First-Order Differential Equations National Chiao Tung University Chun-Jen Tsai 9/16/2019

First-Order Differential Equations

National Chiao Tung University Chun-Jen Tsai 9/16/2019

Solution Curves without Solving DE

 Sometimes, just by looking at the differential equation, we can learn useful information about its solutions:

 The solution curve y = y(x) of a first order DE

dy/dx = f (x, y) on its interval of definition I must possess a tangent line at each point (x, y(x)), and must have no breaks.

 The slope of the tangent line at (x, y(x)) on a solution curve is the value of the first derivative dy/dx at this point.

 A (very small) line segment at (x, y(x)) that has the slope f (x, y) is called lineal element of the solution curve.

Example of Lineal Element

 Consider dy/dx = f (x, y) = 0.2xy, the slope of the lineal element of the solution curve at (2, 3) is f (2, 3) = 1.2.

x y

Lineal element is tangent to solution curve passes (2,3)

(2, 3) solution curve

tangent

x y

Lineal element at (2, 3) (2, 3) slope = 1.2

Direction Field

 The collection of the lineal elements on a rectangular grid on the xy-plane is called a direction field or a slope field of the DE dy/dx = f (x, y).

 A single solution curve on the x–y plane will follow the flow pattern of the slope field.

y 4

2

- 2

x 4

2

- 2

y

x

Increasing/Decreasing of a Solution

 If dy/dx > 0 for all x on the interval of definition I, then the differentiable function y(x) is increasing on I.

 If dy/dx < 0 for all x on the interval of definition I, then the differentiable function y(x) is decreasing on I.

Example: Approximating a Solution

 We can use a slope field to approximate the IVP, dy/dx = sin y, y(0) =－3/2:

1) Define the direction field around y = 0

2) Constraint 1: the solution must pass (0, –3/2)

3) Constraint 2: the slope of the solution curve must be 0 when y

= 0 and y = –

 the solution curve can be

approximated as in the figure.

4 2

-2

y

x

Autonomous First-Order DEs

 A DE in which the independent variable does not appear explicitly is said to be autonomous.

 If x is the independent variable, an autonomous DE can be written as F(y, y') = 0, or dy/dx = f (y).

 Example: If y(t) is a function of time, then the following DE is autonomous and time-independent:

1 y2

dt

dy  

Critical Points

 In dy/dx = f (y), if f(c) = 0, then c is called the critical point of the autonomous DE. A critical point is also refer to as an equilibrium point or a stationary point.

 If c is a critical point of dy/dx = f (y), then y(x) = c is a constant solution of the autonomous equation. This is also called an equilibrium solution.

Example: Autonomous DE

 The DE, dP/dt = (a－bP)P, a > 0, b > 0, is autonomous.

Let (a－bP)P = 0, we have two critical points: 0 and a/b.

 The sign of f(P) = P(a－bP) can be shown in a phase portrait

a/b

0

P-axis

Interval Sign of f(P) P(t) Arrow

(–, 0) minus decreasing down

(0, a/b) plus Increasing Up

(a/b, ) minus decreasing down

Solution Curve Properties

 The solution space can be divided into several regions by equilibrium solutions:

 y(x) is bounded

 f(y) > 0 or f(y) < 0 for all x in a sub region

 y(x) is strictly monotonic

 If y(x) is bounded above or below by a critical point, y(x) approaches this point either as x   or as x  –.

R₃

I

(x₀, y₀) y

R₂ y(x) = c₂

y(x) = c₁

Example: dP/dt = P(a － bP) Revisited

P

a b

0

decreasing

increasing

decreasing

phase line tP-plane

P₀ P₀ P₀ P

R₃

R₂

R₁ t

Example: dy/dx = (y – 1)

²

1

y y

(0, 2)

y = 1 y

y = 1

(0, –1)

x x

increasing

increasing

x = − x = 1

Attractors and Repellers

 The solution curve of a first order DE near a critical point c exhibits one of the following three behaviors:

 Solution curves approach c from either sides. c is called asymptotically stable or an attractor.

 All solution curves starts near c move away from c. c is called unstable critical point or a repeller.

 Solution curves approach c from one side and move away from c from the other side. c is called semistable.

Solution by Integration

 If the DE can be expressed in normal form, f(x, y) = g(x), the equation can be solved by integration.

Since,

Integrating both sides, we have:

where G(x) is the indefinite integral of g(x).

 

^x

dx g dy 

 

^{x dx G}

 

^{x c}

g

y 



 

Example: Solution by Integration

 Solving the initial value problem

By integrating both sides, we have . 3

) 3 2

( )

(x x dx x² x c

y 



   

. 2 )

1 ( ,

3

2  

 x y

dx dy

-6 -5 -4 -3 -2 -1 0 1 2 3 4

-10 -8 -6 -4 -2 0 2 4

C = 2

C = 0 C = –2

C = –6 C = –4

Solution that passes through the initial condition (1, 2) is the curve with C = –2

Family of solution curves

2

^nd

-Order Solution by Integration

 If we have a second-order DE of the special form:

we have

where G is an anti-derivative of g and C₁ is an arbitrary constant. Therefore,

where C₂ is a second arbitrary constant.

),

2 (

2

x dx g

y

d 

, )

( )

( )

(x dx g x dx G x C₁

y  



 





^{( )}



^{( ) ,}

) ( )

(x y x dx G x C₁ dx G x dx C₁x C₂

y 



 



 



 

y

Separable Equations (1/2)

 A first order DE of the form

is said to be separable or to have separable variables.

Divide both side by h(y), the DE becomes

Integrating both sides w.r.t. x, we have

       

^{x h y g x f y , where f (y) 1/ h(y)}

dx g

dy   

 

^g

 

^x

dx y dy

f 



^{( )}



^{dx g}

 

^{x dx C.}

dx x dy y

f 







Separable Equations (2/2)

Cancelling the differential term dx, we have

If the two anti-derivatives

can be found, we have the family of equations F(y(x)) = G(x) + C

that conforms to the differential equation.

 

^{y dy g}

 

^{x dx C.}

f 







    



^

 f y dy G x g x dx y

F( ) and ( )

Example: dy/dx ＝ –6xy, y(0) = 7

 Rearranging the equation, we have dy/y = –6xdx, therefore,

Thus |y| = e^–3x2 e^C1 or y = ±e^C1 e^–3x2. We have, y = C₂e^–3x2, C₂ = ±e^C1  R.

However, y = 0 is also a solution.

Note that as C₂  0, y  0.

Since y(0) = 7, the particular solution is y = 7e^–3x2. .

3 ln

6 /

1

2 C

x y

xdx y

dy

















Example: dy/dx ＝ –x/y, y(4) = –3

 Since



^ydy

=

^–



xdx, we have y²/2 = –x²/2 + c₁. The solution must pass (4, –3), thus, c₁ = 25/2.

 the solution is the lower half-circle of radius 5 centered at (0, 0).

x y

family of solutions:

x²+ y²= C

Losing a Solution

 Some care should be exercised when separating

variables, since the variable divisors could be zero in some cases.

 If r is a zero of h(y), then y = r is a constant solution of the DE. However, y = r may not show up in the family of solutions. Recall that this is called a singular

solution.

Example: dy/dx = y

²

－ 4

 Since y²－4 is separable



_y^2dy_{ 4} ^



^dx



^_^ _y^1/_⁴₂ ^ _y¹_^{/ 4}₂^_^{dy }



^dx

2 1

4 ln 2 1

4 ln

1 y   y   x  c

, 2

2 or 2

, 2 4

ln 2 ₂ ce^4x c e^c

y c y

y x

y   



 

 



x x

ce

y ce₄

4

1 21



 

Example: (e

^2y

– y)cos x = e

^y

sin 2x, y(0) = 0

 Solve the IVP by dividing both sides by e^y cos x, then multiply both sides by dx, we have

x dx dy x

e y e

y y

cos 2

2 sin

 

 



^{ey  yey dy  2}



^{sin x dx}

0 )

0 ( , cos

2  







 ye^ e^ x c y

e^{y y y}

c = 4

Linear First Order DE

 A first-order differential equation of the form:

(1) is said to be a linear equation. When g(x) = 0, the

linear equation is said to be homogeneous, otherwise it is non-homogeneous.

 Dividing both side of (1) by the leading coefficient a₁(x), we have the standard form:

) ( )

( )

( ₀

1 a x y g x

dx x dy

a  

) ( )

(x y f x dy  P 

Solving the 1

^st

-Order Standard Form

 The solution of dy/dx + P(x)y = f(x) can be derived by multiplying both sides of the equation by a special function



(x). We want the function



(x) to satisfy the property:

Thus d



/dx =



P(x) 



= e^P(x)dx.

The function



(x) is called the integrating factor.



^{( )}



^{P(x)y f (x).}

dx y dy

dx d dx

y dy dx x

d

   

 





 



 







Solution by Integrating Factors

 We can solve the DE by multiplying both sides of the standard form by e^P(x)dx, thus:

 

 

^{P x dx} P x e ^{P x dx} y f x e ^{P x dx} dx

e ^{( )} dy ( ) ^{( )} ( ) ^{( )}

 



 ye ^{P x dx} f x e ^{P x dx} dx

d ^{( ) ( )}

) (

c dx e

x f

ye^P(x)dx 



^{( )} ^P(x)dx 

Dropping Integrating Factor Constant

 Note that you do not need to keep the constant when computing the anti-derivative of the integrating factor.

Assume that G(x) is the anti-derivative of P(x), since e^P(x)dx = e^{G(x) + c} = c₁e^G(x),

The constant c₁ = e^c will simply be cancelled out on both side of the differential equation.

Example: Solve dy/dx – 3y = 6

 Solution:

dx x

e e^(3)  ^3

 

^{e x y e x}

dx

d _{3 3}

6 ^

 

x x

x e y e

dx

e^3 dy 3 ^3  6 ^3

c e

y

e^3x  2 ^3x 















 ce x

y 2 ^3x,

General Solution on I

 If P(x) and f(x) in the standard form are continuous on an open interval I, then

is a general solution of dy/dx + P(x)y = f(x).

That is, every solutions on I has the form. In other words, there is no singular solution for the linear 1^st order differential equation on I.



^

 

 ce^ e^ e f x dx

y ^{P(x)dx P(x)dx P(x)dx} ( )

Particular Solution on I

 Given an initial condition y(x₀) = y₀ to the linear first order DE dy/dx + P(x)y = Q(x) on I where P(x) and Q(x) are continuous, the particular solution of the DE has the form:

Note that it is easy to verify that y(x₀) = y₀.











  

 ^



x^x

du u P dt

t

P y e Q t dt

e x

y

t x x

x

0 0

0 ( )

)

( ^{( )} ₀ ^{( )}

Example: (x

²

–9)dy/dx + xy = 0

 Solution:

P(x) is continuous on (–, –3), (–3, 3), and (3, ).

Thus, the integrating factor is:

Therefore,

) 9 ) (

( ,

9 0 ²

2    

 

x x x

P x y

x dx

dy

. 3 , 3 ,

2 9

9 ln 2 / ) 1

9

( ²  ²    

 _ ^

x x

e

e ^{x dx x}

x

. 0

2 9  

 x  y dx

d

. 3 , 3 ,

2 9 y  c x   x

Example: IVP y' + y = x, y(0) = 4

 Since P(x) = 1 and Q(x) = x are continuous on (–, ), we have integrating factor e^dx = e^x:

 

^{ex y xex}

dx

d 















 x ce^ x

y ( 1) ^x,

4 2 0 - 2

x y

c > 0

c < 0 0

transient term

Example: Discontinuous f(x)

 Find a continuous function satisfying

and y(0) = 0.

Solution:

 find c₂ so that









 



 0, 1

1 0

, ) 1

( ),

( x

x x f x f dx y

dy











  _^

1 ,

1 0

, 1

2e x

c

x

y e _x

x

) 1 ( )

(

lim_x₁ y x  y

f(x)

x

y

1 x

Non-elementary Functions

 Some simple function do not possess antiderivatives that are elementary functions, and integrals of this kind of functions are called non-elementary.

 The integrations of non-elementary functions can only be solved by numerical methods.

Example:

dx e ^x



^{ 2}



^{sin x2dx}

Level Curves and Family of Solutions

 In multivariate calculus, for a function of two variables, z = G(x, y), the curves defined by G(x, y) = c (c is a

constant) are called level curves of the function.

Level curves of e^y+ye^–y+e^–y+2cos = c Solutions of IVPs

x y

2

1

-1

-2

-2 -1 1 2

-1

-2

x y

2

1

-2 -1 1 2

c = 2 c = 4

(0, 0)

( /2, 0)

Differentials of Two-Variable Functions

 If z = f (x, y) is a function of two variables with

continuous first partial derivatives in a region R of the xy-plane, then its total differential is:

If f (x, y) = c, we have:

 Given a one-parameter family of curves f (x, y) = c, .

y dy dx f

x dz f



 



 

.

 0



 



 dy

y dx f

x f

Example:

 If x² – 5xy + y³ = c, then taking the differential gives (2x – 5y) dx + (–5x + 3y²) dy = 0.

Question: can we think reversely?

Exact Equations

 A differential expression M(x, y) dx + N(x, y) dy is an exact differential in a region R of the xy-plane if it is the total differential of some function f(x, y).

 A first-order differential equation of the form M(x, y) dx + N(x, y) dy = 0

is said to be an exact equation if the expression on the left-hand side is an exact differential.

Criterion for an Exact Differential

 Theorem: Let M(x, y) and N(x, y) be continuous and

have continuous first partial derivatives in a rectangular region R defined by a < x < b, c < y < d.

Then a necessary and sufficient condition that M(x, y) dx + N(x, y) dy be an exact differential is

x . N y

M



 





Proof of the Necessity

 If M(x, y) dx + N(x, y) dy is exact, there exists some function f such that for all x in R,

Therefore, M(x, y) = f/x, and N(x, y) = f/y, and .

) , ( )

,

( dy

y dx f

x dy f

y x N dx

y x

M 

 



 



.

2

x N y

f x

x y

f x

f y

y M



 



 











 





 



 











 





#

Proof of the Sufficiency (1/2)

 Note that we have

where g(y), shall be a function of y. Since we want

therefore,

If we can prove that g'(y) is a function of y alone, integrating g'(y) w.r.t. y, gives us the solution.

), ( )

, ( )

, ( )

,

(x y f x y M x y dx g y x M

f    







), ( )

, ( )

, ( )

,

( M x y dx g y

y y x N y

x y N

f  



 



 









 

  M x y dx

y y x N y

g ( ) ( , ) ( , )

Proof of the Sufficiency (2/2)

 Since

and M/y = N/x, we have /x[g'(y)] = 0.

Thus, g'(y) is a function of y alone.

In this case, the solution is

y M x

dx N y x y M

y x

x N 

 



 



 







 



 ^{( , )}



^{( , )}

. )

, ( )

, ( )

, ( )

,

(^{x y }



^{M x y dx }



^_^{ N x y } _^_y



^{M x y dx}_^dy

f

#

Observations

 The solution to an exact eq. M(x, y) dx + N(x, y) dy = 0 is

where c is a constant parameter.

 The method of solution can start from f/ y = N(x, y) as well. Then, we have



^

 ( , ) ( )

) ,

(x y N x y dy h x f





 

  N x y dy

y x x M x

h ( ) ( , ) ( , )

, )

, ( )

, ( )

, ( )

,

( M x y dx dy c

y y x N dx

y x M y

x

f  



 







 





  

Example: 2xy dx＋(x

²

– 1)dy = 0

 Solution:

Since M(x, y) = 2xy, N(x, y) = x² – 1, we have

M/y = 2x = N/x so the equation is exact, and there exists f(x, y) such that f/x = 2xy and

f/y = x² – 1.

Integrating the first equation f(x, y) = x²y + g(y)

Take the partial derivative of y, equate it with N(x, y), we have x² + g(y) = x² – 1. Therefore, g(y) = –1, and f (x, y) = x²y – y. The implicit solution is x²y – y = c.

Example: An IVP of Exact Equation

 Solve

Solution:

. 2 ) 0 ( ) ,

1 (

sin cos

2

2 



  y

x y

x x

xy dx

dy

x xdx

x x

h( )   (cos )(sin )   ¹₂cos²





x xy N

y M



 



 

 2

2 2 h (x) cosxsin x xy x xy

f      



 

) ( ) 1

2 ( ) , ( ), 1

( ²

2

2 y x h x

y x f x

y y

f     



 

Example: An IVP (cont.)

The implicit solution is y²(1 – x²) – cos²x = c.

Substitute the initial condition y(0) = 2 into the implicit solution, we have c = 3.

y

x

Integrating Factors for Exactness

 Can we multiply a non-exact equation by an integrating factors



(x, y) to make it exact? That is, can we make



(x, y)M(x, y) dx +



(x, y)N(x, y) dy = 0

an exact differential equation? To achieve this goal,



(x, y) must satisfy

M



_y– N



_x + (M_y – N_x)



= 0.

 In practice, a proper



(x, y) is not easy to find unless it happens to be a function of x or y alone. If



(x, y) =



(x),

 

N N M

dx

d y  x

  Separable equation if (M_y – N_x)/N contains x alone.

Solution by Substitutions

 We can substitute dy/dx = f (x, y) with y = g(x, u), where u is a function of x, to solve for the solution.

By chain rule:

then

We can then solve for du/dx = F(x, u).

If u =



(x) is the solution, then y = g(x,



(x)).

, )

, ( )

,

( dx

u du x g u

x dx g

dy

u

x 



. )

, ( )

, ( ))

, ( ,

( dx

u du x g u

x g u

x g x

f  _x  _u

Homogeneous Equations (1/2)

 If f(tx, ty) = t^f(x, y) for some real number



, then f is said to be a homogeneous equation of degree



.

Example: f (x, y) = x³＋y³ is a homogeneous equation of degree 3.

 Similarly, a first-order DE in differential form M(x, y)dx＋N(x, y)dy = 0

is said to be homogeneous if both M and N are homogeneous function of the same degree.

Homogeneous Equations (2/2)

 The meaning of “homogeneous” here is different from the “homogeneous” in Sec. 2.3.

 If M and N are homogeneous functions of degree



, we have:

M(x, y) = x^M(1, u) and N(x, y) = x^N(1, u), u = y/x and

M(x, y) = y^M(v, 1) and N(x, y) = y^N(v, 1), v = x/y

 We can turn a homogeneous equation into a separable first order DE using substitution with either y = ux or x = vy.

Example: (x

²

＋y

²

)dx ＋(x

²

－xy)dy = 0

 Solution:

M and N are 2^nd-order homogeneous equation.

Let y = ux, then dy = u dx＋x du.

After substitution, we have

 (x²＋u²x²)dx＋(x²－ux²)[u dx＋x du] = 0

 x² (1＋u)dx＋x³ (1－u) du = 0 Therefore

1 0 1 2 1 0

1  





 







 



x du dx

u x

du dx u u

c x

u

u  2ln1 ln  ln



Bernoulli’s Equation

 The DE

where n is any real number, is called Bernoulli’s

equation. Note that for n = 0 and n = 1, it is linear. For any other n, the substitution u = y^1–n reduces any

equation of this form to a linear equation.

yn

x f y

x dx P

dy  ( )  ( )

Example: x dy/dx + y = x

²

y

²

 Solution:

,

substitute with y = u^–1 and dy/dx = –u^–2du/dx.

, the integrating factor on (0, ) is e^–dx/x = x^–1, we have

x^–1u = –x + c  y = 1/(– x² + cx).

1 2

xy x y

dx

dy  

x x u

dx

du   

 1

 

^{x1u  1}

dx d

Another Reduction to Separation

 A DE of the form

can always be reduced to an equation with separable variables by means of the substitution

u = Ax＋By＋C, B ≠ 0.

) (Ax By C dx f

dy   

Example: dy/dx = (–2x + y)

²

– 7, y(0) = 0

 Solution:

Let u =－2x＋y, then du/dx =－2＋dy/dx.

The DE can be reduced to du/dx = u²－9.

dx u du

dx u u

u

du 





 

 

 

 

3 1 3

1 6

1 )

3 )(

3 (

6 1

6

1 ,

3 3 3

ln 3 6

1 _{x c}

e c

u ce c u

u x

u  



 



 

 

x x

ce x ce

y ₆

6

1

) 1

( 2 3



 





. 1 ,

0 )

0

(   

 y c ^x

y

Numerical Methods

 The solution of a DE can be approximated using a tangent line:

Euler’s Method

 One can solve the IVP: y’ = f(x, y), y(x₀) = y₀, numerically using the following procedure:

1. Linearization of y(x) at x = x₀: L(x) = f(x₀, y₀)(x – x₀) + y₀

2. Replace x in the above equation with x₁ = x₀ + h, we have L(x₁) = f (x₀, y₀)(x₀ + h – x₀) + y₀ or y₁ = y₀ + hf (x₀, y₀),

where y₁ = L(x₁)

3. If h  0 then y₁ ~ y(x₁)

4. Use (x₁, y₁) as a new starting point, we have x₂ = x₁ + h = x₀ + 2h, and y(x₂) = y₁ + hf(x₁, y₁)

5. Recursively, we have y_n+1 = y_n + hf(x_n, y_n), where x_n = x₀ + nh, n = 0, 1, 2, …

Error Accumulations

 Numerical solutions are approximations to the exact solution of a DE  approximation errors may become large when x is far away from the initial condition.

exact solution

y

Euler’s method Runge-Kutta

method