Differential Equations 5y6b64
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Chapter 5
Mathematics
CHAPTER 5 Differential Equations A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders.
Order of differential equation Order of differential equation is the order of the highest derivative appearing in it.
Degree of differential equation Degree of a differential equation is the degree of the highest derivative occurring in it, after expressing the equation free from radicals and fractions as far as derivatives are concerned.
Example i)
x
y
ii) Y = x iii) [
+
. / ]
iv)
+
1st order, 1st degree
x =5
1st order, 2nd degree (power of highest derivative) =C
2nd order, 2nd degree 2nd order, 1st degree
Example The differential equation (A) (B) (C) (D)
x
y
= sin hx is
1st order and linear 1st order and non-linear 2nd order and linear 2nd order and non-linear
Solution Option (C)
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Chapter 5
Mathematics
Differential Equations of first Order first Degree Equations of first order and first degree can be expressed in the form (x y y ) y (x y).
or
Following are the different ways of solving equations of first order and first degree: 1. 2. 3. 4.
1.
Variables separable Homogeneous equations Linear equations Exact equations
Variable separable f(x)dx + g(y)dy = 0 ∫ (x) x
∫ (y) y
is the solution
Example Solve the differential equation =
x
Solution =
(
y=
x ) x
= 3 2.
x
x
C
= 2(
x )
C
Homogenous Equation =
( (
) )
To solve a homogeneous equation, substitute y = Vx =V+x
Separate the variable V and x and integrate.
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Chapter 5
Mathematics
Example Solve the differential equation x y dx - (x
y ) dy = 0
Solution = . /
Put y = vx , v+x x
=v+x
= =
dv = ∫
∫
ln y =
∫ vx
C
. /
C
C
3 ln cy =. /
Equations Reducible to homogenous Equation The differential equation:
=
This is non- homogeneous but can be converted to homogeneous equation Case I: If
Substitute x = X + h
y=Y+k
(h and K are constants)
dx=dX dy=dY ( (
=
) )
Solve for h and k =0 =0 = THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750, [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 137
Chapter 5
Case II:
Mathematics
If = =
(say)
( (
) )
Substitute ax +by = t, so that, a+b
=
=
(
( =
) )=
(
)
a
Solve by variable separable method. Example ( (
=
) )
Solution Here a/ 2+3
= b/ , put 2x+3y = t =
(
) =2+
(
)
=
dt = dx 3.
Linear Equations The standard form of a linear equation of first order: + P(x) y = Q(x) , where P and Q are functions of x (x)
(x)y
(x)
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Chapter 5
Commonly known as “L
tz’
y( ∫
(y ∫
y
∫
P) = Q
)=Q ∫
”
∫
Integrating factor, I.F. = ∫
u t
Mathematics
∫
∫
(I F) x
C
y(I F )
∫
(I F) x
C
Example Solve the differential equation (1 + y ) dx = ( t
y
x) y
Solution = x = Q P, Q I. F. = x.
∫
function of y only
= =∫
dy
x.
= ∫t
x.
=t
x=t
B
y
y
d(t
) C
C
u ’ E u t + P y = Q y where, P & Q are functions of x only.
Divide by y y
y
=Q
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Chapter 5
Substitute,
y
(1-n) y y
Mathematics
=z
=
= z =Q (
) z = Q (1-n)
This is linear equation and can be solved easily
Example Solve the differential equation x
y=x y
Solution =x y y y
=x = z, -5y
=
=x z = -5 x ∫
I.F. =
∫
Z x
= ∫x
Z x
= 5 ∫x
y .
x
=
=
=
∫
=
=x
(-5x ) dx x C
Cx / x y = 1
4. Exact Differential Equations M (x, y) dx + N (x, y) dy = 0 The necessary and sufficient condition for the differential equations M dx +N dy = 0 to be exact is = THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750, [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 140
Chapter 5
Mathematics
Solution of exact differential equation x
∫
∫(t
t
t
x ) dy = C
Example Solve the differential equation (secx tan x tan y
) dx + secx se y dy = 0
Solution M = secx. tanx. tany
= secx tanx se y
N = sec x se y So, exact
= secx tan x se y
∫(
∫
xt
xt
y
Sec x tan y -
) x
y =0
=C
Equation Reducible the Exact Equation Integrating factor Sometimes an equation which is not exact may become so on multiplication by some function known as Integrating factor (I.F.). Illustration x dy – y dx = 0
not exact
Multiply by
dy -
Therefore,
,
dx = 0
∫( ) x
exact y
∫
C
0=C Rule 0: Finding by inspection 1. x dy + y dx = d (x y) 2. =d( ) 3.
= d [log (
4.
=-d( )
5.
= d [t
)]
(
)-
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Chapter 5
6.
=d[
(
Mathematics
)-
Rule 1: when M dx + N dy = 0 is homogenous in x and y and M x + N y Rule 2: If the equation
(x, y) y dx +
(x, y) x dy = 0 and M x – N y
Rule 3: If the M dx + N dy = 0 and (
) = f(x), then I.F. =
Rule 4: If the equation M dx + N dy = 0 and
(
0 then I.F. = 0 then I.F. =
∫ ( )
) = f(y) , then I.F. =
∫ ( )
Example (y xy ) dx – (x+x y) dy = 0 Solution M = y xy , N = (x+x y) Mx – Ny = 2xy I.F. = (
0
= y) x
solution log
.
x/ y = 0
exact
xy = C
Equations of first order and higher Degree As will occur in higher degrees, it is convenient to denote form f(x, y, P) = 0. Case I :
by P. Such equations are of the
Equations solvable for P --Where,
,
......
=0
are functions of x and y
[P - (x y)- [P - (x y)- - - - - - [P P = (x y) , P = (x y) -- - - - P =
(x y)- = 0 (x y)
F (x y ) = 0, F (x y ) = 0, - - - - F (x y ) = 0 F (x y ) . F (x y ) - - - - - F (x y ) = 0
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Chapter 5
Mathematics
Example Solve the differential equation
=
Solution Let P = dy/dx P
= (
) -1 = 0
(P + ) ( P - ) = 0 P+
P
=0
d(xy) = 0 ,
y dy – x dx = 0
xy = C
x
y =c
(x y – C) (x
y
)=0
Case II : Equations solvable for y y = f(x, P) - - - - --------------------------------------- - (1) Differentiating P =
= F ( x, P,
)
Let the solution be g (x, p, c) = 0 - - - - - - (2) Eliminating P from (1) and (2). In case elimination of P is not possible, then we may solve (1) and (2) for x and y and obtain, x=F (
) ,y= F (
) as the required solution, where P is parameter
Example Solve the differential equation y=x+at
----(1)
Solution P=1+ dx=(
-----(2) )(
)
, dx =
x = C + a [log (P-1)
0
1dp
log (
)
t
]
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Chapter 5
Mathematics
This equation along with the given equation constitute the required solution in parameter form ‘ ’ wt t t Case III :
Equation solvable for x x = f(y , p) ---------(1) =
= g(y, p,
)
F( y, p, c) = 0 --------(2) Example Solve y = 2px + y Solution y = 2px + y
…………… ( )
x= Differentiating w. r. t. y .
=
/–(
2P = P
y
P
2y
y
P(1+ y (P + y
)
3y
y y
y
) (1+ y
)
(1+ y
y
=0 )=0
)=0
Py = c Eliminate P in equation (1) x =
(
)
y = 2cx + c3
Example The family of straight lines ing through origin is represented by differential equation (A) ydx + xdy = 0
(B) xdy – ydx = 0
(C) xdx + ydy = 0
(D) ydy –xdx=0
Solution: (B) a) d(yx) = 0 yx = c
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Chapter 5
Mathematics
b) d(y /x) = 0 = c y = cx c) y x = c, d) x y = c, Example The equations y – 2x = c represents the orthogonal trajectories of the family. (A) y = a
(B) x
y
(C) xy = 0
(D) x + 2y = a
Solution (D) For orthogonal trajectory Substitute,
=
=0 =0 x + 2y = a
Higher Order Differential Equation Linear Differential Equation with constant Coefficients -------
y=X
The equation can be written as (D
D
(D)
-----
)y = X
{Where, D =
}
=X
(D) = 0 is called A.E. Rules for finding complimentary function Case I
If all the roots of A.E. are real and different (D
) (D
) - - - - - - (D
)y=0
So, the solution is y=C
C
-- - - - -+ C
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Chapter 5
Illustration: D
D
Mathematics
=0
D = 2, 1 y=C Case II:
C
If two roots are equal i.e. (D
) (D
y = (C
)y=0
C x)
Similarly, if y = (C Illustration:
D
=
=
=
C x +C x ) D
=0
(D + 3) = 0 D = 3, 3 y =(C Case III:
C x)
If one pair of roots are imaginary i.e.
Illustration:
=
y=C
(
y=
(C
D
D
, )
D = -1,
x
)
C
x)
=0
2i
y=C
(C
=C Case IV:
(
C
D
=
C
x x
C
C
x) x
If two pairs of root are imaginary i.e. Repeated imaginary root y=
Illustration: D
,(C x
C )
x
, (C x
C )
x ]
=0 (D
D
) (D
D
)=0
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Chapter 5
D= 1 y=
i, 1
i
(C
x
C
Case V: If a pair of root is surd then
Mathematics
,C
x)
(C
√ ,
>0,
(√ )
C
x
C
x)
(√ )]
Rules for finding Particular Integral P.I. =
X=
( )
.X
Case I: When X = P.I. =
put D = a
( )
P.I. = x
( )
P.I. = x
( )
[f(a)
0]
put D = a
[ (a)
0, f(a) = 0]
put D = a
[f(a) = 0,
(a)
0,
(a)
0]
Case II: When X = sin (ax + b) or cos (ax +b) P.I. =
(
=x
=x
(
(
( x
)
)
)
)
( x
( x
put D =
)
)
, (-
put D =
, ’(
put D =
, ’’(
)
)
0]
(
)
’(
(D)
(D)
) = 0]
)
(
) = 0]
Case III : When X = x , m being positive integer P.I. =
( )
= (D) , Case IV : When X =
x = [ (D)-
( )
-
x
x = (D) [1
(D)
-x
V where V is function of x
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Chapter 5
P. I. = =
V
( )
(
Mathematics
)
V, then evaluate
(
)
V as in Case I, II & III
Case V : When X = x V(x) P.I. =
( )
( )
x V(x) = 0
( )
1
( )
V(x)
Case VI : When X is any other function of x P.I. =
( )
X
Factorize f(D) = (D
) (D
fractions and then apply,
) - - - - - - - (D X=
) and resolve
( )
into partial
x on each .
∫
Complete Solution Given differential equation: The complete solution is:
.......
=X
(C)
x
y = C.F + P.
Example Particular integral of (A)
=x
x
x
(B)
(D)
x
Solution (
)
{x
x
}=
(1 D
= [x =
{x
D - - - -) { x x
x
}
] + =
x , Option (C)
Example Solve the differential equation y =x
x
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Chapter 5
Mathematics
Solution (D
D
)y=x
x
(D - ) = 0 C.F. = (C P.I. =
(
C x) )
P.I.
x
x
=
x sin x
=
∫x
=
[∫ x
=
[ x
=
[ ∫x
=
[ x
=
[
x x x x
x-
x x
x
∫
x-
x x
∫ x
x x-
∫
x
x x-
x-
=
(x
x
x)
y = C.F. + P.I. Example Solve the differential equation (D
D
) = 2x
x
Solution D
D
C.F. = C P.I. =
= 0,
D = 1, 3
C
(
)
=2
(2x
(
)
=2 =
x)
(
x
(
)
)
3 x 3
x
3
(
)
(
)
x
x x
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Chapter 5
=
(1+ )
=
.
=
(
=
,
x
x /x
x (
)x x
-
)
x
(
C.S. = C.F. + P.I. = C Note:
Mathematics
x
x)
C
(
is omitted from P.I. since C
)
(
x
x)
is already in C.F.
In the above problem, try to find P.I. using different approach, P.I. =
(x –
) =
(
)
)y
x sin x
note the difference, (overall C.S. is unchanged)
Example Find the P.I. of (D Solution P.I. =
x sin x = 0x
1(
= 0x
sin x
)
1
(
)
=
[ x sin x – 2D
] =
[ x sin x + cos x ]
Example Find the P.I. of (D
D
)y
sin2 x
Solution P.I. =
sin 2x = =
(
)
=
(
)
x (
sin 2x )
Example Find the P.I. of (D
)y = sin 2x + cos 3x
Solution P.I. =
(sin 2x +cos 3x) = x
x
cos 3x =
cos 2x –
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Chapter 5
Mathematics
Example Find the complete solution of
x
x
x
]
Solution D (D+1) = 0 C.F.
=C
C
=C P.I.
=
C
(
)
[x
x
=
(1+D)
=
[ 1 D +D
=
[x
x
=
[x
-
=
-
[x
x
]
D
] [x ( x
)
]
x
Other approach P. I. = (1- D +D
D ) (x
x
)
=(
1+ D -D ) (x
x
)
=(
x
=
x–4
(x
x)
)
x
( x
)
Note: this is because a constant C is present in C.F. (overall C.S is unchanged) C.S. = C
C
x
C u y’ Eu x
E u t
x
Put =
x= .
(H
u
------ -
u t
)
y=X
t = ln x, then if D = =
Dy
x
= Dy
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Chapter 5
(
) =
.
= =
(D
(
Dy
Mathematics
).
D y
D)y
x
= D (D-1) y
x
= D (D-1)(D-2) y
After substituting these differentials, the Cauchy – Euler equation results in a linear equation with constant coefficients. Example Solve the given differential equation x
x
y
x
Solution substitute
x=
x
Dy
x
(D
D
(D
) = 0 , y = (C
P. =
D
t
(
)
y = (C
C
L
’ L
( x
= D(D-1)
)y =t
= ,
C t) D- t = t + 2
x) x
x E u t
)
( x
ax + b = (ax + b)
t = ln x
)
- - - - -- -
y=X
t = ln (ax + b) =aDy
( x
)
=
D(D-1)y
( x
)
=
D(D-1)(D-2)y
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Chapter 5
At u t tut t constant coefficients.
t
t
L
’
Mathematics
u t
ut
u t on with
Example ( x
( x
)
)
y =0
Solution 2x +3 =
, t = ln (2x+3)
4D(D -1)y – 2Dy – 20y = 0 ( D
D
D
)y = 0
D
=0
D = 3.1, -1 .6 y=C
+ C
= C (2x + )
C (2x + )
Partial Differential Equation z = f(x, y) =p,
L
=q,
E u t
= r,
t
= s,
t
(L
’
u t
)
=R Where P, Q, and R are functions of x, y and z (if P, Q, R are independent of z , its called linear otherwise quasi linear equation) Working rule 1.
Form subsidiary equation
=
=
2. 3.
Solve these simultaneous equations giving u = a and v = b as its solution. Write the complete solution as f (u ,v) = 0 , u = f(v)
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Chapter 5
Mathematics
Non- Linear Equations of 1st Order p and q will occur other than in the first degree are called non-linear partial differential equations of the first order. Form I: Equation containing p & q only f(p, q) = 0 Its complete solution z = ax +by + c where a and b connected by relation Since p = q=
f(a, b) = 0
= a,
=b
Form II: f(z, p, q ) = 0, not containing x & y Working rule: 1. 2. 3.
Assume, u = x + ay , substitute p = , and q = a Solve resulting ordinary differential equation (O.D.E.) in z and u Replace u by x + ay
Form III: f (x, p) = F (y, q) . not containing z and containing x and p can be separated from those containing y and q. Assume f (x, p) = F (y, q) = a (x) dz = z
Ψ(y) x
(x) x
z = ∫ (x) x Charpit’
t
y
x
y
Ψ(y) y ∫ Ψ(y) y
(G
t
)
f(x, y, z, p, q ) = 0 -----------------------(1) Since z depends on x and y
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Chapter 5
dz =
x
y
=
x
=
Mathematics
y --------------(2) =
=
=
Homogenous Linear Equation with constant coefficients ------ -
= f( x, y)
this is called homogenous because all
containing derivative is of same order. (D
D
(D D’)
D
-------
D ) = f(x, y) {where D =
D’
}
(x y)
Step I : Finding the C.F. 1. Write A.E. Where m = 2. CF = (y + CF = (y + roots. CF = (y + roots.
----= 0, . Roots are , ---- x) + (y + x) + - - - - - x) + x (y + x) + (y + x) + x (y +
x) +x
,
are distinct x) + - - - - - - ,
(y +
x) + - - - -
,
,
two equal
,
three equal
Step II : Finding P.I. P.I. =
(
)
f (x, y)
1. when f( ax +by ) =
ut , D
D’
-
2. when f( x, y) = sin (mx +ny), put (D 3. when f(x, y) = x y
P.I. =
(
)
DD
D
x y = [ (D D )-
4. when f(x, y) is any function of x and y. P.I. =
(
)
]
x y
f (x, y) , resolve
(
)
into
partial fractions considering (D D ) as a function of D alone and operate each partial fraction on f(x, y) ing that ( f(x, y) = ∫ (x x) x where c, is ) replaced by y + mx after integration. Example Solve the partial differential equation: r - 4s + 4t =
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Chapter 5
Mathematics
Solution = (D
DD
D )z = =0
C.F = P.I. =
m = 2, 2
(y +2x) + x (
(y +2x )
)
D = 2 , D = 1,
D – D’
So, P.I. =
∫
=
x
(
= ∫x
)
x
x
= Hence, C.S. is Z = (y +2x) + x (y +2x) + x
Example Solve the given differential equation y (2xy +
) dx =
dy
Solution (y
x
dy ) + 2x y dx = 0 x x=0
d( )
x x=0
x =0
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Chapter 5
Mathematics
Example Solve the given differential equation Solution (2D
DD
D )z=0 =0
√
m=
=
= 2,
Solution is z = (y 2x) +
(y
x)
Example Solve the given differential equation
= cos x cos 2y
Solution (D
DD ) z = cos x cos 2y = 0;
C.F. =
(y) + (y + x)
P.I. =
cos x cos 2y
= =
m = 0, 1
(x
,
(x
, =
y)
(
,
(x
y)-
y)
(x
) (
)
(x
y)-
y)- =
(x
y)
(x
y)
Example Solve the given differential equation
=2
x y
Solution D
D D =2
x y =0 (
) = 0,
m = 0, 0, 2
C.F. = (y) + x (y) + (y + 2x) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750, [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 157
Chapter 5
P.I. =
(2
x y)
=
(
=
(
)
(
=
(x y
x y
)
x y (
=
=
Mathematics
)(
)
)x y
)
(
)
= Example Solve the given differential equation cos (2x +y) Solution (D
DD
D )z
cos(2x +y) m = -3, 2
C.F. = (y-3x) + (y+2x) P.I.
=
(
)
(
=
(
=
)
( x
=
∫
=
∫
) (
y) =
( x
)
(
=
(
)(
)
)
( x
y)
̅̅̅̅̅̅̅̅x) x
x
=
x
=
x
( x
= ∫x
( x
̅̅̅̅̅̅̅̅x) x
= ∫x
( x
y)
) x
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Chapter 5 (
= x
)
(
)
̅̅̅̅̅̅̅̅)
(
̅̅̅̅̅̅̅̅ y x)
= sin ( x
Mathematics
= sin (2x + y) +
( x
y)
Approach 2 P.I
= = = =
(
)(
(
)
(
)
(
)
=( =
)
(
=
( x
)
. ∫ .
(
)
x) x
(
.
̅̅̅̅̅̅̅̅x) x
( x
∫
)
(
.
y)
)
.
(y
x)
x =
∫
(̅̅̅̅̅̅̅̅x
∫ (y
x) x
x)
Approach 3 P.I.
=
(
)(
( x
)
y)
=
( x
y)
= x
( x
y)
= x =x = =
( x ( (
)
=
)
( x ( x
y) D
( x
y)
y) x y)
Note the difference in answer of approach 1 & approach 2/3 (why?) To find order and degree of differential equation simplify all the decimal and fractional power to 3/2 or ½ to intiger.
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Chapter 5
Mathematics
Example [
. / ]
Square both side [
. / ]
.
/
Some equation of curves 1. y =mx +c
straight line
2. xy = c hyperbola 3. y.ax2 = or ay2 z x
parabola
4. x2 + y2 = a2 circle 5.
edlipse
6.
hyper bola
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Mathematics
CHAPTER 5 Differential Equations A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders.
Order of differential equation Order of differential equation is the order of the highest derivative appearing in it.
Degree of differential equation Degree of a differential equation is the degree of the highest derivative occurring in it, after expressing the equation free from radicals and fractions as far as derivatives are concerned.
Example i)
x
y
ii) Y = x iii) [
+
. / ]
iv)
+
1st order, 1st degree
x =5
1st order, 2nd degree (power of highest derivative) =C
2nd order, 2nd degree 2nd order, 1st degree
Example The differential equation (A) (B) (C) (D)
x
y
= sin hx is
1st order and linear 1st order and non-linear 2nd order and linear 2nd order and non-linear
Solution Option (C)
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Chapter 5
Mathematics
Differential Equations of first Order first Degree Equations of first order and first degree can be expressed in the form (x y y ) y (x y).
or
Following are the different ways of solving equations of first order and first degree: 1. 2. 3. 4.
1.
Variables separable Homogeneous equations Linear equations Exact equations
Variable separable f(x)dx + g(y)dy = 0 ∫ (x) x
∫ (y) y
is the solution
Example Solve the differential equation =
x
Solution =
(
y=
x ) x
= 3 2.
x
x
C
= 2(
x )
C
Homogenous Equation =
( (
) )
To solve a homogeneous equation, substitute y = Vx =V+x
Separate the variable V and x and integrate.
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Chapter 5
Mathematics
Example Solve the differential equation x y dx - (x
y ) dy = 0
Solution = . /
Put y = vx , v+x x
=v+x
= =
dv = ∫
∫
ln y =
∫ vx
C
. /
C
C
3 ln cy =. /
Equations Reducible to homogenous Equation The differential equation:
=
This is non- homogeneous but can be converted to homogeneous equation Case I: If
Substitute x = X + h
y=Y+k
(h and K are constants)
dx=dX dy=dY ( (
=
) )
Solve for h and k =0 =0 = THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750, [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 137
Chapter 5
Case II:
Mathematics
If = =
(say)
( (
) )
Substitute ax +by = t, so that, a+b
=
=
(
( =
) )=
(
)
a
Solve by variable separable method. Example ( (
=
) )
Solution Here a/ 2+3
= b/ , put 2x+3y = t =
(
) =2+
(
)
=
dt = dx 3.
Linear Equations The standard form of a linear equation of first order: + P(x) y = Q(x) , where P and Q are functions of x (x)
(x)y
(x)
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Chapter 5
Commonly known as “L
tz’
y( ∫
(y ∫
y
∫
P) = Q
)=Q ∫
”
∫
Integrating factor, I.F. = ∫
u t
Mathematics
∫
∫
(I F) x
C
y(I F )
∫
(I F) x
C
Example Solve the differential equation (1 + y ) dx = ( t
y
x) y
Solution = x = Q P, Q I. F. = x.
∫
function of y only
= =∫
dy
x.
= ∫t
x.
=t
x=t
B
y
y
d(t
) C
C
u ’ E u t + P y = Q y where, P & Q are functions of x only.
Divide by y y
y
=Q
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Chapter 5
Substitute,
y
(1-n) y y
Mathematics
=z
=
= z =Q (
) z = Q (1-n)
This is linear equation and can be solved easily
Example Solve the differential equation x
y=x y
Solution =x y y y
=x = z, -5y
=
=x z = -5 x ∫
I.F. =
∫
Z x
= ∫x
Z x
= 5 ∫x
y .
x
=
=
=
∫
=
=x
(-5x ) dx x C
Cx / x y = 1
4. Exact Differential Equations M (x, y) dx + N (x, y) dy = 0 The necessary and sufficient condition for the differential equations M dx +N dy = 0 to be exact is = THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750, [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 140
Chapter 5
Mathematics
Solution of exact differential equation x
∫
∫(t
t
t
x ) dy = C
Example Solve the differential equation (secx tan x tan y
) dx + secx se y dy = 0
Solution M = secx. tanx. tany
= secx tanx se y
N = sec x se y So, exact
= secx tan x se y
∫(
∫
xt
xt
y
Sec x tan y -
) x
y =0
=C
Equation Reducible the Exact Equation Integrating factor Sometimes an equation which is not exact may become so on multiplication by some function known as Integrating factor (I.F.). Illustration x dy – y dx = 0
not exact
Multiply by
dy -
Therefore,
,
dx = 0
∫( ) x
exact y
∫
C
0=C Rule 0: Finding by inspection 1. x dy + y dx = d (x y) 2. =d( ) 3.
= d [log (
4.
=-d( )
5.
= d [t
)]
(
)-
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Chapter 5
6.
=d[
(
Mathematics
)-
Rule 1: when M dx + N dy = 0 is homogenous in x and y and M x + N y Rule 2: If the equation
(x, y) y dx +
(x, y) x dy = 0 and M x – N y
Rule 3: If the M dx + N dy = 0 and (
) = f(x), then I.F. =
Rule 4: If the equation M dx + N dy = 0 and
(
0 then I.F. = 0 then I.F. =
∫ ( )
) = f(y) , then I.F. =
∫ ( )
Example (y xy ) dx – (x+x y) dy = 0 Solution M = y xy , N = (x+x y) Mx – Ny = 2xy I.F. = (
0
= y) x
solution log
.
x/ y = 0
exact
xy = C
Equations of first order and higher Degree As will occur in higher degrees, it is convenient to denote form f(x, y, P) = 0. Case I :
by P. Such equations are of the
Equations solvable for P --Where,
,
......
=0
are functions of x and y
[P - (x y)- [P - (x y)- - - - - - [P P = (x y) , P = (x y) -- - - - P =
(x y)- = 0 (x y)
F (x y ) = 0, F (x y ) = 0, - - - - F (x y ) = 0 F (x y ) . F (x y ) - - - - - F (x y ) = 0
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Chapter 5
Mathematics
Example Solve the differential equation
=
Solution Let P = dy/dx P
= (
) -1 = 0
(P + ) ( P - ) = 0 P+
P
=0
d(xy) = 0 ,
y dy – x dx = 0
xy = C
x
y =c
(x y – C) (x
y
)=0
Case II : Equations solvable for y y = f(x, P) - - - - --------------------------------------- - (1) Differentiating P =
= F ( x, P,
)
Let the solution be g (x, p, c) = 0 - - - - - - (2) Eliminating P from (1) and (2). In case elimination of P is not possible, then we may solve (1) and (2) for x and y and obtain, x=F (
) ,y= F (
) as the required solution, where P is parameter
Example Solve the differential equation y=x+at
----(1)
Solution P=1+ dx=(
-----(2) )(
)
, dx =
x = C + a [log (P-1)
0
1dp
log (
)
t
]
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Chapter 5
Mathematics
This equation along with the given equation constitute the required solution in parameter form ‘ ’ wt t t Case III :
Equation solvable for x x = f(y , p) ---------(1) =
= g(y, p,
)
F( y, p, c) = 0 --------(2) Example Solve y = 2px + y Solution y = 2px + y
…………… ( )
x= Differentiating w. r. t. y .
=
/–(
2P = P
y
P
2y
y
P(1+ y (P + y
)
3y
y y
y
) (1+ y
)
(1+ y
y
=0 )=0
)=0
Py = c Eliminate P in equation (1) x =
(
)
y = 2cx + c3
Example The family of straight lines ing through origin is represented by differential equation (A) ydx + xdy = 0
(B) xdy – ydx = 0
(C) xdx + ydy = 0
(D) ydy –xdx=0
Solution: (B) a) d(yx) = 0 yx = c
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Chapter 5
Mathematics
b) d(y /x) = 0 = c y = cx c) y x = c, d) x y = c, Example The equations y – 2x = c represents the orthogonal trajectories of the family. (A) y = a
(B) x
y
(C) xy = 0
(D) x + 2y = a
Solution (D) For orthogonal trajectory Substitute,
=
=0 =0 x + 2y = a
Higher Order Differential Equation Linear Differential Equation with constant Coefficients -------
y=X
The equation can be written as (D
D
(D)
-----
)y = X
{Where, D =
}
=X
(D) = 0 is called A.E. Rules for finding complimentary function Case I
If all the roots of A.E. are real and different (D
) (D
) - - - - - - (D
)y=0
So, the solution is y=C
C
-- - - - -+ C
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Chapter 5
Illustration: D
D
Mathematics
=0
D = 2, 1 y=C Case II:
C
If two roots are equal i.e. (D
) (D
y = (C
)y=0
C x)
Similarly, if y = (C Illustration:
D
=
=
=
C x +C x ) D
=0
(D + 3) = 0 D = 3, 3 y =(C Case III:
C x)
If one pair of roots are imaginary i.e.
Illustration:
=
y=C
(
y=
(C
D
D
, )
D = -1,
x
)
C
x)
=0
2i
y=C
(C
=C Case IV:
(
C
D
=
C
x x
C
C
x) x
If two pairs of root are imaginary i.e. Repeated imaginary root y=
Illustration: D
,(C x
C )
x
, (C x
C )
x ]
=0 (D
D
) (D
D
)=0
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Chapter 5
D= 1 y=
i, 1
i
(C
x
C
Case V: If a pair of root is surd then
Mathematics
,C
x)
(C
√ ,
>0,
(√ )
C
x
C
x)
(√ )]
Rules for finding Particular Integral P.I. =
X=
( )
.X
Case I: When X = P.I. =
put D = a
( )
P.I. = x
( )
P.I. = x
( )
[f(a)
0]
put D = a
[ (a)
0, f(a) = 0]
put D = a
[f(a) = 0,
(a)
0,
(a)
0]
Case II: When X = sin (ax + b) or cos (ax +b) P.I. =
(
=x
=x
(
(
( x
)
)
)
)
( x
( x
put D =
)
)
, (-
put D =
, ’(
put D =
, ’’(
)
)
0]
(
)
’(
(D)
(D)
) = 0]
)
(
) = 0]
Case III : When X = x , m being positive integer P.I. =
( )
= (D) , Case IV : When X =
x = [ (D)-
( )
-
x
x = (D) [1
(D)
-x
V where V is function of x
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Chapter 5
P. I. = =
V
( )
(
Mathematics
)
V, then evaluate
(
)
V as in Case I, II & III
Case V : When X = x V(x) P.I. =
( )
( )
x V(x) = 0
( )
1
( )
V(x)
Case VI : When X is any other function of x P.I. =
( )
X
Factorize f(D) = (D
) (D
fractions and then apply,
) - - - - - - - (D X=
) and resolve
( )
into partial
x on each .
∫
Complete Solution Given differential equation: The complete solution is:
.......
=X
(C)
x
y = C.F + P.
Example Particular integral of (A)
=x
x
x
(B)
(D)
x
Solution (
)
{x
x
}=
(1 D
= [x =
{x
D - - - -) { x x
x
}
] + =
x , Option (C)
Example Solve the differential equation y =x
x
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Chapter 5
Mathematics
Solution (D
D
)y=x
x
(D - ) = 0 C.F. = (C P.I. =
(
C x) )
P.I.
x
x
=
x sin x
=
∫x
=
[∫ x
=
[ x
=
[ ∫x
=
[ x
=
[
x x x x
x-
x x
x
∫
x-
x x
∫ x
x x-
∫
x
x x-
x-
=
(x
x
x)
y = C.F. + P.I. Example Solve the differential equation (D
D
) = 2x
x
Solution D
D
C.F. = C P.I. =
= 0,
D = 1, 3
C
(
)
=2
(2x
(
)
=2 =
x)
(
x
(
)
)
3 x 3
x
3
(
)
(
)
x
x x
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Chapter 5
=
(1+ )
=
.
=
(
=
,
x
x /x
x (
)x x
-
)
x
(
C.S. = C.F. + P.I. = C Note:
Mathematics
x
x)
C
(
is omitted from P.I. since C
)
(
x
x)
is already in C.F.
In the above problem, try to find P.I. using different approach, P.I. =
(x –
) =
(
)
)y
x sin x
note the difference, (overall C.S. is unchanged)
Example Find the P.I. of (D Solution P.I. =
x sin x = 0x
1(
= 0x
sin x
)
1
(
)
=
[ x sin x – 2D
] =
[ x sin x + cos x ]
Example Find the P.I. of (D
D
)y
sin2 x
Solution P.I. =
sin 2x = =
(
)
=
(
)
x (
sin 2x )
Example Find the P.I. of (D
)y = sin 2x + cos 3x
Solution P.I. =
(sin 2x +cos 3x) = x
x
cos 3x =
cos 2x –
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Chapter 5
Mathematics
Example Find the complete solution of
x
x
x
]
Solution D (D+1) = 0 C.F.
=C
C
=C P.I.
=
C
(
)
[x
x
=
(1+D)
=
[ 1 D +D
=
[x
x
=
[x
-
=
-
[x
x
]
D
] [x ( x
)
]
x
Other approach P. I. = (1- D +D
D ) (x
x
)
=(
1+ D -D ) (x
x
)
=(
x
=
x–4
(x
x)
)
x
( x
)
Note: this is because a constant C is present in C.F. (overall C.S is unchanged) C.S. = C
C
x
C u y’ Eu x
E u t
x
Put =
x= .
(H
u
------ -
u t
)
y=X
t = ln x, then if D = =
Dy
x
= Dy
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Chapter 5
(
) =
.
= =
(D
(
Dy
Mathematics
).
D y
D)y
x
= D (D-1) y
x
= D (D-1)(D-2) y
After substituting these differentials, the Cauchy – Euler equation results in a linear equation with constant coefficients. Example Solve the given differential equation x
x
y
x
Solution substitute
x=
x
Dy
x
(D
D
(D
) = 0 , y = (C
P. =
D
t
(
)
y = (C
C
L
’ L
( x
= D(D-1)
)y =t
= ,
C t) D- t = t + 2
x) x
x E u t
)
( x
ax + b = (ax + b)
t = ln x
)
- - - - -- -
y=X
t = ln (ax + b) =aDy
( x
)
=
D(D-1)y
( x
)
=
D(D-1)(D-2)y
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Chapter 5
At u t tut t constant coefficients.
t
t
L
’
Mathematics
u t
ut
u t on with
Example ( x
( x
)
)
y =0
Solution 2x +3 =
, t = ln (2x+3)
4D(D -1)y – 2Dy – 20y = 0 ( D
D
D
)y = 0
D
=0
D = 3.1, -1 .6 y=C
+ C
= C (2x + )
C (2x + )
Partial Differential Equation z = f(x, y) =p,
L
=q,
E u t
= r,
t
= s,
t
(L
’
u t
)
=R Where P, Q, and R are functions of x, y and z (if P, Q, R are independent of z , its called linear otherwise quasi linear equation) Working rule 1.
Form subsidiary equation
=
=
2. 3.
Solve these simultaneous equations giving u = a and v = b as its solution. Write the complete solution as f (u ,v) = 0 , u = f(v)
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Chapter 5
Mathematics
Non- Linear Equations of 1st Order p and q will occur other than in the first degree are called non-linear partial differential equations of the first order. Form I: Equation containing p & q only f(p, q) = 0 Its complete solution z = ax +by + c where a and b connected by relation Since p = q=
f(a, b) = 0
= a,
=b
Form II: f(z, p, q ) = 0, not containing x & y Working rule: 1. 2. 3.
Assume, u = x + ay , substitute p = , and q = a Solve resulting ordinary differential equation (O.D.E.) in z and u Replace u by x + ay
Form III: f (x, p) = F (y, q) . not containing z and containing x and p can be separated from those containing y and q. Assume f (x, p) = F (y, q) = a (x) dz = z
Ψ(y) x
(x) x
z = ∫ (x) x Charpit’
t
y
x
y
Ψ(y) y ∫ Ψ(y) y
(G
t
)
f(x, y, z, p, q ) = 0 -----------------------(1) Since z depends on x and y
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Chapter 5
dz =
x
y
=
x
=
Mathematics
y --------------(2) =
=
=
Homogenous Linear Equation with constant coefficients ------ -
= f( x, y)
this is called homogenous because all
containing derivative is of same order. (D
D
(D D’)
D
-------
D ) = f(x, y) {where D =
D’
}
(x y)
Step I : Finding the C.F. 1. Write A.E. Where m = 2. CF = (y + CF = (y + roots. CF = (y + roots.
----= 0, . Roots are , ---- x) + (y + x) + - - - - - x) + x (y + x) + (y + x) + x (y +
x) +x
,
are distinct x) + - - - - - - ,
(y +
x) + - - - -
,
,
two equal
,
three equal
Step II : Finding P.I. P.I. =
(
)
f (x, y)
1. when f( ax +by ) =
ut , D
D’
-
2. when f( x, y) = sin (mx +ny), put (D 3. when f(x, y) = x y
P.I. =
(
)
DD
D
x y = [ (D D )-
4. when f(x, y) is any function of x and y. P.I. =
(
)
]
x y
f (x, y) , resolve
(
)
into
partial fractions considering (D D ) as a function of D alone and operate each partial fraction on f(x, y) ing that ( f(x, y) = ∫ (x x) x where c, is ) replaced by y + mx after integration. Example Solve the partial differential equation: r - 4s + 4t =
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Chapter 5
Mathematics
Solution = (D
DD
D )z = =0
C.F = P.I. =
m = 2, 2
(y +2x) + x (
(y +2x )
)
D = 2 , D = 1,
D – D’
So, P.I. =
∫
=
x
(
= ∫x
)
x
x
= Hence, C.S. is Z = (y +2x) + x (y +2x) + x
Example Solve the given differential equation y (2xy +
) dx =
dy
Solution (y
x
dy ) + 2x y dx = 0 x x=0
d( )
x x=0
x =0
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Chapter 5
Mathematics
Example Solve the given differential equation Solution (2D
DD
D )z=0 =0
√
m=
=
= 2,
Solution is z = (y 2x) +
(y
x)
Example Solve the given differential equation
= cos x cos 2y
Solution (D
DD ) z = cos x cos 2y = 0;
C.F. =
(y) + (y + x)
P.I. =
cos x cos 2y
= =
m = 0, 1
(x
,
(x
, =
y)
(
,
(x
y)-
y)
(x
) (
)
(x
y)-
y)- =
(x
y)
(x
y)
Example Solve the given differential equation
=2
x y
Solution D
D D =2
x y =0 (
) = 0,
m = 0, 0, 2
C.F. = (y) + x (y) + (y + 2x) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750, [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 157
Chapter 5
P.I. =
(2
x y)
=
(
=
(
)
(
=
(x y
x y
)
x y (
=
=
Mathematics
)(
)
)x y
)
(
)
= Example Solve the given differential equation cos (2x +y) Solution (D
DD
D )z
cos(2x +y) m = -3, 2
C.F. = (y-3x) + (y+2x) P.I.
=
(
)
(
=
(
=
)
( x
=
∫
=
∫
) (
y) =
( x
)
(
=
(
)(
)
)
( x
y)
̅̅̅̅̅̅̅̅x) x
x
=
x
=
x
( x
= ∫x
( x
̅̅̅̅̅̅̅̅x) x
= ∫x
( x
y)
) x
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Chapter 5 (
= x
)
(
)
̅̅̅̅̅̅̅̅)
(
̅̅̅̅̅̅̅̅ y x)
= sin ( x
Mathematics
= sin (2x + y) +
( x
y)
Approach 2 P.I
= = = =
(
)(
(
)
(
)
(
)
=( =
)
(
=
( x
)
. ∫ .
(
)
x) x
(
.
̅̅̅̅̅̅̅̅x) x
( x
∫
)
(
.
y)
)
.
(y
x)
x =
∫
(̅̅̅̅̅̅̅̅x
∫ (y
x) x
x)
Approach 3 P.I.
=
(
)(
( x
)
y)
=
( x
y)
= x
( x
y)
= x =x = =
( x ( (
)
=
)
( x ( x
y) D
( x
y)
y) x y)
Note the difference in answer of approach 1 & approach 2/3 (why?) To find order and degree of differential equation simplify all the decimal and fractional power to 3/2 or ½ to intiger.
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Chapter 5
Mathematics
Example [
. / ]
Square both side [
. / ]
.
/
Some equation of curves 1. y =mx +c
straight line
2. xy = c hyperbola 3. y.ax2 = or ay2 z x
parabola
4. x2 + y2 = a2 circle 5.
edlipse
6.
hyper bola
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750, [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 160
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