first-order differential equations part 2: exact & homogeneous types
TRANSCRIPT
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
Differential of a Function of Two Variables
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.
Differential of a Function of Two Variables
For example:
dy)y3x5(dx)y5x2(0
dyyxy5xy
dxyxy5xx
df
dyy
fdx
x
fdf
then
cyxy5x)y,x(f
2
3232
32
Differential of a Function of Two Variables
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
M(x, y)dx + N(x, y)dy = 0
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
M(x, y)dx + N(x, y)dy = 0
is exact by checking if
x
N
y
M
Solving Exact Equations
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.
Solving Exact Equations
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
Solving Exact Equations
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
Solving Exact Equations
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)
Homogeneous Equations
A first-order DE in differential form
M(x, y)dx + N(x, y)dy = 0
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)
Homogeneous Equations
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
Homogeneous Equations
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
Homogeneous Equations
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