CHAPTER 1 - FIRST ORDER DIFFERENTIAL EQUATIONS · FIRST ORDER DIFFERENTIAL EQUATIONS Differential Equations L.A. Ruivivar (DLSU-Manila) DIFEQUA Math Dept 1 / 41. Differential Equations

CHAPTER 1FIRST ORDER DIFFERENTIAL EQUATIONS

Differential Equations

L.A. Ruivivar (DLSU-Manila) DIFEQUA Math Dept 1 / 41


Definition: A differential equation is an equation that contains afunction and one or more of its derivatives. If the function has onlyone independent variable, then it is an ordinary differentialequation. Otherwise, it is a partial differential equation.The following are examples of differential equations:

(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)




(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)




(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)




(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)




(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)




(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)


Order and Degree

Definition: The order of a differential equation is the order of thehighest ordered derivative that appears in the given equation. Thedegree of a differential equation is the degree of the highestordered derivative treated as a variable.Examples:

(a)∂2u∂x2 +

∂2u∂y2 = 0 is of order 2 and degree 1

(b) (x2 + y2)dx − 2xydy = 0 is of order 1 and degree 1

(c)(

d3xdy3

)2

+ xdxdy− 4xy = 0 is of order 3 and degree 2

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)3

is of order 2 and degree 3


Order and Degree


(a)∂2u∂x2 +



(c)(

d3xdy3

)2


(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)3



Order and Degree


(a)∂2u∂x2 +



(c)(

d3xdy3

)2


(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)3



Order and Degree


(a)∂2u∂x2 +



(c)(

d3xdy3

)2


(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)3



Order and Degree


(a)∂2u∂x2 +



(c)(

d3xdy3

)2


(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)3



Order and Degree


(a)∂2u∂x2 +



(c)(

d3xdy3

)2


(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

)3



Solution of a Differential Equation

Definition: A solution of a differential equation is a functiondefined explicitly or implicitly by an equation that satisfies thegiven equation. The general solution represents all the possiblesolutions of the given equation, while a particular solution is anyone of the possible solutions of a given differential equation.Examples:

(a)dydx

= y , y = Cex where C is an arbitrary constant

(b)dydx

= 3ex , y = 3ex + C where C is an arbitrary constant

(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)

y − 3, (x + 1)2 + (y − 3)2 = c2, where c is an arbitrary

constant.

(e)d2ydt2 + k2y = 0, y = sin kt , where k is a constant




(a)dydx


(b)dydx


(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)


constant.





(a)dydx


(b)dydx


(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)


constant.





(a)dydx


(b)dydx


(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)


constant.





(a)dydx


(b)dydx


(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)


constant.





(a)dydx


(b)dydx


(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)


constant.



Existence and Uniqueness Theorem

The existence of a particular solution satisfying initial conditions ofthe form y(x0) = y0 is guaranteed by the following theorem:Existence and Uniqueness Theorem: Consider a first orderequation of the form

dydx

= f (x , y)

and let T be the rectangular region described by

T = { (x , y) ∈ R2 | |x−x0| ≤ a, |y−y0| ≤ b, a,b are positive constants }.

If f and fy are continuous in T , then there exists a positive numberh and a function y = y(x) such that(a) y = y(x) is a solution of the given equation satisfying y(x0) = y0;

and(b) y = y(x) is unique on the interval |x − x0| ≤ h.


Existence and Uniqueness Theorem

The existence of a particular solution satisfying initial conditions ofthe form y(x0) = y0 is guaranteed by the following theorem:Existence and Uniqueness Theorem: Consider a first orderequation of the form

dydx

= f (x , y)

and let T be the rectangular region described by

T = { (x , y) ∈ R2 | |x−x0| ≤ a, |y−y0| ≤ b, a,b are positive constants }.

If f and fy are continuous in T , then there exists a positive numberh and a function y = y(x) such that(a) y = y(x) is a solution of the given equation satisfying y(x0) = y0;

and(b) y = y(x) is unique on the interval |x − x0| ≤ h.


Exercises

(1) Verify that the given function is a solution of the given differentialequation.(a) y (3) − 3y ′ + 2y = 0, y = e−2x

(b)d2ydt2 + k2y = 0, y = sin kt , where k is a constant

(2) Use antiderivatives to obtain a general or a particular solution toeach of the following equations:

(a)dydx

= x3 + 2x

(b)dydx

= 4 cos 2x

(c)dydx

= 3ex , y = 6 x = 0

(d)dydx

= 4y , y = 3 when x = 0


Separable Differential Equation

A first order ordinary differential equation can generally beexpressed in the form

Mdx + Ndy = 0

where M and N may be functions of two variables x and y . Forthis reason, we call this the general form of a first-order ordinarydifferential equation.If the differential equation can be manipulated algebraically totransform it into an equivalent form

A(x)dx + B(y)dy = 0

where A(x) is a function of x alone and B(y) is a function of yalone, then we say that the variables x and y are separable.


Separable Differential Equation

A first order ordinary differential equation can generally beexpressed in the form

Mdx + Ndy = 0

where M and N may be functions of two variables x and y . Forthis reason, we call this the general form of a first-order ordinarydifferential equation.If the differential equation can be manipulated algebraically totransform it into an equivalent form

A(x)dx + B(y)dy = 0

where A(x) is a function of x alone and B(y) is a function of yalone, then we say that the variables x and y are separable.


Examples

Show that the variables in the differential equation (1− x)y ′ = y2

are separable.Find the general solution of the equation xyy ′ = 1 + y2, if y = 3when x = 2.Find the particular solution of the equation y ′ = −2xy satisfyingy(0) = 1.


Examples




Examples




Exercises

(1) y ′ = y sec x(2) sin x sin y dx + cos x cos y dy = 0(3) y ln x ln y dx + dy = 0(4) (xy + x) dx = (x2y2 + x2 + y2 + 1) dy(5) dx = t(1 + t2) sec2 x dt(6) (e2x + 4)y ′ = y(7) x2 dx + y(x − 1) dy = 0(8) 2xyy ′ = 1 + y2, y = 3 when x = 2(9) 2y dx = 3x dy , y = −1 when x = 2

(10) y ′ = x exp(y − x2), y = 0 when x = 0


Homogeneous Functions

Definition: Let λ be a parameter. A function f (x , y) is said to behomogeneous of degree k , where k is a real number, if f satisfiesthe condition

f (λx , λy) = λk f (x , y)

Example: Show that the function f (x , y) = 4x2 − 3xy + y2 ishomogeneous and determine the degree of homogeneity.


Homogeneous Functions

Definition: Let λ be a parameter. A function f (x , y) is said to behomogeneous of degree k , where k is a real number, if f satisfiesthe condition

f (λx , λy) = λk f (x , y)

Example: Show that the function f (x , y) = 4x2 − 3xy + y2 ishomogeneous and determine the degree of homogeneity.


Two Theorems

Theorem: If M(x , y) and N(x , y) are both homogeneousfunctions of the same degree, then the function

g(x , y) =M(x , y)N(x , y)

is homogeneous of degree zero.

Theorem: If f (x , y) is homogeneous of degree zero, then f is afunction of y/x (or x/y ) alone.


Two Theorems

Theorem: If M(x , y) and N(x , y) are both homogeneousfunctions of the same degree, then the function

g(x , y) =M(x , y)N(x , y)

is homogeneous of degree zero.

Theorem: If f (x , y) is homogeneous of degree zero, then f is afunction of y/x (or x/y ) alone.


Equation With Homogeneous Coefficients

Definition: A first order differential equation of the formM dx + N dy = 0 is said to have homogeneous coefficients if Mand N are homogeneous functions of the same degree.Example: Show that the equation 3(3x2 + y2) dx − 2xy dy = 0has homogeneous coefficients.


Equation With Homogeneous Coefficients

Definition: A first order differential equation of the formM dx + N dy = 0 is said to have homogeneous coefficients if Mand N are homogeneous functions of the same degree.Example: Show that the equation 3(3x2 + y2) dx − 2xy dy = 0has homogeneous coefficients.


Solution of an Equation With HomogeneousCoefficients

Given: Mdx + Ndy = 0(a) Transform the equation into the equivalent form g(y/x) + dy

dx = 0,where g = M/N.

(b) Let y = vx and use this to convert the second equation into theform x dv + (v + g(v)) dx = 0 where the variables are separable.

(c) Solve using the method of separation of variables.(d) Use v = y/x to obtain a solution in terms of x and y .(e) Remark: You may also use the substitution

x = vy ,dx = v dy + y dv .





































Examples

(a) Solve the equation 3(3x2 + y2) dx − 2xy dy = 0(b) Show that the equation

y dx = (x +√

y2 − x2) dy

has homogeneous coefficients and find the general solution usingthe method discussed in this section.


Examples


y dx = (x +√

y2 − x2) dy



Examples


y dx = (x +√

y2 − x2) dy



Exercises

Find the general solution of each of the following equations:(1) 3xy dx + (x2 + y2) dy = 0(2) [x csc(y/x)− y ] dx + x dy = 0(3) xy dx − (x + 2y)2 dy = 0(4) (3x2 − 2xy + 3y2) dx = 4xy dy(5) [x − y tan−1(y/x)] dx + x tan−1(y/x) dy = 0


Exercises

Find the particular solution indicated for each of the followingequations:(1) (x − y) dx + (3x + y) dy = 0 if y = −2 when x = 3

(2) (y −√

x2 + y2)dx − x dy = 0 if y = 1 when x = 0

(3) xy dx + 2(x2 + 2y2) dy = 0 if y = 1 when x = 0(4) (3x2 − 2y2)y ′ = 2xy if y = −1 when x = 0


Exact Differential Equation

Definition: A differential equation of the formM(x , y) dx + N(x , y) dy = 0 is said to be exact if there exists adifferentiable function F (x , y) whose total differential is

dF = M(x , y) dx + N(x , y) dy = 0

Example: Let F (x , y) = x3y − 2x2y2 + 3xy . Then

dF =∂F∂x

dx +∂F∂y

dy

= (3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy

This shows that the first order differential equation

(3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy = 0

is exact, since its left-hand side is exactly dF .




dF = M(x , y) dx + N(x , y) dy = 0


dF =∂F∂x

dx +∂F∂y

dy

= (3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy


(3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy = 0





dF = M(x , y) dx + N(x , y) dy = 0


dF =∂F∂x

dx +∂F∂y

dy

= (3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy


(3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy = 0



Remark

Remark: If M(x , y) dx + N(x , y) dy = 0 is exact, then thereexists a function F such that dF = M(x , y) dx + N(x , y) dy . Thismeans that the general solution is of the form F (x , y) = C, whereC is an arbitrary constant.Example: Consider the simple differential equationy dx + x dy = 0.The left-hand side of this equation is easily seen to be the totaldifferential of the function F (x , y) = xy , so that our differentialequation takes the form dF = 0.Integration then yields the general solution xy = c.


Remark



Remark



Remark



Necessary and Sufficient Condition for Exactness

If M, N,∂M∂y

and∂N∂x

are all continuous functions of x and y , then

the differential equation M(x , y) dx + N(x , y) dy = 0 is exact ifand only if

∂M∂y

=∂N∂x

Example: Show that the equation(6x + y2) dx + y(2x − 3y) dy = 0 is exact.


Necessary and Sufficient Condition for Exactness

If M, N,∂M∂y

and∂N∂x

are all continuous functions of x and y , then

the differential equation M(x , y) dx + N(x , y) dy = 0 is exact ifand only if

∂M∂y

=∂N∂x

Example: Show that the equation(6x + y2) dx + y(2x − 3y) dy = 0 is exact.


Solution of an Exact Differential Equation

The general solution of an exact equation M dx + N dy = 0 is ofthe form F (x , y) = C.Since

∂F∂x

= M,∂F∂y

= N

, the function F can be obtained by integration and the relation

∂M∂y

=∂N∂x

Example: Show that the equation(6x + y2) dx + y(2x − 3y) dy = 0 is exact, and find the generalsolution.




∂F∂x

= M,∂F∂y

= N


∂M∂y

=∂N∂x





∂F∂x

= M,∂F∂y

= N


∂M∂y

=∂N∂x



Additional Examples

(1) Find a particular solution for the differential equation

2xy dx + (x2 + y2) dy = 0

if y = 1 when x = −1.(2) Find a function M(x , y) so that the differential equation

M(x , y) dx +

(xexy + 2xy +

1x

)dy = 0

is exact.


Additional Examples

(1) Find a particular solution for the differential equation

2xy dx + (x2 + y2) dy = 0

if y = 1 when x = −1.(2) Find a function M(x , y) so that the differential equation

M(x , y) dx +

(xexy + 2xy +

1x

)dy = 0

is exact.


Exercises

(1) Solve the following differential equations:(a) (2xy − 3x2) dx + (x2 + y) dy = 0(b) (y2 − 2xy + 6x) dx − (x2 − 2xy + 2) dy = 0(c) (1 + y2 + xy2) dx + (x2y + y + 2xy) dy = 0(d) (xy2 + x − 2y + 3) dx + x2y dy = 2(x + y) dy , y = 1 when x = 1(e) (3y(x2 − 1) dx + (x3 + 8y − 3x) dy = 0, y = 1 when x = 0

(2) Find a function N(x , y) such that the equation(y1/2x−1/2 +

xx2 + y

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

will be exact.


Linear Differential Equation

Definition: A linear differential equation of order one is an equationwhich is expressible in the form

A(x)dydx

+ B(x)y = C(x)

where A, B and C are all functions of the single variable x .Equations of this type may also be written in the form

dydx

+ P(x)y = Q(x).

This equation is linear with respect to y .Examples:(a) The equation y ′ + 3x2y = x2 is linear with P(x) = 3x2 and

Q(x) = x2.(b) Show that the equation 2(y − 4x2) dx + x dy = 0 is linear with

respect to y .




A(x)dydx

+ B(x)y = C(x)


dydx

+ P(x)y = Q(x).



respect to y .




A(x)dydx

+ B(x)y = C(x)


dydx

+ P(x)y = Q(x).



respect to y .




A(x)dydx

+ B(x)y = C(x)


dydx

+ P(x)y = Q(x).



respect to y .


Equations Linear With Respect to x

Remark: An equation which is not linear in y may be linear withrespect to x . In this case, the equation may be written in the form

dxdy

+ P(y)x = Q(y)

In general, an equation containing y dy can not be linear in y , andan equation containing x dx is not linear in x .Example: The equation (x + 4y2)dy + 2y dx = 0 can be written inthe equivalent form

dxdy

+

(1

2y

)x = −2y

which is linear with respect to x .




dxdy

+ P(y)x = Q(y)


dxdy

+

(1

2y

)x = −2y





dxdy

+ P(y)x = Q(y)


dxdy

+

(1

2y

)x = −2y



The Integrating Factor

Definition: A linear equation can be made exact by multiplying itwith an appropriate function v . This function is called anintegrating factor for the linear differential equation (with respect tothe variable y ) of the form v(x) > 0.When multiplied to the above equation, the integrating factor v(x)produces an exact equation of the form

v(x)(

dydx

+ P(x)y)

= Q(x)v(x)


The Integrating Factor

Definition: A linear equation can be made exact by multiplying itwith an appropriate function v . This function is called anintegrating factor for the linear differential equation (with respect tothe variable y ) of the form v(x) > 0.When multiplied to the above equation, the integrating factor v(x)produces an exact equation of the form

v(x)(

dydx

+ P(x)y)

= Q(x)v(x)


Finding the Integrating Factor

v dy + vPy dx = vQ dx(vPy − vQ) dx + v dy = 0

Since the equation is exact, with M = vPy − vQ and N = v , weshould have

∂M∂y

= vP =∂N∂x

= v ′(x)

from which we get P dx =dvv

where the variables have beenseparated.Antidifferentiation gives the integrating factor

v = exp( ∫

P dx)





∂M∂y

= vP =∂N∂x

= v ′(x)



v = exp( ∫

P dx)





∂M∂y

= vP =∂N∂x

= v ′(x)



v = exp( ∫

P dx)


Procedure for Solving a Linear Differential Equation

Write the equation in the formdydx

+ P(x)y = Q(x).

Solve for the integrating factor v = exp( ∫

P dx)

.

Multiply the equation in (1) by the integrating factor v to obtain anexact differential equation.Solve the resulting differential equation.Example: Show that the equation y ′ = x − 2y is linear, and findthe general solution.




+ P(x)y = Q(x).


P dx)

.





+ P(x)y = Q(x).


P dx)

.





+ P(x)y = Q(x).


P dx)

.





+ P(x)y = Q(x).


P dx)

.





+ P(x)y = Q(x).


P dx)

.



Remark

In the preceding example, the left hand side of the equation

e2x dydx

+ 2e2xy = xe2x is of the formddx

(e2xy), so that we get theequation

ddx

(e2xy) = xe2x

which is exact because the solution can be obtained immediatelyby integration.

More generally, if the linear equationdydx

+ Py = Q is multiplied by

the integrating factor v = exp(∫

P dx)

, we obtain

vdydx

+ vPy = Qv where the left hand side is ddx (vy) and the right

hand side is a function of x . This is exact and the solution isobtained by integrating both sides.


Remark


e2x dydx



ddx

(e2xy) = xe2x





P dx)

, we obtain

vdydx




Remark


e2x dydx



ddx

(e2xy) = xe2x





P dx)

, we obtain

vdydx




Equations Linear in x

In general, an equation which contains an expression of the formyk dy is not linear with respect to y . It may, however, be linear withrespect to x if it can be expressed in the form

dxdy

+ P(y)x = Q(y).

In this case, the integrating factor is a function of y obtained from

the expression v = exp( ∫

P(y) dy)

.

Show that the equation 2y(y2 − x) dy = dx is linear with respectto x , and find its general solution.




dxdy

+ P(y)x = Q(y).



P(y) dy)

.





dxdy

+ P(y)x = Q(y).



P(y) dy)

.



Exercises

i Find the general solution of each of the following equations:(a) (x5 + 3y) dx − x dy = 0

(b)dxdy

+1− 3y

yx = 3

(c) 2y dx = (x2 − 1)(dx − dy)(d) (y + 1) dx + (4x − y) dy = 0(e) (y − cos2 x) dx + cos x dy = 0(f) (y − x + xy cot x) dx + x dy = 0


Some Common Exact Differentials

Some exact differentials frequently occur within first order ordinarydifferential equations. Some of these exact differentials are thefollowing:(a) d(xy) = x dy + y dx

(b) d(

xy

)=

y dx − x dyy2

(c) d(y

x

)=

x dy − y dxx2

(d) d(

tan−1 yx

)=

x dy − y dxx2 + y2

These exact differentials are easily recognized because of theirdistinct forms.




(b) d(

xy

)=

y dx − x dyy2

(c) d(y

x

)=

x dy − y dxx2

(d) d(

tan−1 yx

)=






(b) d(

xy

)=

y dx − x dyy2

(c) d(y

x

)=

x dy − y dxx2

(d) d(

tan−1 yx

)=






(b) d(

xy

)=

y dx − x dyy2

(c) d(y

x

)=

x dy − y dxx2

(d) d(

tan−1 yx

)=






(b) d(

xy

)=

y dx − x dyy2

(c) d(y

x

)=

x dy − y dxx2

(d) d(

tan−1 yx

)=






(b) d(

xy

)=

y dx − x dyy2

(c) d(y

x

)=

x dy − y dxx2

(d) d(

tan−1 yx

)=




Solution by Inspection

We consider differential equations in which some terms can begrouped together such that one or more of some commondifferentials can be identified.Using an appropriate change of variables and/or integrating factor,a simple differential equation where the variables are separablecan be obtained.This method of solving a differential equation is called solution byinspection.Examples:(a) y(y3 − x) dx + x(y3 + x) dy = 0(b) y(y2 + 1) dx + x(y2 − 1)dy = 0(c) y(x3 − y5) dx − x(x3 + y5) dy = 0











Exercises

Solve the following differential equations by inspection.(a) y(2xy + 1) dx − x dy = 0(b) (x3y3 + 1) dx + x4y2 dy = 0(c) y(x4 − y2) dx + x(x4 + y2) dy = 0(d) y(x2y2 − 1) dx + x(x2y2 + 1) dy = 0(e) y(2x + y2) dx + x(y2 − x) dy = 0(f) y(x2 + y2 − 1) dx + x(x2 + y2 + 1) dy = 0

(g) y(x3exy − y) dx + x(y + x3exy ) dy = 0


The General Procedure for Determining theIntegrating Factor

Let u be an integrating factor for an equation of the formM dx + N dy = 0.Then the equation uM dx + uN dy = 0 is exact. This means thereexists a function F (x , y) such that

∂F∂x

= uM,∂F∂y

= uN

This means that u must satisfy the equation

u∂M∂y

+ M∂u∂y

= u∂N∂x

+ N∂u∂x

or, equivalently, the equation

u(∂M∂y− ∂N∂x

)= N

∂u∂x−M

∂u∂y




∂F∂x

= uM,∂F∂y

= uN


u∂M∂y

+ M∂u∂y

= u∂N∂x

+ N∂u∂x



)= N

∂u∂x−M

∂u∂y




∂F∂x

= uM,∂F∂y

= uN


u∂M∂y

+ M∂u∂y

= u∂N∂x

+ N∂u∂x



)= N

∂u∂x−M

∂u∂y


Special Cases

If1N

(∂M∂y− ∂N∂x

)= f (x) is a function of x alone, then the

integrating factor is of the form

u = exp∫

f (x) dx

If1M

(∂M∂y− ∂N∂x

)= g(y) is a function of y alone, then the


u = exp∫−g(y) dy


Special Cases

If1N

(∂M∂y− ∂N∂x

)= f (x) is a function of x alone, then the


u = exp∫

f (x) dx

If1M

(∂M∂y− ∂N∂x

)= g(y) is a function of y alone, then the


u = exp∫−g(y) dy


Examples

Solve the following differential equations:(a) 2y(x2 − y + x) dx + (x2 − 2y) dy = 0(b) y2 dx + (3xy + y2 − 1) dy = 0(c) (4xy + y2) dx + (2y − 2x2) dy = 0


Exercises

Find the appropriate integrating factors to solve the followingequations:(a) (x2 + y2 + 1) dx + (x2 − 2xy) dy = 0(b) (4xy + y2) dx + (2y − 2x2) dy = 0(c) (y2 + 2xy − 2y) dx − (2x + 2y) dy = 0(d) (2x2y − xy2 + y) dx + (x − y) dy = 0(e) y(2x − y + 1) dx + x(3x − 4y + 3) dy = 0(f) 2(2y2 + 5xy − 2y + 4) dx + (2x2 + 2xy − x) dy = 0

(g) (2y2 + 3xy − 2y + 6x) dx + x(x + 2y − 1) dy = 0


Coefficients Linear in x and y

We consider equations of the form(a1x + b1y + c1) dx + (a2x + b2y + c2) dy = 0.We will introduce a change of variables which will reduce theabove equation to a homogeneous equation.We consider the associated lines

a1x + b1y + c1 = 0a2x + b2y + c2 = 0

We consider the following cases:Case 1: The lines are intersecting.Case 2: The lines are parallel.




a1x + b1y + c1 = 0a2x + b2y + c2 = 0





a1x + b1y + c1 = 0a2x + b2y + c2 = 0





a1x + b1y + c1 = 0a2x + b2y + c2 = 0





a1x + b1y + c1 = 0a2x + b2y + c2 = 0





a1x + b1y + c1 = 0a2x + b2y + c2 = 0



Case 1: The lines intersect at a point (h, k)

Consider the change of variables x = u + h, y = v + k , so thatdx = du and dy = dv .Our given equation transforms into

(au + b1v) du + (a2u + b2v) dv = 0

which has homogeneous coefficients.Example: Solve the equation(x − 4y − 3) dx − (x − 6y − 5) dy = 0.




(au + b1v) du + (a2u + b2v) dv = 0





(au + b1v) du + (a2u + b2v) dv = 0



Case 2: The associated lines are parallel

In this case, the coefficients of the variables in the equations ofthe two lines are proportional, so there exists a constant k suchthat a2x + b2y = k(a1x + b1y),Letting u = a1x + b1y will transform the equation into anequivalent equation where the variables are separable.Example: Solve the equation(6x − 3y + 2) dx − (2x − y − 1) dy = 0.








Exercises

Solve the following equations:(a) (y − 2) dx − (x − y − 1) dy = 0(b) (x − 4y − 9) dx + (4x + y − 2) dy = 0(c) (x + y − 1) dx + (2x + 2y + 1) dy = 0(d) (x − 3y + 2) dx + 3(x + 3y − 4) dy = 0


CHAPTER 1 - FIRST ORDER DIFFERENTIAL EQUATIONS · FIRST ORDER DIFFERENTIAL EQUATIONS Differential Equations L.A. Ruivivar (DLSU-Manila) DIFEQUA Math Dept 1 / 41. Differential Equations

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