Cs130 day09

First-Order Differential Equations (continued ...)

CS 130: Mathematical Methods in Computer Science Ordinary Differential Equations Nestine Hope S. Hernandez Algorithms and Complexity Laboratory Department of Computer Science University of the Philippines, Diliman nshernandez@dcs.upd.edu.ph Day 9

First-Order Differential Equations (continued ...)

Ordinary Differential Equations

First-Order Differential Equations (continued ...)

Exact Equations Equations with Homogeneous Coefficients Coefficients Linear in Two Variables solved by substitution suggested by the equation integrating factors found by inspection

First-Order Differential Equations (continued ...)

Ordinary Differential Equations

First-Order Differential Equations (continued ...)

Exact Equations Equations with Homogeneous Coefficients Coefficients Linear in Two Variables solved by substitution suggested by the equation integrating factors found by inspection

First-Order Differential Equations (continued ...)

Exact Equations M (x, y)dx + N (x, y)dy = 0 is exact if and only if

∂M ∂y

=

∂N ∂x .

First-Order Differential Equations (continued ...)

Exact Equations M (x, y)dx + N (x, y)dy = 0 is exact if and only if

Example 1. (y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0

∂M ∂y

=

∂N ∂x .

First-Order Differential Equations (continued ...)

Exact Equations M (x, y)dx + N (x, y)dy = 0 is exact if and only if

Example 1. (y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0 2.

x y

y 0 + ln(xy) + 1 = 0

∂M ∂y

=

∂N ∂x .

First-Order Differential Equations (continued ...)

Exact Equations M (x, y)dx + N (x, y)dy = 0 is exact if and only if

Example 1. (y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0 2.

x y

y 0 + ln(xy) + 1 = 0

3. (3xy + y 2 ) dx + (x2 + xy) dy = 0

∂M ∂y

=

∂N ∂x .

First-Order Differential Equations (continued ...)

Exact Equations M (x, y)dx + N (x, y)dy = 0 is exact if and only if

∂M ∂y

=

∂N ∂x .

Example 1. (y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0 2.

x y

y 0 + ln(xy) + 1 = 0

3. (3xy + y 2 ) dx + (x2 + xy) dy = 0 ... (introduce an integrating factor)

First-Order Differential Equations (continued ...)

Exact Equations M (x, y)dx + N (x, y)dy = 0 is exact if and only if

∂M ∂y

=

∂N ∂x .

Example 1. (y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0 2.

x y

y 0 + ln(xy) + 1 = 0

3. (3xy + y 2 ) dx + (x2 + xy) dy = 0 ... (introduce an integrating factor) • If • If

1 N

“

1 M

“

∂M ∂y ∂M ∂y

− −

∂N ∂x

”

∂N ∂x

”

R

f (x)dx

−

R

= f (x), then e

= g(y), then e

is an integrating factor

g(y)dy

is an integrating factor

First-Order Differential Equations (continued ...)

Exact Equations M (x, y)dx + N (x, y)dy = 0 is exact if and only if

∂M ∂y

=

∂N ∂x .

Example 1. (y cos x + 2xey ) + (sin x + x2 ey − 1)y 0 = 0 2.

x y

y 0 + ln(xy) + 1 = 0

3. (3xy + y 2 ) dx + (x2 + xy) dy = 0 ... (introduce an integrating factor) • If • If

1 N

“

1 M

“

∂M ∂y ∂M ∂y

− −

∂N ∂x

”

∂N ∂x

”

R

f (x)dx

−

R

= f (x), then e

= g(y), then e

is an integrating factor

g(y)dy

4. y(x + y + 1) dx + x(x + 3y + 2) dy = 0

is an integrating factor

First-Order Differential Equations (continued ...)

Equations with Homogeneous Coefficients Polynomials in which all terms are of the same degree, such as x2 − 3xy + 4y 2 , x3 + y 3 , x4 y − 5x2 y 3 + 7y 5 , are called homogeneous polynomials.

First-Order Differential Equations (continued ...)

Equations with Homogeneous Coefficients Polynomials in which all terms are of the same degree, such as x2 − 3xy + 4y 2 , x3 + y 3 , x4 y − 5x2 y 3 + 7y 5 , are called homogeneous polynomials.

Suppose that the coefficients M and N in an equation of order one, M (x, y)dx + N (x, y)dy = 0, are both homogeneous and are of the same degree in x and y.

First-Order Differential Equations (continued ...)

Equations with Homogeneous Coefficients Polynomials in which all terms are of the same degree, such as x2 − 3xy + 4y 2 , x3 + y 3 , x4 y − 5x2 y 3 + 7y 5 , are called homogeneous polynomials.

Suppose that the coefficients M and N in an equation of order one, M (x, y)dx + N (x, y)dy = 0, are both homogeneous and are of the same degree in x and y.

Then the ratio M/N is a function of y/x alone.

First-Order Differential Equations (continued ...)

Equations with Homogeneous Coefficients Polynomials in which all terms are of the same degree, such as x2 − 3xy + 4y 2 , x3 + y 3 , x4 y − 5x2 y 3 + 7y 5 , are called homogeneous polynomials.

Suppose that the coefficients M and N in an equation of order one, M (x, y)dx + N (x, y)dy = 0, are both homogeneous and are of the same degree in x and y.

Then the ratio M/N is a function of y/x alone.

Example 1. (x2 − xy + y 2 ) dx − xy dy = 0

First-Order Differential Equations (continued ...)

Equations with Homogeneous Coefficients Polynomials in which all terms are of the same degree, such as x2 − 3xy + 4y 2 , x3 + y 3 , x4 y − 5x2 y 3 + 7y 5 , are called homogeneous polynomials.

Suppose that the coefficients M and N in an equation of order one, M (x, y)dx + N (x, y)dy = 0, are both homogeneous and are of the same degree in x and y.

Then the ratio M/N is a function of y/x alone.

Example 1. (x2 − xy + y 2 ) dx − xy dy = 0 {let y = vx}

First-Order Differential Equations (continued ...)

Equations with Homogeneous Coefficients Polynomials in which all terms are of the same degree, such as x2 − 3xy + 4y 2 , x3 + y 3 , x4 y − 5x2 y 3 + 7y 5 , are called homogeneous polynomials.

Suppose that the coefficients M and N in an equation of order one, M (x, y)dx + N (x, y)dy = 0, are both homogeneous and are of the same degree in x and y.

Then the ratio M/N is a function of y/x alone.

Example 1. (x2 − xy + y 2 ) dx − xy dy = 0 {let y = vx} 2. xy dx + (x2 + y 2 ) dy = 0

First-Order Differential Equations (continued ...)

Equations with Homogeneous Coefficients Polynomials in which all terms are of the same degree, such as x2 − 3xy + 4y 2 , x3 + y 3 , x4 y − 5x2 y 3 + 7y 5 , are called homogeneous polynomials.

Suppose that the coefficients M and N in an equation of order one, M (x, y)dx + N (x, y)dy = 0, are both homogeneous and are of the same degree in x and y.

Then the ratio M/N is a function of y/x alone.

Example 1. (x2 − xy + y 2 ) dx − xy dy = 0 {let y = vx} 2. xy dx + (x2 + y 2 ) dy = 0 {let x = vy}

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

1. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are parallel.

2. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 coincide.

3. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 intersect at only one point, say (h, k).

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

1. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are parallel.

2. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 coincide.

3. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 intersect at only one point, say (h, k).

Consider Case (3): use the substitution x = u + h and y = v + k.

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

1. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are parallel.

2. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 coincide.

3. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 intersect at only one point, say (h, k).

Consider Case (3): use the substitution x = u + h and y = v + k.

Example (x + 2y − 4) dx − (2x + y − 5) dy = 0

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

1. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are parallel.

2. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 coincide.

3. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 intersect at only one point, say (h, k).

Consider Case (3): use the substitution x = u + h and y = v + k.

Example (x + 2y − 4) dx − (2x + y − 5) dy = 0 how about for case (2)?

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

1. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are parallel.

2. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 coincide.

3. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 intersect at only one point, say (h, k).

Consider Case (3): use the substitution x = u + h and y = v + k.

Example (x + 2y − 4) dx − (2x + y − 5) dy = 0 how about for case (2)? - can be reduced to a separable equation

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

1. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are parallel.

2. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 coincide.

3. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 intersect at only one point, say (h, k).

Consider Case (3): use the substitution x = u + h and y = v + k.

Example (x + 2y − 4) dx − (2x + y − 5) dy = 0 how about for case (2)? - can be reduced to a separable equation how about for case (1)?

First-Order Differential Equations (continued ...)

Coefficients Linear in Two Variables Consider the equation (a1 x + b1 y + c1 )dx + (a2 x + b2 y + c2 )dy = 0 where the a’s,b’s,c’s are constants.

1. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are parallel.

2. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 coincide.

3. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 intersect at only one point, say (h, k).

Consider Case (3): use the substitution x = u + h and y = v + k.

Example (x + 2y − 4) dx − (2x + y − 5) dy = 0 how about for case (2)? - can be reduced to a separable equation how about for case (1)? - see next slide

First-Order Differential Equations (continued ...)

by substitution suggested by the equation

Solve the equation (x + 2y − 1) dx + 3(x + 2y) dy = 0

First-Order Differential Equations (continued ...)

by substitution suggested by the equation

Solve the equation (x + 2y − 1) dx + 3(x + 2y) dy = 0

Let v = x + 2y.

First-Order Differential Equations (continued ...)

by substitution suggested by the equation

Solve the equation (1 + 3x sin y) dx − x2 cos y dy = 0

First-Order Differential Equations (continued ...)

by substitution suggested by the equation

Solve the equation (1 + 3x sin y) dx − x2 cos y dy = 0

Let v = sin y.

First-Order Differential Equations (continued ...)

by substitution suggested by the equation

Solve the equation (3 tan x − 2 cos y) sec2 x dx + tan x sin y dy = 0

First-Order Differential Equations (continued ...)

by substitution suggested by the equation

Solve the equation (3 tan x − 2 cos y) sec2 x dx + tan x sin y dy = 0

Let v = tan x.

First-Order Differential Equations (continued ...)

integrating factors found by inspection Recall exact differentials such as d(xy) = xdy + ydx

ydx − xdy x = d y y2 y xdy − ydx d = x x2

y xdy − ydx d arctan = x x2 + y 2

First-Order Differential Equations (continued ...)

integrating factors found by inspection Recall exact differentials such as d(xy) = xdy + ydx

ydx − xdy x = d y y2 y xdy − ydx d = x x2

y xdy − ydx d arctan = x x2 + y 2

Example 1. y dx + (x + x3 y 2 ) dy = 0

First-Order Differential Equations (continued ...)

integrating factors found by inspection Recall exact differentials such as d(xy) = xdy + ydx

ydx − xdy x = d y y2 y xdy − ydx d = x x2

y xdy − ydx d arctan = x x2 + y 2

Example 1. y dx + (x + x3 y 2 ) dy = 0 2. y(x3 − y) dx − x(x3 + y) dy = 0

First-Order Differential Equations (continued ...)

integrating factors found by inspection Recall exact differentials such as d(xy) = xdy + ydx

ydx − xdy x = d y y2 y xdy − ydx d = x x2

y xdy − ydx d arctan = x x2 + y 2

Example 1. y dx + (x + x3 y 2 ) dy = 0 2. y(x3 − y) dx − x(x3 + y) dy = 0 3. 3x2 y dx + (y 4 − x3 ) dy = 0

First-Order Differential Equations (continued ...)

Questions?

See you next meeting!