First-Order Differential Equations Part 2: Exact & Homogeneous Types

First-Order Differential Equations

Part 2:

Exact & Homogeneous Types

Differential of a Function of Two Variables

• Recall that if z = f(x, y) is a function of two variables with continuous first derivatives in a region R of the xy-plane, then its differential is

• In the special case when f(x, y) = c, where c is a constant, then

dyy

fdx

x

fdz

0dyy

fdx

x

f


In other words, given a one-parameter family of functions f(x, y) = c, we can generate a first-order differential equation by computing the differential of both sides of the equality.


For example:

dy)y3x5(dx)y5x2(0

dyyxy5xy

dxyxy5xx

df

dyy

fdx

x

fdf

then

cyxy5x)y,x(f

2

3232

32


Of course, not every 1st-order differential equation written in differential form

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

corresponds to a differential of

f(x, y) = c

Exact Equation• A differential expression

M(x, y)dx + N(x, y)dy

is an exact differential in a region R of the xy-plane if it corresponds to the differential of some function f(x, y) defined in R.

• A first-order differential equation of the form


is said to be an exact equation if the expression on the left-hand side is an exact differential.

Illustration

x2y3dx + x3y2dy = 0 is an exact equation because the left-hand side is an exact differential of (1/3)x3y3:

dyyxdxyxyx3

1d 233233

M(x, y) = x2y3

N(x, y) = x3y2

Illustration

Notice also that for

M(x, y) = x2y3

N(x, y) = x3y2

we have

2223

2232

yx3yxxx

N

and

yx3yxyy

M

Criterion for an Exact Differential

Let M(x, y) and N(x, y) be continuous and have continuous first partial derivatives in a rectangular region R defined by a < x < b, c < y < d. Then a necessary and sufficient condition that M(x, y)dx + N(x, y)dy be an exact differential is

x

N

y

M

Solving Exact Equations

Step 1. Determine if


is exact by checking if

x

N

y

M


Step 2. If the equation is exact, find f(x, y) from either M(x, y) or N(x, y). If we use M(x, y), then assuming y is constant,

)y(gdx)y,x(M)y,x(f

dx)y,x(Mdf

dx)y,x(Mdf

)y,x(Mdx

df

)y,x(Mx

f

g(y) is the “constant” of integration.


Step 3. Differentiate the f(x, y) from step 2 with respect to the other variable y:

But the left side of the equation is N(x, y).

)y(gdx)y,x(Myy

f

)y(gdx)y,x(My

)y,x(N


Step 4. Solve for g’(y).

dx)y,x(My

)y,x(N)y('g

)y('gdx)y,x(My

)y,x(N

)y(gy

dx)y,x(My

)y,x(N

)y(gdx)y,x(My

)y,x(N


Step 5. Integrate the equation g’(y) with respect to y and substitute the resulting g(y) into

The implicit solution of the equation is f(x, y) = c.

)y(gdx)y,x(M)y,x(f

x6xy2ydx

dF

,ttanconsyHolding

x6xy2yx

F

,exactisequationtheSince

y2x2x

2xy2x

x

N

x2y2y

x6xy2y

y

M

0dy)2xy2x(dx)x6xy2y(

:Solution

0dy)2xy2x(dx)x6xy2y(Solve:Example

2

2

2

2

22

22

Be careful about the

signs

Step 1: Check if ∂M/∂y = ∂N/∂x

)y('gxxy2y

F

)y(gx3yxxyyy

F

,Then

)y(gx3yxxyF

dxx6xy2ydF

x6xy2ydx

dF

2

222

222

2

2

Step 2: Find F.

Step 3: Differentiate

F with respect to the other variable.

The g’(y) of the last equation on the previous page

is equal to the encircled quantity

0dyy

Fdx

x

F

0dyNdxM

0dy)2xy2x(dx)x6xy2y( 22

)y('gxxy2y

F 2

In other words,

y2)y(g

2)y('g

2xy2x)y('gxxy2

Ny

F

y

F

y

F

22

Step 4: Solve for

either g’(y) or g’(x), whichever

is applicable.

Hence,

cy2x3yxxy

assolutionthewriteformallyWe

c)y2(x3yxxy

c)y(gx3yxxy

c)y,x(F

222

222

222

Step 5: Integrate either g’(y) or g’(x),

whichever is applicable, then substitute into F.

x2yx3xcosydx

dF

,ttanconsyHolding

x2yx3xcosyx

F

,exactisequationtheSince

x3xcosy2x

ylnxxsiny2

x

N

x3xcosy2y

x2yx3xcosy

y

M

:Solution

e)0(y

0dy)ylnxxsiny2(dx)x2yx3xcosy(Solve:Example

22

22

23

222

322

)y('gxxsiny2y

F

y

)y(gxyxxsiny

y

F

,Then

)y(gxyxxsinyF

dxx2yx3xcosydF

x2yx3xcosydx

dF

,Then

3

232

232

22

22

cyylnyxyxxsinyF

)y(gxyxxsinyF

:FtogSubsitutin

cyylny)y(g

yln)y('g

ylnxxsiny2)y('gxxsiny2

,Hence

)y,x(Ny

F

,But

232

232

33

0yylnyxyxxsinyF

,Thus

0c

0ceelne0e00sineF

,e)0(ySince

0cyylnyxyxxsinyF

232

232

232

0dy]ey3x[dx]yeyx3[x2)6

1y,2xwhen

;0dy])xy1(xy[dx)xy1()5

0dy)1xy(xdx)xyxy()4

0d)1sinr2(cosrdr)cosr2(sin)3

0dy)ysecxx(dx)ytanxy2()2

0dy)yx2y2sinx2y2(cosdx)yx3y2(cos)1

:equationsexactfollowingtheSolve

22 x22x

2222

2

2

22

322

cx2yeyyx)6

5yx3xy5xy)5

cxxy2yx)4

0ccosrsinr)3

0cytanxyx)2

cy2sin2

1yxy2cosx)1

:Answers

3x32

34

222

22

2

23

2

Solution by Substitutions

We usually solve a differential equation by recognizing it as a certain kind of equation (say, separable, linear, exact) and then carrying out a procedure, consisting of equation-specific mathematical steps, that yields a solution of the equation.

Solution by Substitutions

• But it is not uncommon to be stumped by a differential equation because it does not fall into one of the classes of equations previously discussed.

• Now, there are three different kinds of first-order differential equations that are solvable by means of a substitution. – Homogeneous Equations– Bernoulli’s Equation– Reduction to Separation of Variables

Homogeneous Equations

• If a function f possess the property

f ( tx, ty ) = t f ( x, y )

for some real number , then f is said to be a homogeneous function of degree .

• For example,

f(x, y) = x3 + y3

is a homogeneous function of degree 3 since f(tx, ty) = (tx)3 + (ty)3 = t3(x3 + y3)


A first-order DE in differential form


is said to be homogeneous if both coefficient functions M and N are homogeneous equations of the same degree. In other words,

M(tx, ty) = tM(x, y)

N(tx, ty) = tN(x, y)


In addition, if M and N are homogeneous functions of degree , we can also write

M(x, y) = xM(1, u)

N(x, y) = xN(1, u)

M(x, y) = yM(v, 1)

N(x, y) = yN(v, 1)

where y = ux

where x = vy


Either of the substitutions

y = ux

or

x = vy

where u and v are new dependent variables, will reduce a homogeneous equation to a separable first-order differential equation.

0)u,1(uN)u,1(M

du)u,1(N

x

dx

0du)u,1(xNdx)u,1(uN)u,1(M

0xduudx)u,1(Ndx)u,1(M

0dy)u,1(Ndx)u,1(M

0dy)u,1(Nxdx)u,1(Mx

0dy)ux,x(Ndx)ux,x(M:Then

xduudxdy

uxy:Let

0dy)y,x(Ndx)y,x(M:Given

oofPr


In practice, for M(x,y)dx + N(x,y)dy = 0,

we use y = ux if N(x,y) is simpler than M(x,y),

or

we use x = vy if M(x,y) is simpler than N(x,y).

222 y2xy7x4'yx

ofsolutiongeneraltheFind:Example

Solution: Since the equation is homogeneous:

x

dx

4u6u2

du

dx)4u6u2(xdu

dxu2u74x)xduudx(x

dx)ux(2)ux(x7x4)xduudx(x

dx)y2xy7x4(dyx

,Then

xduudxdy

uxyLet

2

2

222

222

222

Solution:

x

dxdu

2u

1

1u

1

2

1

x

dxdu

2u

B

1u

A

2

1x

dx

2u3u

du

2

1x

dx

4u6u2

dux

dx

4u6u2

du

xduudxdy

uxyLet

2

2

2

Solution:

2

2

22

clnx|2

xy|

|1xy|

ln

clnxln|2x

y|ln|1

x

y|ln

cln2xln2|2x

y|ln|1

x

y|ln

clnxln|2x

y|ln|1

x

y|ln

2

1

clnxln|2u|ln|1u|ln2

1

x

dxdu

2u

1

1u

1

2

1

Solution:

2

2

2

2

clnx|

xx2y

|

|xxy

|ln

clnx|2

xy|

|1xy|

ln

|x2y|Cx|xy|

cx|x2y|

|xy|

cx|

xx2y

|

|xxy

|

2

22

2

2

0)1(y0dyxedx)yex(

.problemvalueinitialgiventheSolve:Examplex/yx/y

Solution: Since the equation is homogeneous:

0duxedxueue1

0duxeudxedxe)u(1

0)xduudx(xedxe)u(1x

0)xduudx(xedxe)ux(x

,Then

xduudxdy

uxyLet

uuu

uuu

uu

x/uxx/ux

|c|ln10

|c|lne|1|ln

:0)1(yFor1/0

|c|lne|x|ln

|c|lne|x|ln

0duex

dx

0duex

dx

0duxedxueue1

x/y

u

u

u

uuu

1e|x|ln

:Thusx/y

Note on the substitution

• Although either y = ux or x = vy can be used for every homogeneous differential equation, in practice we try x = vy whenever the function M(x, y) is simpler than N(x,y).

• Also it could happen that after using one substitution, we may encounter integrals that are difficult or impossible to evaluate in closed form; switching substitutions may result in an easier problem.

1y,1xwhen0dyxdx)x16xy7y()7

3y,1xwhen0dy)yx3(dx)y5x16()6

1y,0xwhenxy2'y)y2x3()5

1y,3xwhen0xdydx)yxy()4

0dyx

ytanxdx

x

ytanyx)3

0xdyydxx

ysinxdx)2

0xydydx)yx()1

:equationsogeneoushomfollowingtheSolve

222

22

22

11

2

22

yx|x|ln)x4y(5)7

x3y|x4y|ln)x4y()6

)1y(y2x)5

1y2x)4

xc

yxlnx

x

ytany2)3

0x

y2sinxy2

c

xlnx4)2

c)y2x(x)1

:Answers

22

2

42

221

4222

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First-Order Differential Equations Part 2: Exact & Homogeneous Types