First-Order Differential Equations Part 1. First-Order Differential Equations Types: Variable...

First-Order Differential Equations

Part 1

First-Order Differential Equations

Types:

• Variable Separable

• Linear Equations

• Exact Equations

• Solvable by Substitutions

Variable Separable

• The simplest of all differential equations are those of the first order with separable variables.

• A first-order differential equation of the form

is said to be separable or to have separable variables.

)y(h)x(gdx

dy

Variable Separable

• To solve variable separable first-order differential equations, proceed as follows:

c)x(G)y(H

dx)x(gdy)y(p

dx)x(gdy)y(p

dx)x(g)y(h

dy

)y(h)x(gdx

dy

Let 1/h(y) = p(y)

H(y) and G(x) are antiderivatives of

p(y) and g(x), respectively.

ExampleSolve:

0dy)x1(dx)y1( 22 Solution:

c)xarctan()yarctan(

x1

dx

y1

dy

x1

dx

y1

dy

)x1)(y1(

10dy)x1(dx)y1(

0dy)x1(dx)y1(

22

22

2222

22

Alternate Solution:

xy1

yxc

)xy1(cxy

xccxyycx1

xcy

)xarctan(tancarctantan1

)xarctan(tancarctantany

carctan)xarctan(tan)yarctan(tan

carctan)xarctan()yarctan(

x1

dx

y1

dy

x1

dx

y1

dy

)x1)(y1(

10dy)x1(dx)y1(

0dy)x1(dx)y1(

22

22

2222

22

ExampleSolve:

2

3)0(y,0dyx1dxy1 22

Solution:

Cxarcsinyarcsin

dxx1

1dy

y1

1

0dxx1

1dy

y1

1

x1y1

10dyx1dxy1

22

22

22

22

Example

Initial Boundary Condition:2

3)0(y

Solving for C:

3C

C03

C)0arcsin(2

3arcsin

Cxarcsinyarcsin

Example

Initial Boundary Condition:2

3)0(y

Solving for C:

3xarcsinyarcsin

Cxarcsinyarcsin

Example

0dyx1dxy1 22

22

x12

3

2

xy)

2

3)(

1

x1()

2

1(xy

)3

sin()xcos(arcsin)3

cos()xsin(arcsiny

)3

xsin(arcsin)ysin(arcsin

3xarcsinyarcsin

Alternate Form of Final Answer:

Linear Equations

A first-order differential equation of the form

is said to be a linear equation in the dependent variable y.

When g(x) = 0, the linear equation is said to be homogeneous; otherwise, it is nonhomogeneous.

)x(gy)x(adx

dy)x(a 01

Linear Equations

We can divide both sides of the equation by the lead coefficient a1(x):

)x(fy)x(Pdx

dy

)x(a

)x(gy)x(a

)x(a

dx

dy

)x(gy)x(adx

dy)x(a

11

0

01

Linear Equations

Standard form of a linear 1st-order DE:

This differential equation has the property that its solution is the sum of two solutions:

y = yc + yp

)x(fy)x(Pdx

dy

Linear Equations

Now, yc is a solution of the associated homogeneous equation

and yp is a particular solution of the nonhomogeneous equation:

0y)x(Pdx

dy

)x(fy)x(Pdx

dy

Linear EquationsProof:

)x(f)x(f0

)x(fy)x(Pdx

dyy)x(P

dx

dy

)x(fy)x(Pdx

dyy)x(P

dx

dy

)x(fyy)x(Pyydx

d

)x(fy)x(Pydx

d

?

pp

cc

?

pp

cc

?

pcpc

Now, the previous homogeneous equation is also separable:

dx)x(P)ykln(

dx)x(Pklnyln

dx)x(Py

dy

0dx)x(Py

dy

0y)x(Pdx

dy

dx)x(P1

1c

dx)x(Pc

dx)x(P

dx)x(P

dx)x(P

eywhere

cyy

ceyyLet

cey

ek

1y

eyk

Now, let yp = u(x)y1:

)x(f)x(udx

dy

)x(f)x(udx

dy0)x(u

)x(f)x(udx

dyy)x(P

dx

dy)x(u

)x(fy)x(u)x(P)x(udx

dy

dx

dy)x(u

)x(fy)x(u)x(Py)x(udx

d

)x(fy)x(Pydx

d

1

1

111

111

11

pp

Separating variables and integrating gives

dx)x(feu

dxe

)x(fu

dx)x(y

)x(fu

dx)x(y

)x(fdu

)x(f)x(udx

dy

dx)x(P

dx)x(P

1

1

1

Now, going back to yp = uy1:

Now, going back to y = yc + yp:

dx)x(Pdx)x(P

p

1p

edx)x(fey

uyy

dx)x(feecey

yyy

dx)x(Pdx)x(Pdx)x(P

pc

Remember this special term called the “integrating factor”:

We can use the integrating factor as follows:

dx)x(Pe

dx)x(fecye

edx)x(feecey

dx)x(Pdx)x(P

dx)x(Pdx)x(Pdx)x(Pdx)x(P

General Solution

RECALL AGAIN

Standard form of a linear 1st-order DE:

)x(fy)x(Pdx

dy

Left-hand side of the standard form, to be used for deriving the solution

(see next slide)

A Simpler Derivation

This derivation hinges on the fact that the left hand side of the 1st-order differential equation (in standard form) can be recast into the form of the exact derivative of a product by multiplying both sides of the equation by a special function (x).

Left side of standard form of 1st order, linear DE

multiplied by .

dx

dyy

dx

dy)x(P

dx

dy

Left-hand side of the standard form of a 1st

order linear D.E.

Derivative of a product of two variables

ydx

dy)x(P

dx

dy

?


The derivation then involves solving the encircled elements as follows:

1cdx)x(P

1

e)x(

cdx)x(P||ln

Pdxd

Pdx

d


Continuing, we have:

dx)x(P2

dx)x(Pc

cdx)x(P

ec)x(

ee)x(

e)x(

1

1


Even though there are infinite choices of (x), all produce the same result. Hence, to simply, we let c2 = 1 and obtain the integrating factor.

dx)x(P

dx)x(P

dx)x(P2

e)x(

e)1()x(

ec)x(


• This is what we have derived so far. We multiply both sides of the standard form of the 1st-order equation by the integrating factor (x).

• We can then integrate both sides of the resulting equation and solve for y, resulting in a one-parameter family of solutions.


dx)x(Pdx)x(Pdx)x(P

dx)x(Pdx)x(P

dx)x(Pdx)x(P

dx)x(Pdx)x(P

dx)x(Pdx)x(Pdx)x(P

ecdx)x(feey

cdx)x(feye

dx)x(feyed

)x(feyedx

d

)x(fey)x(Pedx

dye

)x(fy)x(Pdx

dy

Solving a Linear, 1st-Order DE

1. Put the differential equation in standard form.

2. From the standard from, identify P(x) and then find the integrating factor

3. Multiply the standard form equation by the integrating factor.

dx)x(Pe

Solving a Linear, 1st-Order DE

4. The left hand side of the resulting equation is automatically the derivative of the integrating factor and y:

5. Integrate both sides of this last equation.

)x(feyedx

d dx)x(Pdx)x(P

ExampleFind the general solution of:

Solution:

Step 1:

0xdydx)y3x( 5

4

5

5

xyx

3

dx

dy

0x

y3x

dx

dy

xdx

10xdydx)y3x(

ExampleSolution:

Step 2:

Integrating Factor:

4xyx

3

dx

dy

P(x): include the negative

sign if present

3xln3dx)x

3(dx)x(P

xeee

ExampleSolution:

Step 3:

Step 4:

Recall:

4333 xxyx

3x

dx

dyx

dx

duv

dx

dvuuv

dx

d

xyx3dx

dyx 43

Example

Solution:

Step 3:

Step 4:

Thus:

xyx3dx

dyx 43

xyxdx

d 3

Derivative of y

Derivative of x –3

ExampleSolution:

Step 5:

35

35

32

3

23

3

3

3

Cxxy2

x'c2xy2

x2'c2

xyx

'c2

xyx

xdxyxd

xdxyxd

xyxdx

d

Solve the following variable separable differential equations.

1.

2.

3.

4.

5.

Exercises

ar,0when;dsinrdr)ra2( 322

0y,0xwhen);xyexp(x'y 2

0dye)1y(dxxy x3

ycosxcos'y 2

xdtsec)t1(tdx 22

Answers:

c)1t(x2sinx2)5

cx2sinx2|ytanysec|ln4)4

cy2

1y2)1x(e)3

a

rlnracosr)2

)e1ln(2lny)1

22

2x

222

x2

Solve the following 1st-order, linear differential equations.

1.

2.

3.

4.

5.

0xdydx)y3x( 5

xcotyxcsc'y 0xdydx)xcotxyxy(

)dydx)(1x(ydx2 2 3x2y'y)3x2(

Exercises

|3x2|ln3x2y2)5

)c|1x|ln2x)(1x(y)1x()4

cxsinxcosxxsinxy)3

xcosxsincy)2

cxxy2)1 35

Answers:

First-Order Differential Equations Part 1. First-Order Differential Equations Types: Variable...

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