝑥| + 𝑐1 ; ∫ 2 (𝑢)2
let 𝑢 = 𝑣 2 + 1;
𝑑𝑢 𝑑𝑣
= 2𝑣
1
− 2𝑢 +𝑐2 = ln|𝑥| + 𝑐1 1
− 2[(𝑢)2+1] = ln|𝑥| + 𝐶 1
−
= ln|𝑥| + 𝐶
𝑦 2
2[(𝑥) +1] 𝑥2 2[𝑦2 +𝑥2 ]
−
(3)
= ln|𝑥| + 𝐶
𝑥 2 𝑑𝑦 + (𝑦 2 + 𝑥𝑦)𝑑𝑥 = 0; 𝑦(0) = 2 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥
=−
=−
𝑦2 +𝑥𝑦
Write the equation in form of
𝑥2
(𝑣𝑥)2 +𝑥(𝑣𝑥) 𝑥2
=−
= −𝑣2 − 𝑣 𝑑𝑣
𝑣 + 𝑥 𝑑𝑥 = −𝑣 2 − 𝑣
𝑥2 (𝑣2 +𝑣) 𝑥2
𝑑𝑦 𝑑𝑥
= 𝑓(𝑥, 𝑦)
Substitute y = vx on RHS and simplify 𝑑𝑦 𝑑𝑥
= 𝐹(𝑣) 𝑤ℎ𝑒𝑟𝑒 𝑣 =
Substitution
8
𝑑𝑦 𝑑𝑥
=𝑣+𝑥
𝑑𝑣 𝑑𝑥
,
𝑦 𝑥
đ?‘‘đ?‘Ł
đ?‘Ľ đ?‘‘đ?‘Ľ = −đ?‘Ł 2 − 2đ?‘Ł = −đ?‘Ł(đ?‘Ł + 2) 1
Separate the variables
1
âˆŤ đ?‘Ł(đ?‘Ł+2) đ?‘‘đ?‘Ł = − âˆŤ đ?‘Ľ đ?‘‘đ?‘Ľ
Integrate both sides
By partial fraction decomposition 1
1
1 1
1
1
= 2 (đ?‘Ł) − 2 (đ?‘Ł+2) đ?‘Ł(đ?‘Ł+2)
1
1
âˆŤ(đ?‘Ł + đ?‘Ł+2) đ?‘‘đ?‘Ł = − âˆŤ đ?‘Ľ đ?‘‘đ?‘Ľ
2 1
1
(ln đ?‘Ł + đ?‘™đ?‘›|đ?‘Ł + 2|) + đ?‘?2 = − ln|đ?‘Ľ| + đ?‘?1
2
1 ln(đ?‘Ł(đ?‘Ł + 2)) + ln|đ?‘Ľ| 2 đ?‘Ś
=đ?‘?
đ?‘Ś
đ?‘™đ?‘› ((đ?‘Ľ ) (đ?‘Ľ + 2)) + 2đ?‘™đ?‘›đ?‘Ľ = 2đ?‘? đ?‘Ś
Substitute đ?‘Ł =
đ?‘Ś
đ?‘™đ?‘› ((đ?‘Ľ ) (đ?‘Ľ + 2) đ?‘Ľ 2 ) = đ??ś
đ?‘Ś đ?‘Ľ
Simplify the answer
đ?‘™đ?‘›[(đ?‘Ś)(đ?‘Ś + 2đ?‘Ľ)] = đ??ś; đ?‘Ś(đ?‘Ś + 2đ?‘Ľ) = đ?‘’ đ?‘? đ?‘Ś(đ?‘Ś + 2đ?‘Ľ) = đ??´ Given y(0) =2; 2(2 + 2(0)) = đ??´
Apply the initial value problem
đ??´=4 Therefore, the particular solution is đ?‘Ś(đ?‘Ś + 2đ?‘Ľ) = 4
Exercise 2 Solve the following differential equations. 1.
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
=
3đ?‘Ľ 2 +đ?‘Ś 2
2. (đ?‘Ľ 2 + 3đ?‘Ľđ?‘Ś + đ?‘Ś 2 )đ?‘‘đ?‘Ľ − đ?‘Ľ 2 đ?‘‘đ?‘Ś = 0
đ?‘Ľđ?‘Ś
đ?‘‘đ?‘Ś
3. đ?‘Ľđ?‘Ś đ?‘‘đ?‘Ľ + 4đ?‘Ľ 2 + đ?‘Ś 2 = 0
4.
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
=
đ?‘Ś 2 +đ?‘Ľâˆšđ?‘Ľ 2 +đ?‘Ś 2 đ?‘Ľđ?‘Ś
; y(1)=2
Answers 1.
3.
đ?‘Ś 2đ?‘Ľ
1 4
− 3đ?‘™đ?‘›đ?‘Ľ = đ??ś 4đ?‘Ľ 2 +2đ?‘Ś 2
�� (
đ?‘Ľ2
2.
) = −đ?‘™đ?‘›đ?‘Ľ + đ?‘?
4.
9
−
đ?‘Ľ đ?‘Ś+đ?‘Ľ
= ln|đ?‘Ľ| + đ??ś 2
√1 + (đ?‘Ś) = đ?‘™đ?‘›đ?‘Ľ + √5 đ?‘Ľ
Solving Linear Differential Equations A first order differential equation is said to be linear if it can written in the form đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ đ?‘ƒ(đ?‘Ľ)đ?‘Ś = đ?‘„(đ?‘Ľ)
where đ?‘ƒ(đ?‘Ľ) and đ?‘„(đ?‘Ľ) are function of x and constant.
Here are some examples of linear first order differential equations: 1) 2)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ 3đ?‘Ľđ?‘Ś = đ?‘Ľ
where đ?‘ƒ(đ?‘Ľ) = 2đ?‘Ľ đ?‘Žđ?‘›đ?‘‘ đ?‘„(đ?‘Ľ) = đ?‘Ľ
− (đ?‘Ąđ?‘Žđ?‘›đ?‘Ľ)đ?‘Ś = đ?‘ đ?‘–đ?‘›đ?‘Ľ where đ?‘ƒ(đ?‘Ľ) = −đ?‘Ąđ?‘Žđ?‘›đ?‘Ľ đ?‘Žđ?‘›đ?‘‘ đ?‘„(đ?‘Ľ) = sin đ?‘Ľ
A linear differential equation can be solved by multiplying by a function known as an integrating factor, đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ . In order to use this technique, firstly we must write the differential equation in the standard form,
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ đ?‘ƒ(đ?‘Ľ)đ?‘Ś = đ?‘„(đ?‘Ľ).
Here are the steps to find the solution for linear first order differential equation: 1. Rewrite the equation in the standard form đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ đ?‘ƒ(đ?‘Ľ)đ?‘Ś = đ?‘„(đ?‘Ľ)
2. Find the integrating factor, đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ 3. Multiply both sides by the integrating factor
đ?‘‘đ?‘Ś + đ?‘ƒ(đ?‘Ľ)đ?‘Ś] = đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ đ?‘„(đ?‘Ľ) đ?‘‘đ?‘Ľ
đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ [
3. Simplify both sides of the equations đ?‘‘ đ?‘‘đ?‘Ľ
(đ?‘’âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ đ?‘Ś) = đ?‘’âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ đ?‘„(đ?‘Ľ)
5. Integrate both sides with respect to đ?‘Ľ đ?‘‘
âˆŤ đ?‘‘đ?‘Ľ (đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ đ?‘Ś) = âˆŤ đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ đ?‘„(đ?‘Ľ) đ?‘‘đ?‘Ľ đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ đ?‘Ś = âˆŤ đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ đ?‘„(đ?‘Ľ) đ?‘‘đ?‘Ľ + đ??ś 6.
Simplify the answer.
10
Example 3 Solve (1)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ 2đ?‘Ľđ?‘Ś = đ?‘Ľ đ?‘‘đ?‘Ś
đ?‘Ľ
(2) (1 + đ?‘Ľ 2 ) đ?‘‘đ?‘Ľ + 2đ?‘Ľđ?‘Ś = 1+đ?‘Ľ2 đ?‘‘đ?‘Ś
đ?œ‹
(3) sin đ?‘Ľ đ?‘‘đ?‘Ľ − đ?‘Śđ?‘?đ?‘œđ?‘ đ?‘Ľ =;
đ?‘Ś (2 ) = 3
Solution (1)
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ 2đ?‘Ľđ?‘Ś = đ?‘Ľ
The given equation is already in the standard form of 2� ��� �(�) = �
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ đ?‘ƒ(đ?‘Ľ)đ?‘Ś = đ?‘„(đ?‘Ľ) , where đ?‘ƒ(đ?‘Ľ) =
The integrating factor is
đ?‘’ âˆŤ đ?‘ƒ(đ?‘Ľ)đ?‘‘đ?‘Ľ = đ?‘’ âˆŤ 2đ?‘Ľ đ?‘‘đ?‘Ľ = đ?‘’đ?‘Ľ
2 +đ?‘? 2
= đ?‘’ đ?‘Ľ đ?‘’đ?‘? = đ??´đ?‘’ đ?‘Ľ
đ?‘™đ?‘’đ?‘Ą đ??´ = đ?‘’ đ?‘?
2
Multiply the integrating factor to the both sides of the standard linear equation đ?‘‘đ?‘Ś
2
2
đ??´đ?‘’ đ?‘Ľ (đ?‘‘đ?‘Ľ + 2đ?‘Ľđ?‘Ś) = đ??´đ?‘’ đ?‘Ľ (đ?‘Ľ) 2
đ?‘‘đ?‘Ś + 2đ?‘Ľđ?‘Ś) đ?‘‘đ?‘Ľ
đ?‘’đ?‘Ľ ( đ?‘‘
= đ?‘Ľđ?‘’ đ?‘Ľ
2
(đ?‘Śđ?‘’đ?‘Ľ ) = đ?‘Ľđ?‘’đ?‘Ľ đ?‘‘đ?‘Ľ
2
2
Integrate both sides đ?‘‘
2
2
âˆŤ đ?‘‘đ?‘Ľ (đ?‘Śđ?‘’ đ?‘Ľ )đ?‘‘đ?‘Ľ = âˆŤ đ?‘Ľđ?‘’ đ?‘Ľ đ?‘‘đ?‘Ľ 2
1
2
1
��
let � = � 2 ; �� = 2�
đ?‘Śđ?‘’ đ?‘Ľ = 2 âˆŤ đ?‘’ đ?‘˘ đ?‘‘đ?‘˘ đ?‘Śđ?‘’ đ?‘Ľ = 2 đ?‘’ đ?‘˘ + đ?‘? The general solution is 1
đ?‘Ś = 2 + đ?‘?đ?‘’ −đ?‘Ľ
2
11
(1 + đ?‘Ľ 2 )
(2)
đ?‘‘đ?‘Ś + 2đ?‘Ľđ?‘Ś đ?‘‘đ?‘Ľ
đ?‘Ľ
= 1+đ?‘Ľ2
Write the given differential in the form of n the standard form of đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+(
2đ?‘Ľ
1+đ?‘Ľ 2
đ?‘Ľ
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ đ?‘ƒ(đ?‘Ľ)đ?‘Ś = đ?‘„(đ?‘Ľ)
2đ?‘Ľ
) đ?‘Ś = (1+đ?‘Ľ 2 )2 where đ?‘ƒ(đ?‘Ľ) = 1+đ?‘Ľ2 and đ?‘„(đ?‘Ľ) =
Find the integrating factor 2đ?‘Ľ
đ?‘’ âˆŤ1+đ?‘Ľ2
đ?‘‘đ?‘Ľ
= � ��|1+�
2 |+đ?‘?
2
= đ?‘’ đ?‘™đ?‘›|1+đ?‘Ľ | đ?‘’ đ?‘? = đ??´(1 + đ?‘Ľ 2 );
đ??´ = đ?‘’đ?‘?;
Multiply the integrating factor to đ?‘‘đ?‘Ś
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+(
2đ?‘Ľ
đ?‘Ľ
1+đ?‘Ľ 2
) đ?‘Ś = (1+đ?‘Ľ 2)2
2đ?‘Ľ
đ??´(1 + đ?‘Ľ 2 ) [đ?‘‘đ?‘Ľ + (1+đ?‘Ľ2) đ?‘Ś] = đ??´(1 + đ?‘Ľ 2 ) [ (1 + đ?‘Ľ 2 ) [
đ?‘‘đ?‘Ś 2đ?‘Ľ + (1+đ?‘Ľ2) đ?‘Ś] đ?‘‘đ?‘Ľ
= (1 + đ?‘Ľ 2 ) [
đ?‘Ľ (1+đ?‘Ľ2 )
đ?‘Ľ (1+đ?‘Ľ2 )
2
2
]
]
Simplify the equation đ?‘‘ đ?‘‘đ?‘Ľ
[đ?‘Ś(1 + đ?‘Ľ2 )] =
đ?‘Ľ 1+đ?‘Ľ2
Integrate both sides đ?‘‘
đ?‘Ľ
âˆŤ đ?‘‘đ?‘Ľ [đ?‘Ś(1 + đ?‘Ľ 2 )] = âˆŤ 1+đ?‘Ľ 2 đ?‘‘đ?‘Ľ; 1
let � = 1 + � 2 ;
1
đ?‘Ś(1 + đ?‘Ľ 2 ) = 2 âˆŤ đ?‘˘du 1
đ?‘Ś(1 + đ?‘Ľ 2 ) = 2 ln đ?‘˘ + đ?‘? The general solution is 1
đ?‘Ś(1 + đ?‘Ľ 2 ) = 2 ln|1 + đ?‘Ľ 2 | + đ?‘?
12
�� ��
= 2đ?‘Ľ
đ?‘Ľ (1+đ?‘Ľ2 )
2
(3)
đ?‘‘đ?‘Ś
đ?œ‹
sin đ?‘Ľ đ?‘‘đ?‘Ľ − đ?‘Śđ?‘?đ?‘œđ?‘ đ?‘Ľ = sin 2đ?‘Ľ; đ?‘Ś (2 ) = 3 Write the given differential in the form of n the standard form of đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
đ?‘?đ?‘œđ?‘ đ?‘Ľ
−(
sin 2đ?‘Ľ
)đ?‘Ś = sin đ?‘Ľ
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ đ?‘ƒ(đ?‘Ľ)đ?‘Ś = đ?‘„(đ?‘Ľ) sin 2đ?‘Ľ
cos đ?‘Ľ
where đ?‘ƒ(đ?‘Ľ) = − sin đ?‘Ľ and đ?‘„(đ?‘Ľ) = sin đ?‘Ľ
sin đ?‘Ľ
Find the integrating factor is cos đ?‘Ľ
1
đ?‘’ âˆŤ − sin đ?‘Ľ đ?‘‘đ?‘Ľ = đ?‘’ − âˆŤđ?‘˘ đ?‘‘đ?‘˘
Let � = sin �;
�� ��
= cos đ?‘Ľ
−1
= đ?‘’ −đ?‘™đ?‘›|đ?‘˘|+đ?‘? = đ?‘’ ln(sin x) . đ?‘’ đ?‘? 1
= đ??´ (sin đ?‘Ľ ) Multiply the integrating factor to đ?‘‘đ?‘Ś
1
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
cos đ?‘Ľ
cos đ?‘Ľ
−( 1
)đ?‘Ś = sin đ?‘Ľ
sin 2đ?‘Ľ sin đ?‘Ľ
sin 2đ?‘Ľ
đ??´ (sin đ?‘Ľ) [đ?‘‘đ?‘Ľ − ( sin đ?‘Ľ ) đ?‘Ś] = đ??´ (sin đ?‘Ľ) [ sin đ?‘Ľ ] đ?‘‘đ?‘Ś
1
cos đ?‘Ľ
1
sin 2đ?‘Ľ
(sin đ?‘Ľ) [đ?‘‘đ?‘Ľ − ( sin đ?‘Ľ ) đ?‘Ś] = (sin đ?‘Ľ) [ sin đ?‘Ľ ] Simplify the equation đ?‘‘ 1 [đ?‘Ś (sin đ?‘Ľ)] đ?‘‘đ?‘Ľ đ?‘‘ đ?‘‘đ?‘Ľ
[đ?‘Ś (
=
2 sin đ?‘Ľ cos đ?‘Ľ đ?‘ đ?‘–đ?‘›2 đ?‘Ľ
1
)] = sin đ?‘Ľ
2 cos đ?‘Ľ sin đ?‘Ľ
Integrate both sides đ?‘‘
1
âˆŤ đ?‘‘đ?‘Ľ [đ?‘Ś (sin đ?‘Ľ )] đ?‘‘đ?‘Ľ = âˆŤ
2 cos đ?‘Ľ sin đ?‘Ľ
đ?‘‘đ?‘Ľ;
let � = sin � ;
�� ��
= cos đ?‘Ľ
1
1
đ?‘Ś (sin đ?‘Ľ) = 2 âˆŤ đ?‘˘du 1
đ?‘Ś (sin đ?‘Ľ) = 2 ln đ?‘˘ + đ?‘? đ?‘Ś = sin đ?‘Ľ (2 ln|sin đ?‘Ľ| + đ?‘?) đ?œ‹
Find the value of c by applying the initial value problem, đ?‘Ś ( 2 ) = 3 đ?œ‹
đ?œ‹
3 = (sin 2 ) (2 ln |sin 2 | + đ?‘?) đ?‘?=3 The particular solution is y = sin đ?‘Ľ (2 ln|sin đ?‘Ľ| + 3)
13
Exercise 3 Solve the following differential equations. đ?‘‘đ?‘Ś
1. đ?‘Ľ 4 đ?‘‘đ?‘Ľ + 4đ?‘Ľ 3 đ?‘Ś = sin 2đ?‘Ľ 3.
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
đ?‘‘đ?‘Ś
2. đ?‘Ľ đ?‘‘đ?‘Ľ = đ?‘Ś + 2đ?‘Ľ 3 + đ?‘Ľ 2 − đ?‘Ľ
– � tan � = 4 cos �
4.
đ?‘‘đ?‘„ đ?‘‘đ?‘Ą
đ?‘„
+ 25+đ?‘Ą = 16; đ?‘„(0) = 0
Answers 1
1. đ?‘Śđ?‘Ľ 4 = − 2 cos 2đ?‘Ľ + đ?‘?
2.
3. đ?‘Ś = sec đ?‘Ľ(sin 2đ?‘Ľ + 2đ?‘Ľ + đ?‘?)
4.
đ?‘Ś = đ?‘Ľ 3 + đ?‘Ľ 2 − đ?‘Ľ ln(đ?‘Ľ) + đ??śđ?‘Ľ đ?‘„ = 8(25 + đ?‘Ą) − 5000(25 + đ?‘Ą)−1
14
Case 5: 𝒇(𝒙) is the multiplication of function in Case 1, 2 and 3 The table below shows the general form of particular integral for multiplication cases in 𝑓(𝑥). Expressions of 𝒇(𝒙)
Particular integral 𝒚𝒑 𝑦𝑝 = 𝑥 𝑟 (𝐴𝑛 𝑥 𝑛 + 𝐴𝑛−1 𝑥 𝑛−1 +. . . +𝐴0 )𝑒 𝑏𝑥
𝒇(𝒙) = (𝒂𝒏 𝒙𝒏 + 𝒂𝒏−𝟏 𝒙𝒏−𝟏 +. . . +𝒂𝟎 )𝒆𝒃𝒙
𝒇(𝒙) = (𝒂𝒏 𝒙𝒏 + 𝒂𝒏−𝟏 𝒙𝒏−𝟏 +. . . +𝒂𝟎 ) 𝐜𝐨𝐬(𝒙) 𝑦𝑝 = 𝑥 𝑟 (𝐴𝑛 𝑥 𝑛 + 𝐴𝑛−1 𝑥 𝑛−1 +. . . +𝐴0 ) cos(𝑏𝑥) +𝑥 𝑟 (𝐵𝑛 𝑥 𝑛 + 𝐵𝑛−1 𝑥 𝑛−1 +. . . +𝐵0 ) sin(𝑏𝑥)
or 𝒇(𝒙) = (𝒂𝒏 𝒙𝒏 + 𝒂𝒏−𝟏 𝒙𝒏−𝟏 +. . . +𝒂𝟎 ) 𝐬𝐢𝐧(𝒙) 𝒇(𝒙) = 𝑨𝒆𝒂𝒙 𝐜𝐨𝐬(𝒃𝒙)
𝑦𝑝 = 𝑥 𝑟 [𝐵 cos(𝑏𝑥) + 𝐶 sin(𝑏𝑥)]𝑒 𝑎𝑥
or 𝒇(𝒙) = 𝑨𝒆𝒂𝒙 𝐬𝐢𝐧(𝒃𝒙)
𝒇(𝒙) = (𝒂𝒏 𝒙𝒏 + 𝒂𝒏−𝟏 𝒙𝒏−𝟏 +. . . +𝒂𝟎 ) 𝐜𝐨𝐬(𝒙) 𝑦𝑝 = 𝑥 𝑟 (𝐴𝑛 𝑥 𝑛 + 𝐴𝑛−1 𝑥 𝑛−1 +. . . +𝐴0 ) 𝑒 𝑏𝑥 cos(𝑏𝑥) or
+𝑥 𝑟 (𝐵𝑛 𝑥 𝑛 + 𝐵𝑛−1 𝑥 𝑛−1 +. . . +𝐵0 )𝑒 𝑏𝑥 sin(𝑏𝑥)
𝒇(𝒙) = (𝒂𝒏 𝒙𝒏 + 𝒂𝒏−𝟏 𝒙𝒏−𝟏 +. . . +𝒂𝟎 ) 𝐬𝐢𝐧(𝒙) Table 6.2: General form for the particular integral
Example 6.8.5 Find the solution for (a) 𝑦" + 4𝑦′ + 3𝑦 = 16𝑥𝑒 𝑥 (b) 𝑦" + 4𝑦′ + 3𝑦 = 𝑥𝑒 −𝑥 (c) (d)
𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2
−2
𝑑𝑦 𝑑𝑥
+ 3𝑦 = 𝑒𝑥 sin(2𝑥)
+ 𝑦 = 𝑥 cos(𝑥)
Solution (a)
Let 𝑦" + 4𝑦′ + 3𝑦 = 16𝑥𝑒 𝑥
Finding the complementary solution for (6.113), 𝑚2 + 4𝑚 + 3 = 0; 15
(đ?‘š + 3)(đ?‘š + 1) = 0; đ?‘š = −3, −1. Therefore, đ?‘Śđ?‘? (đ?‘Ľ) = đ??´đ?‘’ −3đ?‘Ľ + đ??ľđ?‘’ −đ?‘Ľ . Since đ?‘“(đ?‘Ľ) has no similar term as as term in đ?‘Śđ?‘? (đ?‘Ľ), the particular integral is đ?‘Śđ?‘? (đ?‘Ľ) = (đ??śđ?‘Ľ + đ??ˇ)đ?‘’ đ?‘Ľ ; đ?‘Śâ€˛đ?‘? (đ?‘Ľ) = đ??śđ?‘Ľđ?‘’ đ?‘Ľ + đ??śđ?‘’ đ?‘Ľ + đ??ˇđ?‘’ đ?‘Ľ ; đ?‘Ś"đ?‘? (đ?‘Ľ) = đ??śđ?‘Ľđ?‘’ đ?‘Ľ + 2đ??śđ?‘’ đ?‘Ľ + đ??ˇđ?‘’ đ?‘Ľ . Substitute (6.114), (6.115) and (6.116) into (6.113) yields đ??śđ?‘Ľđ?‘’ đ?‘Ľ + 2đ??śđ?‘’ đ?‘Ľ + đ??ˇđ?‘’ đ?‘Ľ + 4đ??śđ?‘Ľđ?‘’ đ?‘Ľ + 4đ??śđ?‘’ đ?‘Ľ + 4đ??ˇđ?‘’ đ?‘Ľ + 3đ??śđ?‘Ľđ?‘’ đ?‘Ľ + 3đ??ˇđ?‘’ đ?‘Ľ = 16đ?‘Ľđ?‘’ đ?‘Ľ ; 8đ??śđ?‘Ľđ?‘’ đ?‘Ľ + 6đ??śđ?‘’ đ?‘Ľ + 8đ??ˇ đ?‘Ľ = 16đ?‘Ľđ?‘’ đ?‘Ľ ; (8đ??śđ?‘Ľ + 6đ??ś + 8đ??ˇ)đ?‘’ đ?‘Ľ = 16đ?‘Ľđ?‘’ đ?‘Ľ Comparing the coefficient of the similar terms on both sides in (6.117), 8đ??ś = 16; 6đ??ś + 8đ??ˇ = 0; Solving (6.118) and (6.119) yields đ??ś = 2 and đ??ˇ = −3â „2. Therefore, the particular integral is 3
đ?‘Śđ?‘? (đ?‘Ľ) = (2đ?‘Ľ − 2) đ?‘’ đ?‘Ľ (b)
Let đ?‘Ś" + 4đ?‘Śâ€˛ + 3đ?‘Ś = đ?‘Ľđ?‘’ −đ?‘Ľ .
The complementary solution is the same as in the previous example which is đ?‘Śđ?‘? = đ??´đ?‘’ −3đ?‘Ľ + đ??ľđ?‘’ −đ?‘Ľ . Since there is a similar term in đ?‘Śđ?‘? and đ?‘“(đ?‘Ľ), then the particular integral must be adjusted by neglecting đ?‘&#x; = 0. If đ?‘&#x; = 1 then đ?‘Śđ?‘? = đ?‘Ľ(đ??śđ?‘Ľ + đ??ˇ)đ?‘’ −đ?‘Ľ . There is no similar term exists in đ?‘Śđ?‘? and đ?‘Śđ?‘? . Therefore, the particular integral is đ?‘Śđ?‘? (đ?‘Ľ) = đ??śđ?‘Ľ 2 đ?‘’ −đ?‘Ľ + đ??ˇđ?‘Ľđ?‘’ −đ?‘Ľ ; đ?‘Śâ€˛đ?‘? (đ?‘Ľ) = −đ??śđ?‘Ľ 2 đ?‘’ −đ?‘Ľ + 2đ??śđ?‘Ľđ?‘’ −đ?‘Ľ − đ??ˇđ?‘Ľđ?‘’ −đ?‘Ľ + đ??ˇđ?‘’ −đ?‘Ľ ; đ?‘Ś"đ?‘? (đ?‘Ľ) = 2đ??śđ?‘’ −đ?‘Ľ − 4đ??śđ?‘Ľđ?‘’ −đ?‘Ľ + đ??śđ?‘Ľ 2 đ?‘’ −đ?‘Ľ − 2đ??ˇđ?‘’ −đ?‘Ľ + đ??ˇđ?‘Ľđ?‘’ −đ?‘Ľ . Substitute (6.121), (6.122) and (6.123) into (6.120) yields 2đ??śđ?‘’ −đ?‘Ľ − 4đ??śđ?‘Ľđ?‘’ −đ?‘Ľ + đ??śđ?‘Ľ 2 đ?‘’ −đ?‘Ľ − 2đ??ˇđ?‘’ −đ?‘Ľ + đ??ˇđ?‘Ľđ?‘’ −đ?‘Ľ + 4(−đ??śđ?‘Ľ 2 đ?‘’ −đ?‘Ľ + 2đ??śđ?‘Ľđ?‘’ −đ?‘Ľ − đ??ˇđ?‘Ľđ?‘’ −đ?‘Ľ + đ??ˇđ?‘’ −đ?‘Ľ ) + 3(đ??śđ?‘Ľ 2 đ?‘’ −đ?‘Ľ + đ??ˇđ?‘Ľđ?‘’ −đ?‘Ľ ) = đ?‘Ľđ?‘’ −đ?‘Ľ Comparing the coefficient of the similar terms on both sides in (6.124), 4đ??ś = 1; 16
2đ??ś + 2đ??ˇ = 0;
Solving (6.125) and (6.126) yields đ??ś = 1â „4 and đ??ˇ = −1â „4. Therefore, the particular integral is 1
1
đ?‘Śđ?‘? (đ?‘Ľ) = 4 đ?‘Ľ 2 đ?‘’ −đ?‘Ľ − 4 đ?‘Ľđ?‘’ −đ?‘Ľ . The general solution for (6.120) is đ?‘Ś = đ?‘Śđ?‘? (đ?‘Ľ) + đ?‘Śđ?‘? (đ?‘Ľ); 1
1
= đ??´đ?‘’ −3đ?‘Ľ + đ??ľđ?‘’ −đ?‘Ľ + 4 đ?‘Ľ 2 đ?‘’ −đ?‘Ľ − 4 đ?‘Ľđ?‘’ −đ?‘Ľ . (c)
Let đ?‘‘2đ?‘Ś đ?‘‘đ?‘Ľ 2
−2
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
+ 3đ?‘Ś = đ?‘’đ?‘Ľ sin(2đ?‘Ľ).
Finding the complementary solution for (6.113), đ?‘š2 − 2đ?‘š + 3 = 0; 2
đ?‘š=
−(−2)Âąâˆš(−2) −4(1)(3) 2(1)
;
đ?‘š = 1 Âą √2đ?‘–. Therefore, đ?‘Śđ?‘? (đ?‘Ľ) = đ?‘’ đ?‘Ľ [đ??´ cos(√đ?‘Ľ) + đ??ľ sin(√2 đ?‘Ľ)]. Since đ?‘“(đ?‘Ľ) has no similar term as a term in đ?‘Śđ?‘? (đ?‘Ľ), the particular integral is đ?‘Śđ?‘? (đ?‘Ľ) = đ??śđ?‘’ đ?‘Ľ sin(2đ?‘Ľ) + đ??ˇđ?‘’ đ?‘Ľ cos(2đ?‘Ľ) ; đ?‘Śâ€˛đ?‘? (đ?‘Ľ) = đ??śđ?‘’ đ?‘Ľ sin(2đ?‘Ľ) + 2đ??śđ?‘’ đ?‘Ľ cos(2đ?‘Ľ) + đ??ˇđ?‘’ đ?‘Ľ cos(2đ?‘Ľ) − 2đ??ˇđ?‘’ đ?‘Ľ sin(2đ?‘Ľ) ; đ?‘Ś"đ?‘? (đ?‘Ľ) = −3đ??śđ?‘’ đ?‘Ľ sin(2đ?‘Ľ) + 4đ??śđ?‘’ đ?‘Ľ cos(2đ?‘Ľ) − 3đ??ˇđ?‘’ đ?‘Ľ cos(2đ?‘Ľ) − 2đ??ˇđ?‘’ đ?‘Ľ sin(2đ?‘Ľ). Substitute (6.128), (6.129) and (6.130) into (6.127) yields [−2đ??ś sin(2đ?‘Ľ) − 2đ??ˇ cos(2đ?‘Ľ) + 2đ??ˇ sin(2đ?‘Ľ)]đ?‘’ đ?‘Ľ = đ?‘’ đ?‘Ľ sin(2đ?‘Ľ) Equating the coefficient of similar terms in (6.131) on both sides yields −2đ??ś + 2đ??ˇ = 1; −2đ??ˇ = 0; Solving (6.132) and (6.133) yields đ??ś = −1â „2 and đ??ˇ = 0. Therefore, the particular integral is 1
đ?‘Śđ?‘? (đ?‘Ľ) = (− 2 sin(2đ?‘Ľ)) đ?‘’ đ?‘Ľ The general solution for (6.127) is đ?‘Ś = đ?‘Śđ?‘? (đ?‘Ľ) + đ?‘Śđ?‘? (đ?‘Ľ); = đ?‘’ đ?‘Ľ [đ??´ cos(√đ?‘Ľ) + đ??ľ sin(√2 đ?‘Ľ)] 17
(d)
Let đ?‘‘2đ?‘Ś đ?‘‘đ?‘Ľ 2
+ đ?‘Ś = đ?‘Ľ cos(đ?‘Ľ)
(6.138)
Finding the complementary solution for (6.134), đ?‘š2 + 1 = 0; đ?‘š = Âąđ?‘–. Therefore, đ?‘Śđ?‘? (đ?‘Ľ) = đ??´ cos(đ?‘Ľ) + đ??ľ sin(đ?‘Ľ). For đ?‘&#x; = 0, there exists a similar term in đ?‘Śđ?‘? (đ?‘Ľ). Therefore, đ?‘&#x; = 1 which leads to đ?‘Śđ?‘? (đ?‘Ľ) = (đ??śđ?‘Ľ 2 + đ??ˇđ?‘Ľ) cos(đ?‘Ľ) + (đ??¸đ?‘Ľ 2 + đ??šđ?‘Ľ) sin(đ?‘Ľ) ; (6.135) đ?‘Śâ€˛đ?‘? (đ?‘Ľ) = (đ??š + đ??¸đ?‘Ľ 2 − đ??ˇđ?‘Ľ − đ??śđ?‘Ľ 2 ) sin(đ?‘Ľ) + (2đ??śđ?‘Ľ + đ??ˇ + đ??¸đ?‘Ľ 2 + đ??šđ?‘Ľ) cos(đ?‘Ľ) ;(6.136) đ?‘Ś"đ?‘? (đ?‘Ľ) = (2đ??¸ − đ??¸đ?‘Ľ 2 − đ??šđ?‘Ľ − 2đ??ˇ − 4đ??śđ?‘Ľ) sin(đ?‘Ľ) + (2đ??ś − đ??śđ?‘Ľ 2 − đ??ˇđ?‘Ľ + 4đ??¸đ?‘Ľ + 2đ??š) cos(đ?‘Ľ). (6.137) Substitute (6.135), (6.136) and (6.137) into yields (2đ??ś + 4đ??¸đ?‘Ľ + 2đ??š) cos(đ?‘Ľ) + (2đ??¸ − 2đ??ˇ − 4đ??śđ?‘Ľ) sin(đ?‘Ľ) = đ?‘Ľ cos(đ?‘Ľ)(6.138) Equating the coefficient of similar terms in (6.138) on both sides yields 2đ??ś + 2đ??š = 0; 4đ??¸ = 1;
(6.139) (6.140)
−4đ??ś = 0; 2đ??¸ − 2đ??ˇ = 0;
(6.141) (6.142)
Solving (6.139), (6.140), (6.141) and (6.142) yields đ??ś = 0, đ??ˇ = 1â „4, and đ??š = 0. Therefore, the particular integral is 1
1
đ?‘Śđ?‘? (đ?‘Ľ) = 4 đ?‘Ľ 2 sin(đ?‘Ľ) + 4 đ?‘Ľ cos(đ?‘Ľ) The general solution for (6.127) is đ?‘Ś = đ?‘Śđ?‘? (đ?‘Ľ) + đ?‘Śđ?‘? (đ?‘Ľ); 1
1
= đ??´ cos(đ?‘Ľ) + đ??ľ sin(đ?‘Ľ) + 4 đ?‘Ľ 2 sin(đ?‘Ľ) + 4 đ?‘Ľ cos(đ?‘Ľ)
18
Exercise 6.8.5 Solve the following equations 1.𝑦" − 4𝑦′ + 4𝑦 = 𝑥𝑒 2𝑥 3.
𝑑2𝑦 𝑑𝑥 2
−4
𝑑𝑦 𝑑𝑥
2. 𝑦" − 2𝑦′ + 5𝑦 = 𝑒 𝑥 cos(2𝑥) 2𝑥
+ 4𝑦 = (𝑥 + 1)𝑒 4.
𝑑2𝑦 𝑑𝑥 2
+2
𝑑𝑦 𝑑𝑥
1
+ 5𝑦 = 𝑒𝑥 sin(𝑥) 𝑦(0) = 0, 𝑦 ′ (0) = 65
Answers 1
1
1. (𝐴 + 𝐵𝑥 + 6 𝑥 3 ) 𝑒 2𝑥 2. 𝑒 𝑥 [𝐴 cos(2𝑥) + 𝐵 sin(2𝑥) + 4 𝑥 sin(2𝑥)] 1
1
1
1
4
7
3. (𝐴 + 𝐵𝑥 + 2 𝑥 2 + 6 𝑥 3 ) 𝑒 2𝑥 4. 𝑒 −𝑥 [65 cos(2𝑥) + 65 sin(2𝑥) + 𝑒 −𝑥 [65 cos(𝑥) + 65 sin(𝑥) +]]
Miscellaneous Solve the following differential equations. 1. 2. 3.
𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2
− −
𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥
+3
2
𝑑 𝑦
− 2𝑦 = 1 − 3𝑥 − 2𝑦 = 4𝑥2 − 5𝑥
𝑑𝑦
= 6𝑥
𝑑𝑥 𝑑𝑦
4. 2 𝑑𝑥2 − 5 𝑑𝑥 + 2𝑦 = 𝑒 2𝑥 5. 6. 7. 8. 9.
𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2
− 9𝑦 = 12 cos(3𝑥) − 9𝑦 = 13 sin(2𝑥) − cos(2𝑥) −4 −2 +2
𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥
+ 3𝑦 = 4𝑒3𝑥 + 3𝑥 + 8 + 2𝑦 = 𝑥2 𝑒
𝑥
+ 5𝑦 = 𝑒𝑥 sin(𝑥)
19
Answers 3
5
1. 𝑦 = 𝐴𝑒 2𝑥 + 𝐵𝑒 −𝑥 + 2 𝑥 − 4 9
17
2. 𝑦 = 𝐴𝑒 2𝑥 + 𝐵𝑒 −𝑥 − 2𝑥 2 + 2 𝑥 − 4 2
3. 𝑦 = 𝐴 + 𝐵𝑒 −𝑥 + 𝑥 2 − 3 𝑥 1
4. 𝑦 = 𝐴𝑒 0.5𝑥 + 𝐵𝑒 2𝑥 + 3 𝑥𝑒 2𝑥 5. 𝑦 = 𝐴 sin(3𝑥) + 𝐵 cos(3𝑥) + 2𝑥 sin(3𝑥) 1
6. 𝑦 = 𝐴𝑒 3𝑥 + 𝐵𝑒 −3𝑥 − sin(𝑥) + 3 cos(3𝑥) 7. 𝑦 = 𝐴𝑒 3𝑥 + 𝐵𝑒 𝑥 + 2𝑥𝑒 3𝑥 + 𝑥 + 4 8. 𝑦 = [4 sin(𝑥) + 2 cos(2𝑥) + 𝑥 2 − 2]𝑒 𝑥 7
4
9. 𝑦 = 𝑒 −𝑥 [𝐴 sin(3𝑥) + 𝐵 cos(3𝑥)] + 𝑒 𝑥 [65 sin(𝑥) − 65 cos(𝑥)]
20