Limit and Continuity - Webs

Chapter 2 First-order Differential Equations Chapter 1: Introduction to Differential Equations Overview Overview I. Separable variables II. Linear equations III. Exact Equations IV. Solution by substitutions I. Separable variables Learning Learning Objective Objective At the end of this section you should be able to identify and solve a separable DE. I. Separable variables

Definition Definition A first-order DE of the form dy g ( x ) h( y ) dx is said to be separable or to have separable variables. I. Separable variables Examples: Examples: dy 1) y 2 xe3 x 4 y dx Indeed, is separable dy y 2 xe3 x 4 y dx

y 2 xe 3 x e 4 y xe3 x y 2 e 4 y dy 2) y sin x dx is not separable Indeed, it cannot be put on the product form I. Separable variables Remark: Remark: By dividing by h( y ) , the separable DE, can be written in the form: p ( y )dy g ( x)dx where 1 p( y )

h( y ) dy g ( x ) h( y ) dx I. Separable variables Method Method of of solution: solution: From the form : p ( y )dy g ( x)dx we have to integrated both sides. p( y)dy g ( x)dx to obtain P( y ) G ( x) c One-parameter family of implicit or explicit solutions. I. Separable variables

Examples: Examples: dy 1) y 2e3 x dx y 2 dy e3 x dx 2 3x y dy e dx 1 1 3x e c y 3 y 1 1 3x e c

3 I. Separable variables Examples: Examples: 2) (1 x)dy ydx 0 (1 x)dy ydx dy dx y 1 x dy dx y 1 x ln y ln x 1 c2 ln y ln x 1 ln c1 ln y ln c1 x 1 y c1 x 1 y c1 x 1 y c x 1 I. Separable variables

Examples: Examples: dy x 3) dx y y 2 x 2 C y 4 3 y (4) 3 ydy xdx ydy xdx 1 2 1 2 y x c 2 2 y 2 x 2 2c .

32 4 2 C C 25 So, the solution of the IVP: y 2 x 2 25 I. Separable variables Examples: Examples: dy 4) y cos x e y sin 2 x dx y 0 0 sin 2. x ye dy dx cos x sin 2 x y ye dy cos x dx y

y ye dy 2 sin xdx u y dv e y dy du dy v e y y y y ye dy ye e dy y y y ye

dy ye e c y y ye dy y 1 e c 2 sin xdx 2 cos x I. Separable variables

Examples: Examples: y 2 cos x y 1 e c y 0 0 . 2 cos 0 0 1e 0 c c 1 Solution of IVP: 2 cos x 1 y 1e y I. Separable variables Exercise Exercise I:I: Solve the following DE by separation of variables: dy

sin 5 x dx dy 2 x 1 dx dy e 3 x 2 y dx . dy 2 y 3 dx 4 x 5 3x dx e dy 0 2 dx x dy 0 2 dx 1 2 y

dy y sin x e y 2 1 sin xdx 1 cos x dy, y 0 0 dy y 2 1 2 , y 2 2 dx x 1 II. Linear equations Learning Learning Objective Objective At the end of this section you should be able to identify and solve a linear DE. II. Linear Equations

Definition Definition A first-order DE of the form dy a 1 x a0 x y g ( x) dx is said to be linear equation in the dependent variable y. II. Linear Equations Remark Remark By dividing both sides by a 1 x , a linear equation can be written in the standard form: dy P x y f ( x) dx II. Linear Equations Definition

Definition dy a 1 x a0 x y g ( x ) dx Values of x that will make a 1 x 0 , are called singular points of the equation. Example: Example: x 1 dy x 1 3 y ln x dx is a singular point. II. Linear Equations Definition Definition dy P x y f ( x) dx P x dx

The function x e is defined as the integrating factor. Remark: Remark: P x dx ' x P x dx e P ( x) ( x ) ' II. Linear Equations Examples: Examples: Find the integrating factor for : 1) dy 3 y 0 dx P ( x) 3, f(x) 0 P x dx 3 dx

x e e e 3 x II. Linear Equations Examples: Examples: Find the integrating factor for : dy 4 y x 5e x dx x dy 2) x 4 y x 6e x dx 4 P ( x ) , f(x) x 5e x x 4 dx

x P x dx 1 4 ln x ln x 4 x e e e e 4 x Singular point : 0 II. Linear Equations Method Method of of solution solution

a1 x y' a0 x y g(x) 1) Write the standard form : 2) Find the integrating factor : y' P x y f ( x) P x dx x e II. Linear Equations Method Method of of solution solution 3) Multiply the Standard form by the integrating factor : Standard form : y ' P x y f ( x) (x)y' (x)P x y (x)f ( x) (x)y ' ' (x)y (x)f ( x )

(x)y ' (x)f ( x) II. Linear Equations Method Method of of solution 4) Integrate bothsolution sides of the last equation : (x)y ' (x)f ( x) ' (x)y dx (x)f ( x)dx (x)y (x)f ( x)dx 1 y

(x)f ( x)dx (x) II. Linear Equations Examples: Examples: Solve the following linear DE : 1) x dy 4 y x 5e x dx x dy 4 y x 6e x dx 4 P( x ) x 1 x 4 x

5 x f(x) x e II. Linear Equations Examples: Examples: 1) 4 5 x x ( x ) f ( x ) dx x x e

dx xe dx xe x x x e dx ( x 1 ) e c 1 1 x

y (x)f ( x ) dx ( x 1 ) e c 4 (x) x x 4 ( x 1)e x cx 4 II. Linear Equations

Examples: Examples: x y ' 2 y 0 x 9 dy xy 0 2) x 9 dx 2 P( x) x x2 9 f(x) 0

x 1 2x P x dx x 2 9 dx dx x e e 2 x 2 9 e e 2 ln x 9 1 2 x2 9 e

1 ln x 2 9 2 II. Linear Equations Examples: Examples: 2) c 1 1 y (x)f ( x)dx 0 dx (x) (x) (x) c x2 9

Valid on ( , ) \ 3 II. Linear Equations Exercise-II: Exercise-II: Solve the following linear DE : 1) x dy y x 2 sin x dx 4) x dy 3 x 1 y e 3 x dx dy 2) x 4 y x 3 x dx

5) xy y e x , y 1 2 dy 3) cos x sin x y 1 dx 6) x 1 dy y ln x, y 1 10 dx III. Exact equations Learning Learning Objective Objective At the end of this section you should be able to identify and solve an exact ODE. III. Exact equations Definition Definition Consider a function of two variables:

f x, y x f x, y y f x, y is the partial derivative of f regarding ( y is considered as a constant). x is the partial derivative of f regarding ( x is considered as a constant). y III. Exact equations Definition Definition Consider a function of two variables: its differential is :

f x, y f f df dx dy x y III. Exact equations Example: Example: f x, y x 2 y 4 4 f x, y 2xy x f x, y x 2 4 y 3 4 x 2 y 3 y df x, y 2 xy 4 dx 4 x 2 y 3dy III. Exact equations Examples: Examples:

2 x 5 y 2 x 5 y 5 y 2 x 5 y 2 x 2 x 5 y e e e e e e 2e 2e x x x 1 ln xy (ln x ln y ) ln x ln y x x x x x

III. Exact equations Definition Definition A differential expression : M x, y dx N x, y dy is an exact differential if it corresponds to the differential of some function f x, y that means f M x, y x f N x, y y III. Exact equations

Example Example 2 2 xydx x dy Indeed is an exact differential 2 f ( x, y ) x y f 2 xy x f 2 x y III. Exact equations Definition Definition A DE of the form: M x, y dx N x, y dy 0

is an exact equation if the left side is an exact differential. In that case, the DE is equivalent to An implicit solution will be f c df 0 III. Exact equations Example Example x 2 y 3 dx x 3 y 2 dy 0 Indeed is an exact equation f ( x, y ) 13 x 3 y 3 x y

1 3 x 3 y 3 x 2 y 3 1 3 x3 y 3 x 3 y 2 III. Exact equations Theorem Theorem A necessary and sufficient condition that be an exact differential is .

M N y x M x, y dx N x. y dy III. Exact equations Example Example x 2 y 3 dx x 3 y 2 dy 0 Indeed M ( x, y ) x 2 y 3 N ( x, y ) x 3 y 2 M 2 3 x y 3 x 2 y 2 y N 3 2 x y 3 x 2 y 2 x

exact equation III. Exact equations Method Method of of solution solution Example: 2 xy dx x 2 1 dy 0 Step 1: Check exactitude . M ( x, y ) 2 xy N ( x, y ) x 2 1 Exact DE

M N y x ? M 2 x y N 2 x x there exists f such that f 2 xy x III. Exact equations Method Method of of

solution solution Step 2: integrate regarding x f 2 xy x . f x dx 2 xydx f ( x, y ) x 2 y g ( y ) g ( y) constant for x but not for y III. Exact equations Method Method of of

solution solution Step 3: Differentiate regarding y f ( x, y ) x 2 y g ( y ) . f ( x, y ) x 2 g ' ( y ) y x 2 1 x 2 g ' ( y ) g ' ( y ) 1 III. Exact equations Method Method of of solution solution Step 4: Integrate regarding y g ' ( y ) 1 .

g ' ( y)dy dy g ( y ) y c III. Exact equations Method Method of of solution Step 5: Solutionsolution An implicit solution is f ( x, y ) 0 x 2 y y c 0 . An explicit solution is y C 2 x 1

defined when x 1 III. Exact equations Example: Example: Solve the following ODEs: 1) e 2 y y cos( xy) dx 2 xe2 y x cos( xy) 2 y dy 0 M ( x, y ) e 2 y y cos( xy ) N ( x, y ) 2 xe 2 y x cos( xy) 2 y M x, y 2e 2 y cos( xy ) xy sin( xy ) y N x, y 2e 2 y cos( xy) xy sin( xy) x

exact equation III. Exact equations Example: Example: f e 2 y y cos( xy ) x f ( x, y ) xe 2 y sin( xy) g ( y ) f 2 xe 2 y x cos( xy ) g ' ( y ) y N ( x, y ) 2 xe 2 y x cos( xy) 2 y 2 xe 2 y x cos( xy) g ' ( y ) g ' ( y ) 2 y III. Exact equations Example: Example: g ' ( y ) 2 y g ( y) y 2 c Implicit solution xe 2 y sin( xy ) y 2 c 0

III. Exact equations Example: Example: dy xy 2 cos x sin x 2) , y( 0 ) 2 2 dx y 1 x xy 2 cos x sin x dx y 1 x 2 dy 0 M ( x, y ) xy 2 cos x sin x N ( x, y ) y 1 x 2 M x, y 2 xy

y N x, y 2 xy x exact equation III. Exact equations Example: Example: f xy 2 cos x sin x x 1 2 2 1 2 f ( x, y ) x y sin x g ( y ) 2 2 f x 2 y g ' ( y) y N ( x, y ) y yx 2 x 2 y g ' ( y ) g ' ( y ) y

III. Exact equations Example: Example: g ' ( y ) y 2 y g ( y ) c 2 Family of implicit solutions 1 2 2 1 2 y2 x y sin x c 0 2 2 2 y( 0 ) 2 Solution of the IVP: c 2

1 2 2 1 2 y2 x y sin x 2 0 2 2 2 III. Exact equations Exercise-IIIa: Exercise-IIIa: Determine whether the given DE is exact. If it is, solve it . 3. 2 x 1 dx 3 y 7 dy 0 2 x y dx x 6 y dy 0 3 5 x 4 y dx 4 x 8 y dy 0 4. sin y

5. 2 xy 1. 2. y sin x dx cos x x cos y y dy 0 2 3dx 2 x 2 y 4dy 0 III. Exact equations Remark: Remark:make makeexact exactsome somenon-exact non-exact DEs DEs M ( x, y )dx N ( x, y )dy 0 where M y is non exact if

M y N x M N , Nx y x There are cases where the equation can be made exact. How? III. Exact equations Remark: Remark:make makeexact exactsome somenon-exact non-exact DE DE M ( x, y )dx N ( x, y )dy 0 M y N x compute

M y Nx N If the result is a function of the sole variable : x p ( x ) dx then find the integrating factor : x e M y Nx N p (x) III. Exact equations Remark: Remark: make make exact exact some some non-exact non-exact DE DE the DE by the integrating factor :

Multiply ( x) M ( x, y )dx ( x) N ( x, y )dy 0 ( x) M ( x, y ) ( x) M y y ( x) N ( x, y) ' ( x) N ( x) N x p( x) ( x) N ( x) N x x M y Nx N ( x) N x ( x)( M y N x N x ) ( x) N ( x) M y Now Exact! III. Exact equations Remark: Remark: make make exact exact some some non-exact non-exact DE

2 DE Example xy dx x dy 0 Example M x, y xy N x, y x 2 M y Nx N M y x M y N x N x 2 x

x ( 2 x) 3 p x 2 x x x e 3 dx x e 3 ln x 1 3 x III. Exact equations

Remark: Remark: make make exact exact some some non-exact non-exact DE 1 DE Example x Example xy x2 dx 3 dy 0 3 x x M x, y y x2

1 x N x, y x3 y 1 dx dy 0 2 x x My Nx 1 x2 1 x2 Exact!

III. Exact equations Remark: Remark: make make exact exact some some non-exact non-exact DE DE M ( x, y )dx N ( x, y )dy 0 M y N x If M y Nx N compute is NOT a function of the unique variable x M y Nx Nx M y N

p (x) M If the result is a function of the sole variable : then find the integrating factor : y Nx M y M p ( y ) dy y e p( y ) III. Exact equations Remark: Remark: make make exact exact some some non-exact non-exact

DE 2 2 DE Example xy dx 2 x 3 y 20 dy 0 Example M x, y xy N x, y 2 x 2 3 y 2 20 M y Nx N Nx M y M M y x N x 4 x

x 4x 3x p x 2 2 2 2 2 x 3 y 20 2 x 3 y 20 4 x x 3x 3 p y xy xy y M y N x III. Exact equations Remark: Remark: make make exact exact some some non-exact non-exact

DE DE Example Example y e y 3 3 dy y e 3 ln y e ln y 3

y3 xy dx y 3 2 x 2 3 y 2 20 dy 0 xy 4 dx 2 x 2 y 3 3 y 5 20 y 3 dy 0 M x, y xy 4 N 2 x 2 y 3 3 y 5 20 y 3 M y 4 xy 3 N x 4 xy 3 Exact! III. Exact equations Exercise-IIIb:

Exercise-IIIb: Solve the given D.E by finding an appropriate integrating factor. 1. 2 y 2. y x y 1 dx x 2 y dy 0 2 3x dx 2 xy dy 0 2 dy 0 3. 6 xy dx 4 y 9 x

4. 2 cos x dx 1 sin x dy 0 y IV. Solution by substitutions Learning Learning Objective Objective At the end of this section you should be able to solve Homogeneous and Bernoullis DEs. IV. Solution by substitutions Bernoulli Bernoulli Equation Equation Definition Definition :: A DE in the form a dy

n P x y f x y dx real number is said a Bernoulli equation. wheren is IV. Solution by substitutions Bernoulli Bernoulli Equation Equation Example Example :: dy 2 3 xy x y dx IV. Solution by substitutions Bernoulli Bernoulli Equation Equation

Substitution: u y1 n du du dy 1 n 1 dy n dy 1 n y 1 n y dx dy dx dx dx dy 1 n du y dx 1 n dx IV. Solution by substitutions Bernoulli Bernoulli Equation Equation dy

P x y f x y n dx y n du P x y f x y n 1 n dx du (1 n) y1 n P x (1 n) f x dx du (1 n) P x u (1 n) f x dx IV. Solution by substitutions Bernoulli Bernoulli Equation Equation Example Example :: n 2 dy x y x 2 y 2 dx 1 P( x)

x f ( x) x du (1 n) P x u (1 n) f x dx du 1 u x dx x IV. Solution by substitutions Bernoulli Bernoulli Equation Equation du 1 u x dx x Example Example :: P x

1 x 1 x e x dx e ln x e ln x 1 f x x 1 x x 1

1 1 2 x ( c x ) cx x u (x)f ( x)dx x ( x)dx (x) x IV. Solution by substitutions Bernoulli Bernoulli Equation Equation Example Example ::

u cx x 2 1 y u cx x 2 1 IV. Solution by substitutions Bernoulli Bernoulli Exercise ExerciseIVb IVb:: Solve the given Bernoulli equation by using an appropriate substitution. dy 1 1. x y 2 dx y dy 2. y e x y 2 dx dy

3. y xy 3 1 dx IV. Solution by substitutions Homogeneous Homogeneous DE DE Definition Definition :: If a function f has the property that f tx, ty t n f x, y , for some real number n , then f is said to be a homogeneous function of degree n . IV. Solution by substitutions Homogeneous Homogeneous DE DE Example1 Example1::

f x, y x 2 3 xy 5 y 2 f tx, ty (tx) 2 3(tx)(ty ) 5(ty ) 2 f tx, ty t 2 x 2 3t 2 xy 5t 2 y 2 f tx, ty t 2 ( x 2 3xy 5 y 2 ) f tx, ty t 2 f ( x, y ) f is a homogeneous function of degree 2 . IV. Solution by substitutions Homogeneous Homogeneous DE DE Example Example22:: f x, y 3 x 2 y 2 f tx, ty 3 (tx ) 2 (ty ) 2 f tx, ty 3 t 2 ( x 2 y 2 ) 3 f tx, ty t f tx, ty t

2 3 23 2 2 ( x y ) t 2 33 (x2 y 2 ) f ( x, y ) f is a homogeneous function of degree 2/3 . IV. Solution by substitutions Homogeneous Homogeneous DE

DE Example Example33:: f x, y x 3 y 3 1 f tx, ty (tx ) 3 (ty ) 3 1 f tx, ty t 3 x 3 t 3 y 3 1 We cant factorize by a power of t f is not homogeneous IV. Solution by substitutions Homogeneous Homogeneous DE DE Example Example44:: x f x, y 4 2y

tx f tx, ty 4 2ty x f tx, ty 4 f ( x, y ) 2y f is homogeneous of degree 0 IV. Solution by substitutions Homogeneous Homogeneous DE DE Definition Definition :: A DE of the form M x, y dx N x, y dy 0 is said to be homogeneous if both M and N are homogeneous functions of the same degree. IV. Solution by substitutions Homogeneous Homogeneous DE

DE Example Example:: 2 xydx y 2 dy 0 is homogeneous. Indeed 2 M ( x, y ) 2 xy and N ( x, y ) y are homogeneous of degree 2 IV. Solution by substitutions Homogeneous Homogeneous DE DE Method Methodof ofsolution: solution: x y dx xdy 0 1) Homogeneity : M ( x, y ) x y

and N ( x, y ) x are homogeneous of degree 1 2) Substitution y ux dy udx xdu x ux dx x( xdu udx) 0 IV. Solution by substitutions Homogeneous Homogeneous DE DE Method Methodof ofsolution: solution: x ux dx x( xdu udx) 0 xdx x 2 du 0 dx xdu 0

dx du x c ln u x IV. Solution by substitutions Homogeneous Homogeneous DE DE Method Methodof ofsolution: solution: c y ln u x x c y x ln x IV. Solution by substitutions

Homogeneous Homogeneous DE DE Method Methodof ofsolution: solution: 2) Substitution (2nd option) x y dx xdy 0 vy y (vdy ydv) vydy 0 v 1(vdy ydv) vdy 0 v 1 ydv v 2 dy 0 x vy dx vdy ydv IV. Solution by substitutions Homogeneous Homogeneous DE DE Method Methodof ofsolution:

solution: 2) Substitution v 1 ydv v 2 dy 0 (v 1) dy dv 2 v y dy 1 1 v v 2 dv y 1 ln v ln c ln y v IV. Solution by substitutions Homogeneous Homogeneous DE DE Method Methodof ofsolution: solution: 2) Substitution

x v y 1 ln v ln c ln y v y ln x ln y ln c ln y x y c ln c ln x ln x x c y x ln x IV. Solution by substitutions Homogeneous Homogeneous DE DE Exercise ExerciseIVa IVa::

Solve the given homogeneous equation by using an appropriate substitution. 1. xdx y 2 x dy 0 2. y 2 yx dx x 2 dy 0 dy y x 3. dx y x 4. ydx x xy dy 0 End Chapter 2

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