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Chapter 19 Ordinary Differential Equations I (Part B)
19.4 First order linear equations: use of an integrating factor
19.4.1 Exact equations I
A first order linear differential equation is an exact equation when it takes the form a ( x) dy dx
- a ' (x ) y = g ( x ) (^) or a ( x) y ' + a ' (x ) y = g (x) (1)
where a^ (^ x)^ is a function of the independent variable (x in the above equation).
- From the general form (^) a 1 ( x) y ' + a 0 ( x) y = g ( x) : ◦ We can see that the coefficient a 0 (^ x)^ of y is the derivative of the coefficient a 1 (^ x)^ of dy dx . ◦ The equation (1) can easily be rewritten as d dx [a 1 (x )⋅y ] = g ( x) (^) (2) and easily integrated to lead to a solution a 1 (^ x)^ y^ =^ G^ (x^ )^ +^ c or y(^ x)^ =^ G ( x) a 1 ( x)
c a 1 ( x)
Example 19.14 Solving an exact equation
Solve the first order linear DE x 2 dy dx
Solution
- We recognize an exact equation in the form of a^ dy dx
- a ' y = g (x ) Where a 1 ( x) = x 2 , a 0 (^ x)^ =^2 x^ =^ a 1 '^ (^ x)^ and g (x ) = cos x
- We rewrite the equation as d dx [ x 2 y] = cos x (^) ► (^) d [ x 2 y ] = cos x dx
- Integration on both side : (^) x 2 y = sin x + c
- Explicit general solution : y(^ x)^ =^ sin x x
2 +^
c x
2 I^ :^ x^ ≠^0
See Examples 19.15 in textbook (Example 19.13 in 5
th
edition)
19.4.3 First order linear equations
From the general form of a linear DE: an (x ) d n y dx n
- an− 1 ( x ) d n− 1 y dx n− 1
- ⋯ + a 1 ( x) dy dx
- a 0 (x) y = g ( x ) We get the first order linear DE in the dependent variable y. a 1 ( x) dy dx
- a 0 ( x) y = g (x )
- If g (x ) = 0 , then the equation is homogeneous.
- If g (x ) ≠ 0 , then the equation is non-homogeneous. This equation can be rewritten in standard form dy dx
- P ( x ) y = Q ( x ) (^) (5)
- with P^ (^ x)^ =^ a 0 (x ) a 1 ( x) and Q^ (x^ )^ =^ g ( x) a 1 (x )
- P and Q are function of the independent variable only or simply constants.
Examples of standard equations
General form Standard form P Q
x dy dx
- 3 x 2 y = 7 x 3 dy dx
- 3 x y = 7 x 2 P ( x ) = 3 x (^) Q ( x) = 7 x 2 x dy dx − 2 y = 4 x e − x dy dx − 2 y x = 4 e −x P ( x ) = − 2 x Q ( x) = 4 e −x 2 x 2 dy dx − 10 x 2 y = 2 x 2 sin x dy dx − 5 y = sin x P = − 5 Q ( x) = sin x − 6 dx dt − 48 t x = − 18 t 2
- 30 t dx dt
- 8 t x = 3 t 2 − 5 t (^) P (t ) = 8 t (^) Q (t ) = 3 t 2 − 5 t L di dt
- R i = E (t ) di dt
R L i = E (t) L P = R L Q (t ) = E (t ) L Note : In textbook, when it says the apply voltage in a RL circuit is a ramp, it means E (t ) = t.
1) Homogeneous DE
The homogeneous DE dy dx
- P ( x ) y = (^0) is separable ► dy y = −P (x ) dx By integrating we get ln y = −[∫ P ( x) dx (^) ] + c ► (^) y H =^ c^ e −∫ P ( x) dx (8) with (^) y 1 =^ e −∫ P ( x )dx we can rewrite (8) as ► yH (^ x^ )^ =^ c^ y 1 (^ x)^ (9)
- Verification of the solution (9) in the homogeneous DE dy dx + P ( x ) y = (^0) ► dyH dx + P (x) yH = 0 ► c dy 1 dx + P ( x) c y 1 = 0 or dy 1 dx
- P (x ) y 1 = 0 (will be used to determine y^ p )
2) Non-homogeneous DE
To find a particular solution of equation (5), we will use a procedure called Variation of Parameters
- Basic idea : To find a function ν ( x ) that will be used to multiply y 1 (^ x)^ from equation (9) so that the product y^ p (^ x^ )^ =^ ν^ (^ x )⋅ y 1 (^ x )^ is a solution of equation (5). dy dx
- P (x ) y = Q ( x ) ► dy (^) p dx
- P ( x) y (^) p = Q ( x) ► d [ ν y 1 ] dx
- P (x) ν y 1 = Q (x ) ► ν dy 1 dx
- y 1 d ν dx
- P (x ) ν y 1 = Q ( x) ► ν ( dy 1 dx
- P (x ) y 1 )
- y 1 d ν dx = Q( x) = 0 (see verification of equation (9) previous page) ► y 1 d ν dx = Q ( x) (^) (10)
From the resulting solution (13), we can develop the method to solve the DE (5) Multiplying the equation (13) by (^) e ∫ P^ (^ x^ )^ dx^ we get e ∫ P^ (^ x)^ dx^ y = c + (^) ∫e ∫ P^ (^ x^ )^ dx^ Q (x) dx Then deriving both sides d dx [ (^) e ∫ P^ (^ x^ )^ dx^ y ]^ = e ∫P^ (x^ )^ dx^ Q ( x ) (^) ( d [ uv ] = uv ' + u ' v (^) ) e ∫ P^ (^ x)^ dx^ ⋅ dy dx
- y ⋅ d dx [e ∫ P^ (^ x^ )^ dx^ ] = e ∫ P^ (^ x^ )^ dx^ Q ( x ) e ∫ P^ (^ x)^ dx^ ⋅ dy dx
- e ∫ P^ (^ x^ )^ dx^ ⋅ d dx [∫ P ( x ) dx ]⋅ y = e ∫ P^ (^ x^ )dx^ Q( x) e ∫ P^ (^ x)^ dx^ ⋅ dy dx
- P( x) e ∫ P^ (^ x^ )^ dx ⋅y = e ∫ P^ (^ x^ )^ dx^ Q (x ) (^) (14) Equation (14) is obviously the DE dy dx
- P (x ) y = Q ( x ) (^) multiplied by the expression e ∫ P^ (^ x)^ dx^ .
Solving a linear first order differential equation
By multiplying equation (5) by a function μ we get
μ dy dx
- μ P y = μ Q (^) (15) with μ = e ∫ P^ (^ x^ )^ dx^ we can recognize that μ^ dy dx
- μ P y = d dx [μ y] as previously demonstrated with equation (13) to (14). Equation (15) can then be rewritten as d dx [μ y] = μ Q (^) (16) This exact equation can be solved by integration to get μ^ y^ =^ ∫μ^ Q^ dx^ (17) And The solution of the non-homogeneous DE dy dx
- P (x ) y = Q ( x ) is given by (^) y( x) =
μ ∫μ^ Q^ dx^ (18) where μ^ =^ e ∫ P^ (^ x^ )^ dx^ is called an integrating factor.
General Solution
The general solution of dy dx
- P ( x ) y = Q (x ) is y(^ x)^ =^ yH +^ yP =^ c^ e −∫ P ( x ) dx
- e −∫ P ( x ) dx ⋅∫e ∫ P^ (^ x^ )^ dx^ Q ( x) dx
Example 19.4.4.1 Solving a linear DE
Solve dy dx − 3 y = 6
Solution
- Already in standard form
- P ( x) = − (^3) μ = e∫^ P ( x) dx = e ∫−^3 dx^ = e − 3 x μ = e − 3 x Textbook method Prof method
- e − 3 x dy dx − 3 e − 3 x y = 6 e − 3 x ► d dx [e − 3 x y] = 6 e − 3 x ∫d^ [e − 3 x y ] = (^) ∫ 6 e − 3 x dx ► (^) e − 3 x y = − 2 e − 3 x
- c 3') y^ =^ 1 μ ∫μ^ Q^ dx^ =^ 1 e − 3 x ∫^ e − 3 x ⋅ 6 dx = e 3 x ⋅ (^6) ∫e − 3 x dx = e 3 x ⋅ 6 {− 1 3 e − 3 x
- c}
- Solution : (^) y = − 2 + ce 3 x on interval I^ :^ −∞^ <^ x^ <^ ∞
Note :
- For the DE a 1 (^ x)^ dy dx + a 0 ( x) y = g ( x) (^) ,
◦ when a 1 , a 0 and g are constant, the DE is then said to be autonomous.
- First order autonomous DE generate constant solution.
- In Example 19.4.4.1, dy dx = 3 ( y+ 2 ) (^) ► (^) y( x) = − 2 is a constant solution. (and part of the general solution)
Constant of Integration (integrating factor)
From example 19.4.4.1,
- (^) μ = e ∫ P^ (^ x)^ dx^ = e − 3 x+c 1 = e c 1 e − 3 x = c 2 e − 3 x
- c 2 e − 3 x dy dx − 3 c 2 e − 3 x y = 6 c 2 e − 3 x ► c 2 [e − 3 x dy dx − 3 e − 3 x y] = c 2 [ 6 e − 3 x ]
∴ N o needs to introduce a constant of integration when solving for the integrating factor.
Example 19.4.4.3 General Solution
Solve (^ x 2 − 9 ) dy dx
- Standard form : dy dx
x x 2 − 9 y = (^0) ► P^ (^ x)^ =^ x ( x 2 − 9 ) Q (x ) = 0
- Integrating factor : (^) μ = e ∫ xdx/(^ x (^2) − 9 ) = e 1 2 ∫ 2 x dx/( x^2 − 9 ) = e 1 2 ln|x^2 − 9 |
► μ = √ x
2 − 9 Note : P is continuous on (−∞^ ,−^3 )^ , (−^3 ,^3 )^ and (^3 ,^ ∞)^ , but μ is undefined on (− 3 , 3 ) (square root of negative values)
- y =
μ ∫μ^ Q^ dx^ :^ y^ =^
√ x
2 − 9 ∫√^ x 2 − 9 ⋅ 0 dx =
√ x
2 − 9 ∫^0 dx^ =^ c
√ x
2 − 9
- General solution: y(^ x)^ =^ c
√ x
2 − 9 on I : x > 3 or I : x < − 3
Example 19.4.4.4 An Initial-Value Problem
dy dx
- y = x with (^) y( 0 ) = 4 Solution
- Already in standard form : P ( x) = 1 , Q (x ) = x Both are continuous on (−∞^ ,^ ∞)
- Integrating factor : μ = e ∫dx^ = e x ► μ = e x
- y =
μ ∫μ^ Q^ dx^ :^ y^ =^
e x ∫ e x ⋅ x dx = e −x [ x e x − e x
- General solution : (^) y( x) = x − 1 + c e − x
- Initial value : y( 0 ) = 4 ► (^4) = 0 − 1 + c e 0 ► c = 5
- Solution of the IVP : (^) y( x) = x − 1 + 5 e −x on I : −∞ < x < ∞
Example 19.4.4.5 An Initial-Value Problem and piece-wise function
Solve dy dx
- y = f ( x ) , y ( 0 ) = (^0) where f^ (^ x)^ =^
1, 0 < x ⩽ 1 0, x > 1
Solution
We first solve for y( x) on [0, 1 ] (A.), and in a second step (B.), we solve for (1,∞). A. For 0 < x ⩽ 1 ► dy dx
- P ( x) = 1 ► (^) μ = e∫ dx = e x ► (^) μ = e x
- y^ =^
e
x ∫
e x ⋅ 1 dx = e −x [e x
- c] (^) ► y( x) = 1 + c 1 e −x
- y( 0 ) = 0 ► c 1 =^ −^1 ► (^) y( x) = 1 − e − x B. For x > 1 ► dy dx
- (^) P ( x) = 1 ► (^) μ = e∫ dx = e x (^) ► μ = e x
- y^ =^
e
x ∫e
x ⋅ 0 dx = e −x [c 2 ] (^) ► y( x) = c 2 e − x
From expressions found in A. and B. Intermediate solution : y(^ x)^ =^
1 − e −x , 0 < x ⩽ 1 c 2 e −x , x > 1 I have no initial value to use to find c 2 but … C. By definition of continuity at a point, it is possible to determine c 2_._ At x = 1 ► y(^1 )^ =^1 −^ e − 1 = c 2 e − 1 ► c 2 =^ e−^1 Solution of the IVP : y(^ x)^ =^
1 − e −x , 0 < x ⩽ 1 (e− 1 )e − x , x > 1 I : 0 < x < ∞
