First-Order Differential Equations: Solutions, Methods, and Applications, Lecture notes of Differential Equations

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Chapter 2

Chapter 2 First-order Differential Equations

2.1 Solution curves without the solution

Lineal Elements : a line segment representing a slope at a specific point of a curve The differential equation we are going to consider is dy/dx = f(x, y). For example, dy/dx = y. When y = 2, each of the elements on the line y = 2 has slope 2 and its midpoint on the lineal element as shown in the figure.

Isoclines and Direction Fields Any member of the family f(x, y) = c is called isocline , which literally means a curve along which the inclination of the tangents is the same. As the parameter c is varied, we obtain a collection of isoclines on which the lineal elements are judiciously constructed. The totality of these lineal elements is variously called a direction field , slope field , or lineal element field.

2.2 Separable Variables

Solution by Integration dy dx

= g x( )

can be solved by integration. If g(x) is a continuous function, then

y( x)= ∫ g(x)dx=G(x)+c.

Separable Equation dy dx

g x h y

= (^ )

if h y( ) ≠ 0

Method of the solution:

∫ h(y)dy+^ c^1 =∫ g(x)dx+c^2

∫ h(^ y)dy=^ ∫ g(x)dx+c^2 −c^1 = ∫g(x)dx+c

Note: A one-parameter family of solutions is obtained by integrating both sides of h(y)dy = g(x)dx

y ce ce

x = (^) x

4 4 In addition, y = 2, -2 is also a (singular) solution, too. The solution curves are shown in the following graph.

由上圖可以看出解函數可能有發散點。事實上該點並非是方程式的 singular point， 因為從 f(x, y) 看不出它的存在。在該點處的導函數趨向正無窮大，y 雖趨向正負無 窮大，但因 f(x, y) 含有 y 的正平方項，故仍能符合方程式。The solution subject to the initial condition is y = -2.

Autonomous First-order DEs 從方程式中看出重要的 Boundary A differential equation in which the independent variable does not appear explicitly is said to be autonomous. For the first-order autonomous DE can be written in

normal form as f(y) dx

dy (^) =.

Critical Points : The zeros of f(y) in the first-order autonomous DE. A critical point is also called an equilibrium point or stationary point. If c is a critical point of the first-order autonomous DE, then y(x) = c is a constant solution of the equation. The corresponding solution is called an equilibrium solution.

Example: The differential equation dP dt

= P a( −bP)

is an autonomous DE. Please find the solutions. Solution: dP P a bP

dt ( − )

Hence,

P t aP bP a bP e at

0 0 0 where P(0) = Po. In addition, the equilibrium solutions are P = 0 and P = a/b

Graph of P(t)

As t → ∞, P t a b

As t → −∞ , P t( ) → 0 Now, differentiating the solution twice gives

⎟⎛^ −

= ⎛^ −

2 b

P a b

2 b P P a dt

d P 2 2

2

If Po < a/b, then the inflection point is when P = a/2b. If Po > a/b, P t( ) → ±∞ as

⎟⎟⎠

bP a

ln bP a

t^1 o

o

Note:由方程式僅能看出重要的y boundary，對於 (^) ⎟⎟ ⎠

bP a

ln bP a

t^1 o

0 o 等重要resonant

point(因y值趨向正或負無窮大)，仍需等到解找到時才可看出。尤其是針對第三個case 而言(P 0 < 0或 > a/b時才有共振點)，此發散點並不是方程式的 singular point (因為 f(x, y) 及 ∂f /∂y均連續)，該點的導函數趨向負無窮大，y 雖趨向正負無窮大，但因 f(x, y) 含有 y 的負平方項，故仍能符合方程式。

Phase portrait : We can judge the sign of dP/dt from the function f(P) i.e., the increasing trend if dP/dt > 0 and the decreasing trend if dP/dt < 0. The separation of those regions is the critical points.

Phase Line : The vertical line in the left of the above figure is called the phase line in which the critical points are marked.

Example: An autonomous DE

(y 1 )^2 dx

dy (^) = −

The general solution which can be derived by the method of Section 2.2 is

x c

y 1 1

where c is an arbitrary constant. That is, the function is a one-parameter family of solutions.

2.3 Linear Equations

First-Order Linear Equation

a x dy dx 1 (^ )^ +^ a^0 (^ x y)^ =^ g x(^ ) or^ dy dx

• P x y( ) =f x( )

In general, y is called the output or response while g(x) or f(x) is called input or driving function.

The Property : The solution is the sum of the two solutions, y = y (^) c + yp where yc is a solution of dy dx

• P x y( ) = 0

and yp is a particular solution of the original equation.

Method of solution :

1. Integrating Factor : (1) Find an integrating factor μ(x) such that y dx

d dx

[ (x)y(x)] (x)dy dx

(x)P(x)y (x)f(x) d dx

μ(x )dy+μ =μ ≡ μ =μ + μ is a derivative of some function. That is, d dx

μ (^) = P x( )μ ( x)

The integrating factor is

μ (x)=e^ ∫P(x)dx

(2) The above equation is the same as

e y e f(x) dx

d ∫ P^ (x)dx = ∫P(x)dx

Integrating both sides of the equation gives

y = yc+yp=cexp(−∫ P(x)dx)+exp(−∫P(x)dx) ∫ exp(∫P(x)dx)f(x)dx

2. Variation of Parameter : (1) Finding the homogeneous solution yc dy dx

c+ P x y( ) (^) c= 0

Separation of the variables gives

y c= cexp(− ∫P(x)dx)

(2) Changing the parameter into a dependent variable u(x) and using it as a trial particular solution yp

y p = u(x)exp(−∫P(x)dx)=u(x)y 1 (x)

(3) Substituting yp into the original equation

f(x) dx

P(x)y y du dx

u dy

That is,

y du dx 1 =^ f^ (^ x) so that

= ∫y(x)dx

u f(x) 1 Thus

y = yc+yp=cexp(−∫ P(x)dx)+exp(−∫P(x)dx) ∫ exp(∫P(x)dx)f(x)dx

Note: y = yc + yp is called a general solution of the equation.

Example 1: Solve ( ) xy 0 dx

x 2 + 9 dy+ =

Solution: dy dx

x x

• y

is a linear differential equation. The integrating factor is

ln^ (x^9 )^ x 9 2

exp^1 x 9

(x ) exp xdx^22 2 ⎥⎦= +

= ⎡^ +

so that

x 9 y 0 dx

d (^2) = ⎥⎦

The solution is

y c x

Example 2: Find a continuous solution satisfying dy dx

• y = f x( ) , where ⎩

0 , x 1

1 , 0 x 1 f(x)

and the initial condition y(0) = 0. Solution: The integrating factor is μ(x) = ex. Since f(x) is not continuous, we solve the equation in two parts.

(1) On x ٛ [0, 1], the equation becomes

d dx

e yx^ = ex, and the solution is y = 1 + c 1 e -x Since the initial condition, we find c 1 = -1. (2) On the interval of x > 1, the equation becomes d dx

e yx^ = 0 , and the solution is y = c 2 e -x.

df = M(x, y)dx + N(x, y)dy

[Exact Equation] if M(x, y)dx + N(x, y)dy is an exact differential, then M(x, y)dx + N(x, y)dy = 0 is an exact equation.

[Criterion for an Exact Differential] Let M(x, y) an N(x, y) be continuous and have continuous first partial derivatives in a rectangular region 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 ∂ ∂

M x y y

N x y x

Statement: ∂ ∂

M x y y

N x y x

( , ) (^) = ( , )⇔ M(x, y)dx +N(x, y)dy is an exact differential

Proof: (1) Necessity ( ⇐ ):

df = M(x, y)dx + N(x, y)dy = ∂ ∂

f x y x

dx f x y y

( , ) (^) + ( , )dy

Therefore,

M x y f x y x

( , ) = ∂(^ ,^ )

and N x y f x y y

( , ) = ∂ (^ ,^ )

and we conclude that ∂ ∂

M x y y

f x y y x

f x y x y

N x y x

(2) Sufficiency ( ⇒ ): the same procedure to get the solution of an exact equation

Method of Solution: Given the equation M(x, y)dx + N(x, y)dy = 0 with the condition of ∂ ∂

M

y

N

x

to find the solution, we assume that ∂ ∂

f x y x

( , ) (^) = M x y( , ) and ∂ ∂

f x y y

( , ) (^) = N x y( , ).

The problem becomes to find the function of f(x, y). From the former equation, we find that

f (x,y)=∫ M(x,y)dx+g(y)

g(y) is an "integration constant" with respect to x. Differentiating the above equation with respect to y, we find that

∂^ ∫^ + =

∂ (^) M(x,y)dx g'(y) N(x,y) y Hence, the function g(y) can be found by

M(x,y)dxdy y

g( y) ∫ N(x,y)dy ∫ ∫

Therefore, f(x, y) is found and the solution of the exact equation is f(x, y) = c.

Note: In the above equation, although the right-hand formula is denoted as a function of x and y, actially it is a function of y only since

∂ (^) M(x,y)dy 0 y

N(x,y)dy x

M(x,y)dxdy x y

N(x,y)dy x

g(y) x Please pay attention to the difference of the two results:

1. G(x,y)dx G(x,y) g(y) x

2. G(x,y)dx G(x,y) x

Example: Solve (e 2y^ - ycos xy)dx + (2xe 2y^ -x cos xy + 2y)dy = 0 Solution: ∂ ∂

M

y

e xy xy xy N x

= 2 2 y+ sin − cos =

The differential is exact. We try to find a function f(x, y) such that

f (x,y)= ∫ M(x,y)dx+g(y)=xe −sinxy+g(y)

2 y

From N(x, y), we find that g has to satisfy 2 xe^2 y^ − x cos xy + g '( y ) = 2 xe^2 y− x cos xy + 2 y or g'(y) = 2y Therefore, g(y) = y^2. The solution of the equation is xe 2 y − sinxy + y 2 =c

[How to solve a differential M(x, y)dx + N(x, y)dy = 0] 結合 Integrating factor 及 Exact differential 的概念 (1) As an exact differential (2) General case : Similar to the first-order linear DE, we can define one integrating factor μ(x, y) such that μ(x, y)M(x, y)dx + μ(x, y)N(x, y)dy = 0 is exact instead of the derivative in the case of the linear DE. Then we can get a partial differential equation of μ(x, y). ( N) x

( M)

y

μ ∂

μ = ∂ ∂

∂ (^) i.e., μ yM +^ μM^ y =^ μx N +^ μNx The above partial differential equation attempts to find all the integrating factors. In fact, it is enough for us to find one. Therefore, we may try to find one under some special conditions. (3) We may suppose that μ is a function of x only, i.e., μ (x) , then μ

μ= N

M N

dx

d (^) y x

This happens to have solutions if (My – Nx )/N is a function of x only. (4) We may suppose that μ is a function of y only, i.e., μ (y) , then

2 dx^1 x

u u

• −^ du=

we obtain 2 ln x − u −^1 − lnu =c x = 2c 1 ye1/2xy where ec^ was replaced by c 1.

[Reduction to Separation of Variables] 一階微分方程，加入適當的條件即可變為 Separable A DE of the form f(Ax By C) dx

dy (^) = + +

can always be reduced to an equation with separable variables by means of the substitution u = Ax + By + C, B≠0.

Example: Solve dy/dx = (-5x + y) 2 - 4. Solution: Define u = -5x + y and transform the equation into du dx

= u^2 − 9

which is separable. The family of solutions is

u ce ce

x = (^) x

6 6

( ) (^) and the singular solution is u = -

so that

y ce ce

x

x = (^) x

6 6

( ) (^) and y = 5x - 3

Homogeneous Function If f(tx, ty) = t nf(x, y) for some real number n, then f(x,y) is said to be a homogeneous function of degree n.

Example: f x y( , ) = x 3 +y^3 homogeneous of degree 3/

f x y( , ) = x 3 −y not homogeneous

Let t = 1/x or 1/y, we can get f(x, y) = xn^ f(1, y/x) or yn^ f(x/y, 1). If an equation in the differential form M(x, y)dx + N(x, y)dy = 0 has the property that both of M(x, y) and N(x, y) are homogeneous we say that it has homogeneous coefficients or is a homogeneous equation.

Method of Solution : Let y = ux or x = vy to separate variables. Since u or v is an independent variable, dy = udx + xdu if we take y = ux as an example. The above equation becomes M(x, ux)dx + N(x, ux)(udx + xdu) = 0 xn^ M(1, u)dx + xn^ N(1, u)(udx + xdu) = 0 which gives dx x

N u du M u uN u

Then we can integrate the equation to solve it.

Example: Solve ( 2 xy−y)dx −xdy= 0 Solution: Let y = ux du u u

dx 2 2 x

1 2^0

/ +^ =

If t = u1/2, we can simplify it as dt t

dx − x

The solution is ln t − 1 + ln x =lnc. That is xy − x =c Note: y = 0 or x = 0 is a singular solution.

Another form of the homogeneous differential equation: dy dx

M x y N x y

M y x N y x

= − (^ ,^ )= −

That is,

⎟ ⎠

x

F y dx

dy

is also a homogeneous differential equation.

Example: Solve x dy dx

= y + xe y x/^ subject to y(1) = 1.

Solution: dy dx

y x

= + e y x/

Let y = ux, then the equation becomes

u x du dx

• = u +e u

e du dx x

− u =

Hence the solution is − e −u^ + c =ln x or − e −y x/^ + c =lnx Since y = 1 when x = 1, we get c = e-1. Therefore, the solution of the initial-value problem is e −^1 − e −y x/^ =lnx

Bernoulli's Equation dy dx

• P x y( ) =f x y( ) n

where n is any real number. For n =0 and n = 1, the equation is linear in y. Now for y ≠ 0, the equation can be written as

y dy dx

−n (^) + P x y( ) 1 − n=f x( ).

x yn (x) yo x (^) of(t,yn 1 (t))dt

The repetitive use of the above equation is known as Picard's method of iteration. As n approaches to infinity, the solution becomes exact, i.e., y(x) = lim ( ) n n

y x →∞

Example: Solve y' = y - 1, y(0) = 2 Solution: Let yo(x) = 2 and we obtain

y (x) 2 1 dt 2 x

x

1 =^ +∫ 0 = +

y (x) 2 x( 1 t)dt 2 x x^2

2 =^ +∫ 0 + = + +

y x x x^ x^ x n

x n k

n k

k

n ( ) !!

=

2 3

0 As n approaches infinity, we obtain y = 1 + ex^ which is an exact solution of the given initial-value problem.

微分數值遞迴法 :Euler's Method For an IVP dy/dx = f(x, y) subject to y(x 0 ) = y 0 we may choose a step size h to be reasonable small and use a local linear approximation ( linearization ) to calculate y(x) at x. That is xn = x 0 + nh yn+1 = yn + f(xn, yn)h

Absolute Error : |True value – Approximation |⎢ (Percentage) Relative Error : Absolute error/ | True value | x 100%

Remark:

1. Truncation Errors for Euler's Method Taylor's expansion y x y a y^ a^ x a y^ a^ x a y^ a k

x a y^ c k

x a

k k k k ( ) ( ) ' (^ ) !

( ) ''(^ )

( ) (^ )

( ) ( ) = + − + − + + − +

1 2 1

2 L^11

where c is some constant between a and x. In the Euler method, y x( (^) n ) y x( (^) n ) hf ( x (^) n , y (^) n) h^ y c !

+ 1 =^ +^ + "( )

2 2

The truncation error is

Δ = h^ y c

2 2!

"( ) where xn < c < xn+

Unfortunately, the value of c is usually unknown. If |y"(x)| has an upper bound on the range we are interested in, then

Δ ≤ h^ M

2 2!

where M (^) x c x y x n n

max "( ) 1 In the above analysis, we assumed that the value of yn was exact in the calculation of yn+1, but it is not because it contains local truncation error from the previous step. The total error in yn+1 is an accumulation of the errors in each of the previous steps. This total error is called the global truncation error. In general it can be shown that if a method for the numerical solution of a differential equation has a local truncation error O(hα+1), then the gobal truncation error is O(hα).

2. Improved Euler Method (Heun's formula)

y (^) n + 1 = y (^) n+ h f^ x^ n^ y^ n^ +f^ x^ n^ +^1 yn+^1 2

where y (^) n* +^1 = y (^) n +hf x( (^) n , yn)

3. Truncation Errors for the Improved Euler's Method The local truncation error for the improved Euler's method is O(h^3 ) while the global truncation error is O(h^2 ).

2.7 Supplement – 其它型式 (not normal form)

[ 聯立方程組 ]

一個低階的 DE 與另一高階方程式聯立。如果低階的解並不完全都是高階

DE 的解，則可對低階進行逐次微方，直到與高階者同階數，然後依照兩同階方程

則只能降至與低階成比例。因此，若低階是二階 DE，則問題實際上有可能是求解

Example: 求解聯立方程組

⎪

y 0 x

y" y'^1 x

y^1

y 0 2 x 1

y'^4 2 x 1

4 x

5 sin

y" ( 3 )

Solution 將第一式對 x 再微分得

y 0 ( 2 x 1 )

y'^8 ( 2 x 1 )

8 x

5 sin y" 2 x 1

4 x

5 sin y (^3 ) 2 2 = −

然後減去第二式並 normalization 得到

Q’-2PP’=-8ce -8x Q-P^2 =ce -8x (2P’) 2 =-64ce -8x c=-1，R=-e-2x，P=e -4x。

Note 1: Equations of Ricatti and Clairaut Ricatti's Equation dy dx

= P x( ) + Q x y( ) +R x y( ) 2

If y 1 is a particular solution of the above equation, we substitute y = u +y 1 into the equation to get a differential equation of u:

( Q(x) 2 y 1 R(x))u R(x)u^2 dx

du (^) − + =

This is a Bernoulli's equation with n = 2, and we can let w = u -1^ to get

( Q 2 yR) w R dx

dw

• • 1 =−

which is a linear equation.

Example: Solve dy dx

= 2 − 2 xy +y^2

Solution: Guess y = 2x is a particular solution. The equation of w is dw dx

• 2 xw= − 1

The integrating factor is exp(x 2 ), and the solution of w is

e w= − ∫ e dt+c

x^2 t^2

Hence, u is

e dt c

u e 2

2 t

x

A solution of y is u+2x. In many cases, a solution of a Ricatti's equation cannot be expressed in terms of elementary functions.

Clairaut's Equation y = xy' + f(y') Differentiating the equation with respect to x gives (x + f'(y'))y" = 0 and the solutions are y" = 0 or x + f'(y') = 0. For y" = 0, then y’ = c, and substituting it back into the original equation gives the solution y = cx + f(c) where c is an arbitrary constant. For x + f'(y') = 0, we can let y' = t, and find a solution in the parametric form x = -f'(t), y = f(t) - tf'(t) If we can find the parametric solution in the family solution, then we do have f(t) - tf'(t) = -cf'(t) + f(c) for all t Differentiating with t leads to (c - t)f"(t) = 0

The above equation is true if and only if f"(t) = 0 for all t. That is, if f"(t) = 0, the parametric solution is one of the family solution. Otherwise, the parametric solution is a sigular solution, which cannot be found from the family solution.

Example: Solve ( y)^2 2

y =xy+^1

Solution: f(t) = t 2 /2 and the solution is

y = cx + 1 c 2

2

The singular solution is

x = -t, y = − 1 t 2

2

That is,

y = − 1 x 2

2