1.3 De nitions, Classi cation, and Terminologygmarple/Sept8.pdfdt2 = ty (b) dQ dt = 1 1+t Q + 2sint...

Agenda

Sections 1.3, 2.1

Reminders

Lab 1 prelab due 9/12 or 9/14

WebHW due 9/15

Office hours Tues, Thurs1-2 pm (5852 East Hall)

MathLab office hourSun 7-8 pm (MathLab)

(Gary Marple) September 8th, 2017 Math 216: Introduction to Differential Equations

§1.3 Definitions, Classification, and Terminology

Objectives

Be able to give the order of a differential equation

Be able to decide if a DE is linear or nonlinear

Determine if a linear DE is homogeneous

Be able to setup an initial value problem (IVP)


Definition: Order of a DE

The order of a DE is the order of the highest derivative,ordinary or partial, that appears in the equation.

Example

What is the order of each DE?

(a) ut = kuxx

(b) ydx + xdy = 0

(c) ∂2u∂t2 = c2 ∂2u

∂x2

(d) x ′x (4) + x sin y = cos y


Definition: Order of a DE

The order of a DE is the order of the highest derivative,ordinary or partial, that appears in the equation.

Example

What is the order of each DE?

(a) ut = kuxx 2

(b) ydx + xdy = 0 1

(c) ∂2u∂t2 = c2 ∂2u

∂x2 2

(d) x ′x (4) + x sin y = cos y 4


Definition: Linear DE

An nth order ordinary differential equation (ODE) is said to belinear if it can be written in the form

a0(t)y (n) + a1(t)y (n−1) + · · ·+ an(t)y = g(t).

The functions a0, a1, . . . , an are called the coefficients of theequation. An ODE that is not linear is called nonlinear.

Example

Which ODE’s are linear?

(a) (1 + y)d2ydt2 + t dy

dt+ y = et

(b) d2ydt2 + sin (t + y) = sin t

(c) d3ydt3 + t + (cos2 t)y = t3

(d) dydt

+ ty 2 = 0


Definition: Linear DE

An nth order ordinary differential equation (ODE) is said to belinear if it can be written in the form

a0(t)y (n) + a1(t)y (n−1) + · · ·+ an(t)y = g(t).

The functions a0, a1, . . . , an are called the coefficients of theequation. An ODE that is not linear is called nonlinear.

Example

Which ODE’s are linear?

(a) (1 + y)d2ydt2 + t dy

dt+ y = et

(b) d2ydt2 + sin (t + y) = sin t

(c) d3ydt3 + t + (cos2 t)y = t3

(d) dydt

+ ty 2 = 0


Definition: Homogeneous DE

A linear ODE of the form

a0(t)y (n) + a1(t)y (n−1) + · · ·+ an(t)y = g(t)

is called homogeneous if g(t) = 0 for all t. Otherwise, theequation is nonhomogeneous.

Example

Which DE’s are homogeneous?

(a) d2ydt2 = ty

(b) dQdt

= −(

11+t

)Q + 2 sin t

(c) ddx

[p(x)dy

dx

]= r(x)y

(d) y ′ + sin t = y


Definition: Homogeneous DE

A linear ODE of the form

a0(t)y (n) + a1(t)y (n−1) + · · ·+ an(t)y = g(t)

is called homogeneous if g(t) = 0 for all t. Otherwise, theequation is nonhomogeneous.

Example

Which DE’s are homogeneous?

(a) d2ydt2 = ty

(b) dQdt

= −(

11+t

)Q + 2 sin t

(c) ddx

[p(x)dy

dx

]= r(x)y

(d) y ′ + sin t = y


Definition: Initial Value Problem (IVP)

An initial value problem is a DE

y ′ = f (t, y)

along with a point (t0, y0) in the domain of f called the initialcondition.

Example

A cup of coffee has a temperature of 200◦F when freshlypoured and is left in a room at 70◦F. One minute later, thecoffee has cooled to 190◦F.

(a) Write an IVP that models the temperature of the coffee.

(b) How long will it take for the coffee to reach 170◦F?


(a). Newton’s law of heating and cooling says that

u′ = −k(u − T )

We know that T = 70◦F and the initial temperature of thecoffee is 200◦F, so the IVP is

u′ = −k(u − 70), u(0) = 200◦F

(b). We know that the general solution to the DE isu = T + ce−kt . We can find c using the initial condition.

200 = 70 + ce0 =⇒ c = 130

We can find k since we know the temperature after 1 minute.

190 = 70 + 130e−k·1 =⇒ k = ln (13/12)



u′ = −k(u − T )


u′ = −k(u − 70), u(0) = 200◦F


200 = 70 + ce0 =⇒ c = 130


190 = 70 + 130e−k·1 =⇒ k = ln (13/12)



u′ = −k(u − T )


u′ = −k(u − 70), u(0) = 200◦F


200 = 70 + ce0 =⇒ c = 130


190 = 70 + 130e−k·1 =⇒ k = ln (13/12)



u′ = −k(u − T )


u′ = −k(u − 70), u(0) = 200◦F


200 = 70 + ce0 =⇒ c = 130


190 = 70 + 130e−k·1 =⇒ k = ln (13/12)


(b). To find the time when the temperature reaches 170◦F, letu = 170 and solve for t.

170 = 70 + 130e−t ln (13/12)

170 = 70 + 130

(12

13

)t

ln(10/13) = t ln(12/13)

t =ln(10/13)

ln(12/13)≈ 3.278 min



170 = 70 + 130e−t ln (13/12)

170 = 70 + 130

(12

13

)t

ln(10/13) = t ln(12/13)

t =ln(10/13)

ln(12/13)≈ 3.278 min



170 = 70 + 130e−t ln (13/12)

170 = 70 + 130

(12

13

)t

ln(10/13) = t ln(12/13)

t =ln(10/13)

ln(12/13)≈ 3.278 min



170 = 70 + 130e−t ln (13/12)

170 = 70 + 130

(12

13

)t

ln(10/13) = t ln(12/13)

t =ln(10/13)

ln(12/13)≈ 3.278 min


§2.1 Separable Equations

Objectives

Be able to recognize a separable DE

Use separation of variables to solve a separable DE

Understand the difference between an explicit and animplicit solution

Solve word problems requiring separation of variables


Definition: Separable DE

A DE of the form y ′ = f (x , y) is separable if f (x , y) can bewritten as the product of a function that only depends on xtimes another function that only depends on y . That is,

dy

dx= f (x , y) = p(x)q(y).

Example

Which DE’s are separable?

(a) sin (2x)dx + 3ydy = 0

(b) x2y ′ = y − xy

(c)√x2 − y 2 + y = xy ′

(d) y ′ = 1 + x + y 2 + xy 2


Definition: Separable DE

A DE of the form y ′ = f (x , y) is separable if f (x , y) can bewritten as the product of a function that only depends on xtimes another function that only depends on y . That is,

dy

dx= f (x , y) = p(x)q(y).

Example

Which DE’s are separable?

(a) sin (2x)dx + 3ydy = 0

(b) x2y ′ = y − xy

(c)√x2 − y 2 + y = xy ′

(d) y ′ = 1 + x + y 2 + xy 2


Suppose we have a separable DE of the form

dy

dx= p(x)q(y).

If we assume that q(y) is nonzero for the y values we’reinterested in, we can divide both sides by q(y).

1

q(y)

dy

dx= p(x)

Integrating both sides with respect to x gives∫1

q(y)

dy

dxdx =

∫p(x)dx =⇒

∫1

q(y)dy =

∫p(x)dx .



dy

dx= p(x)q(y).


1

q(y)

dy

dx= p(x)


q(y)

dy

dxdx =

∫p(x)dx =⇒

∫1

q(y)dy =

∫p(x)dx .



dy

dx= p(x)q(y).


1

q(y)

dy

dx= p(x)


q(y)

dy

dxdx =

∫p(x)dx =⇒

∫1

q(y)dy =

∫p(x)dx .


Example

Show that the general solution to u′ = −k(u − T ) isu = T + ce−kt .

We begin with the DE

du

dt= −k(u − T ).

We separate the dependent and independent variables to get

du

u − T= −kdt.

(Note that the integration will be easier if the −k term stayson the right side.)


Example



du

dt= −k(u − T ).


du

u − T= −kdt.



Example



du

dt= −k(u − T ).


du

u − T= −kdt.



Integrating both sides gives∫du

u − T= −

∫kdt.

Let w = u − T , so dw = du and the integral becomes∫dw

w= −

∫kdt.

Evaluating the integrals and subbing u − T in for w gives

ln |u − T | = −kt + c ,

where c is our constant of integration. Exponentiating bothsides gives

u − T = e−kt+c .



u − T= −

∫kdt.


w= −

∫kdt.


ln |u − T | = −kt + c ,


u − T = e−kt+c .



u − T= −

∫kdt.


w= −

∫kdt.


ln |u − T | = −kt + c ,


u − T = e−kt+c .



u − T= −

∫kdt.


w= −

∫kdt.


ln |u − T | = −kt + c ,


u − T = ±e−kt+c .


Using our rules for exponents, we get

u = T ± ece−kt .

Since ±ec could be any arbitrary number (what about 0?), wecan say it’s just some arbitrary constant and replace it with c .That is,

u = T + ce−kt


Using our rules for exponents, we get

u = T ± ece−kt .

Since ±ec could be any arbitrary number (what about 0?), wecan say it’s just some arbitrary constant and replace it with c .That is,

u = T + ce−kt


Definition: Explicit and Implicit Solutions

An explicit solution is a solution where the dependent variabley is isolated on one side. For example, y = −x/2. An implicitsolution is a solution where the dependent variable is notisolated. For example, sin (x + ey ) = 3y .

Example

y 2 − 2y = x3 + 2x2 + 2x + c is an implicit expression for thegeneral solution of the DE

dy

dx=

3x2 + 4x + 2

2(y − 1).

Is it possible to find an explicit expression for the solution thatsatisfies y(0) = −1? If so, what is it?


Yes, y can be isolated because the general solution is aquadratic equation in terms y , and any quadratic equation canbe solved using the quadratic formula. To obtain an explicitexpression, let’s first find c by plugging in the initial condition.

(−1)2 − 2(−1) = 03 + 2 · 02 + 2 · 0 + c =⇒ c = 3

The implicit expression that satisfies the initial condition isthen

y 2 − 2y = x3 + 2x2 + 2x + 3.

To simplify the expression, let h(x) = x3 + 2x2 + 2x + 3, sowe have

y 2 − 2y − h(x) = 0.



(−1)2 − 2(−1) = 03 + 2 · 02 + 2 · 0 + c =⇒ c = 3


y 2 − 2y = x3 + 2x2 + 2x + 3.


y 2 − 2y − h(x) = 0.



(−1)2 − 2(−1) = 03 + 2 · 02 + 2 · 0 + c =⇒ c = 3


y 2 − 2y = x3 + 2x2 + 2x + 3.


y 2 − 2y − h(x) = 0.



(−1)2 − 2(−1) = 03 + 2 · 02 + 2 · 0 + c =⇒ c = 3


y 2 − 2y = x3 + 2x2 + 2x + 3.


y 2 − 2y − h(x) = 0.


The quadratic equation gives us

y =2±√

4 + 4h(x)

2=⇒ y = 1±

√x3 + 2x2 + 2x + 4

To satisfy the initial condition, we must have

y = 1−√x3 + 2x2 + 2x + 4


The quadratic equation gives us

y =2±√

4 + 4h(x)

2=⇒ y = 1±

√x3 + 2x2 + 2x + 4

To satisfy the initial condition, we must have

y = 1−√x3 + 2x2 + 2x + 4


Example

Water leaks out of a cylindrical bucket at a rate proportionalto the square root of the water’s depth. If the water has aninitial depth of 1 meter and half of the water leaks out in 1minute, how long before the bucket is empty?

The DE being described is

dV

dt= −k

√h,

where V is the volume of water in the bucket and h is theheight of the water. Since V = Ah, where A is the area of thebottom of the bucket, we get that

dV

dt= A

dh

dt.


Example



dV

dt= −k

√h,


dV

dt= A

dh

dt.


Example



dV

dt= −k

√h,


dV

dt= A

dh

dt.


Plugging this into the DE gives

dh

dt= −B

√h,

where B = k/A. This is a separable DE, so we’ll separatevariables to get ∫

dh√h

= −∫

Bdt.

Integrating both sides gives

2√h = −Bt + c ,

where c is our constant of integration. To find c , we can plugin our initial condition.

2√

1 = −B · 0 + c =⇒ c = 2



dh

dt= −B

√h,


dh√h

= −∫

Bdt.


2√h = −Bt + c ,


2√

1 = −B · 0 + c =⇒ c = 2



dh

dt= −B

√h,


dh√h

= −∫

Bdt.


2√h = −Bt + c ,


2√

1 = −B · 0 + c =⇒ c = 2



dh

dt= −B

√h,


dh√h

= −∫

Bdt.


2√h = −Bt + c ,


2√

1 = −B · 0 + c =⇒ c = 2


Since we know that h(1) = 1/2, we can find B .

2

√1

2= −B · 1 + 2 =⇒ B = 2−

√2

When the bucket is empty, h = 0, so we can now solve for thetime.

2√

0 = −(2−√

2)t + 2 =⇒ t =2

2−√

2≈ 3.41 minutes


Since we know that h(1) = 1/2, we can find B .

2

√1

2= −B · 1 + 2 =⇒ B = 2−

√2

When the bucket is empty, h = 0, so we can now solve for thetime.

2√

0 = −(2−√

2)t + 2 =⇒ t =2

2−√

2≈ 3.41 minutes


Thanks!


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1.3 De nitions, Classi cation, and Terminologygmarple/Sept8.pdfdt2 = ty (b) dQ dt = 1 1+t Q + 2sint...

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