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 1st order, 1st degree
ii) Y = x
+ 1
st order, 2nd degree (power of highest derivative)
iii) [ . / ] = C 2 nd order, 2nd degree
iv) + x = 5 2nd order, 1st degree
Example
The differential equation
x
y = sin hx is
(A) 1st order and linear
(B) 1st order and non-linear
(C) 2nd order and linear
(D) 2nd order and non-linear
Solution Option (C)
Differential Equations of first Order first Degree
Equations of first order and first degree can be expressed in the form (x y y ) or y (x y).
Following are the different ways of solving equations of first order and first degree: 1. Variables separable 2. Homogeneous equations 3. Linear equations 4. Exact equations 1. Variable separable f(x)dx + g(y)dy = 0
∫ (x) x ∫ (y) y is the solution
Example
Solve the differential equation
= x Solution = ( x ) y = x x x = C 3 = 2( x ) C 2. Homogenous Equation = ( ) ( )
To solve a homogeneous equation, substitute y = Vx
= V + x
Example
Solve the differential equation x y dx - (x y ) dy = 0 Solution = . / Put y = vx , = v + x v + x = x = dv = ∫ ∫ ∫ C vx C ln y = . / 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) d x = d X d y = d Y
=
( )
( ) Solve for h and k
= 0 = 0 =
Case II: If = (say) = ( ) ( )
Substitute ax +by = t, so that, a + b = = ( ) ( ) = = ( ) a
Solve by variable separable method. Example = ( ) ( ) Solution
Here a/ = b/ , put 2x+3y = t
2 + 3 = ( ) = 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)
Commonly known as “L tz’ u t ”
Integrating factor, I.F. = ∫ ∫ y( ∫ P) = Q ∫ (y ∫ ) = Q ∫ y ∫ ∫ (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 function of y only I. F. = ∫ = x. = ∫ dy x . = ∫ t y d(t ) x . = t C x = t y C B u ’ E u t
+ P y = Q y where, P & Q are functions of x only.
Divide by y y
y
Substitute, y = z (1-n) y = y = 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 = x y = z, -5y = = x z = -5 x I.F. = ∫ = ∫ = ∫ = = x Z x = ∫ x (-5x ) dx Z x = 5 ∫ x x y 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
=
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
∫( x t x t y ) x ∫ y = 0 Sec x tan y - = 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 - dx = 0 exact Therefore, ∫( ) x ∫ 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 ( ) -
6.
= d [ ( ) -
Rule 1: when M dx + N dy = 0 is homogenous in x and y and M x + N y 0 then I.F. =
Rule 2: If the equation (x, y) y dx + (x, y) x dy = 0 and M x – N y 0 then I.F. =
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 (
) = 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 0 I.F. = = ( y) x . x/ y = 0 exact solution log xy = C
Equations of first order and higher Degree As
will occur in higher degrees, it is convenient to denote
by P. Such equations are of the
form f(x, y, P) = 0.
Case I : Equations solvable for P
- - - = 0
Where, , . . . are functions of x and y [P - (x y)- [P - (x y)- - - - [P - (x y)- = 0 P = (x y) , P = (x y) -- - - - P = (x y)
F (x y ) = 0, F (x y ) = 0, - - - - F (x y ) = 0 F (x y ) . F (x y ) - - - F (x y ) = 0
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 + a t ----(1) Solution P = 1 + ---(2) d x = ( )( ) , dx = 0 1dp x = C + a [log (P-1) log ( ) t ]
This equation along with the given equation constitute the required solution in parameter form ‘ ’ w t 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 3y y y P y 2y y = 0 P(1+ y ) y (1+ y ) = 0 (P + y ) (1+ y ) = 0 Py = c Eliminate P in equation (1) x = ( ) y = 2cx + c 3 Example
The family of straight lines passing 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
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 - - - )y = X {Where, D = }
(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
Illustration: D D = 0 D = 2, 1 y = C C
Case II: If two roots are equal i.e. = (D ) (D ) y = 0 y = (C C x ) Similarly, if = = y = (C C x +C x ) Illustration: D D = 0 (D + 3) = 0 D = 3, 3 y =(C C x)
Case III: If one pair of roots are imaginary i.e. = , = y = C ( ) C ( ) y = (C x C x) Illustration: D D D = 0 D = -1, 2i y = C (C x C x ) = C C x C x Case IV: If two pairs of root are imaginary
i.e. Repeated imaginary root , y = ,(C x C ) x (C x C ) x ] Illustration: D = 0
D = 1 i, 1 i
y = (C x C x) (C x C x )
Case V: If a pair of root is surd √ , > 0 ,
then ,C (√ ) C (√ )]
Rules for finding Particular Integral
P.I. = X = ( ) .X Case I: When X = P.I. = ( ) put D = a [f(a) 0] P.I. = x ( )
put D = a [ (a) 0, f(a) = 0]
P.I. = x
( )
put D = a [f(a) = 0, (a) 0, (a) 0]
Case II: When X = sin (ax + b) or cos (ax +b)
P.I. = ( ) ( x ) put D = , (- ) 0 ] = x ( ) ( x ) put D = , ’( ) ( ) = 0] = x ( ) ( x ) put D = , ’’( ) ’( ) ( ) = 0]
Case III : When X = x , m being positive integer P.I. = ( ) x = [ (D)- x
= (D) ,
( )
x = (D) [1 (D) (D) (D) - x
P. I. = ( ) V = ( ) 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 ) - - - - (D ) and resolve
( ) into partial
fractions and then apply, X = ∫ x on each terms.
Complete Solution
Given differential equation:
. . . = X The complete solution is: y = C.F + P.
Example Particular integral of = x x (A) x (B) (C) x (D) x Solution ( ) { x x } = (1 D D - - - -) { x x } = [x x ] = { x + = x , Option (C) Example
Solve the differential equation
Solution (D D ) y = x x (D - ) = 0 C.F. = (C C x) P.I. = ( ) x x P.I. = 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 = 0, D = 1, 3 C.F. = C C P.I. = ( ) (2x x) = 2 ( ) ( ) x 3 ( ) ( ) x = 2 x 3 x = ( ) x 3 x
= (1+ ) x x = . / x x = ( )x ( ) x = , x - ( x x) C.S. = C.F. + P.I. = C C ( ) ( x x) Note: is omitted from P.I. since C is already in C.F.
In the above problem, try to find P.I. using different approach,
P.I. = (x – ) = ( ) note the difference, (overall C.S. is unchanged)
Example
Find the P.I. of (D )y x sin x Solution P.I. = x sin x = 0x 1 ( ) sin x = 0x 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 = ( ) ( ) sin 2x = = x ( ) 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 –
Example
Find the complete solution of
x x Solution D (D+1) = 0 C.F. = C C = C C P.I. = ( ) [x x - = (1+D) [x x ] = [ 1 D +D D ] [x x ] = [x x ( x ) ] = [x - = x Other approach P. I. = (1- D +D D ) (x x ) = ( 1+ D -D ) (x x ) = ( x x) (x x ) ( x ) = x – 4
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 E u t (H u u t ) x x - - - - y = X Put x = t = ln x, then if D = = . = Dy x = Dy
( ) = . ( ). = Dy D y = (D 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 = t = ln x x Dy x = D(D-1) (D D D ) y = t (D ) = 0 , y = (C C t) P. = ( ) t = , t = t + 2 y = (C C x) x x L ’ L E u t ( x ) ( x ) - - - - -- - y = X ax + b = t = ln (ax + b) (ax + b) = a D y ( x ) = D(D-1)y ( x ) = D(D-1)(D-2)y
A t u t tut t t t L ’ u t u t u t on with constant coefficients. Example ( x ) ( x ) y = 0 Solution 2x +3 = , t = ln (2x+3) 4D(D -1)y – 2Dy – 20y = 0 ( D D )y = 0 D D = 0 D = 3.1, -1 .6
y = C + C
= C (2x + ) C (2x + )
Partial Differential Equation
z = f(x, y) = p , = q , = r, = s, t L E u t t (L ’ u t ) = RWhere 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. Solve these simultaneous equations giving u = a and v = b as its solution. 3. Write the complete solution as f (u ,v) = 0 , u = f(v)
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 f(a, b) = 0 Since p =
= a,
q =
= b
Form II:
f(z, p, q ) = 0, not containing x & y Working rule:
1. Assume, u = x + ay , substitute p =
, and q = a
2. Solve resulting ordinary differential equation (O.D.E.) in z and u 3. Replace u by x + ay
Form III:
f (x, p) = F (y, q) . Terms not containing z and terms containing x and p can be separated from those containing y and q.
Assume f (x, p) = F (y, q) = a (x) Ψ(y) dz = x y x y z (x) x Ψ(y) y z = ∫ (x) x ∫ Ψ(y) y Charpit’ t ( G t ) f(x, y, z, p, q ) = 0 ---(1) Since z depends on x and y
dz = x y x y ---(2) = = = = =
Homogenous Linear Equation with constant coefficients
- - - - = f( x, y) this is called homogenous because all terms containing derivative is of same order.
(D D D - - - D ) = f(x, y) {where D =
D’ }
(D D’) (x y)
Step I : Finding the C.F. 1. Write A.E.
- - - = 0,
Where m = . Roots are , - - - - -
2. CF = (y + x) + (y + x) + - - - , are distinct
CF = (y + x) + x (y + x) + (y + x) + - - - , , two equal roots.
CF = (y + x) + x (y + x) +x (y + x) + - - - - , , three equal roots.
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 DD D ] 3. when f(x, y) = x y P.I. =
( ) x y = [ (D D
x y
4. when f(x, y) is any function of xand y. P.I. =
( ) 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) remembering that ( ) f(x, y) = ∫ (x x) x where c, is replaced by y + mx after integration.
Example
Solution = (D DD D )z = = 0 m = 2, 2 C.F = (y +2x) + x (y +2x ) P.I. = ( ) 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
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; m = 0, 1 C.F. = (y) + (y + x) P.I. = cos x cos 2y = , (x y) (x y)- = , (x y) (x y)- = , ( ) ( ) (x 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)
P.I. = (2 x y) = ( ) x y = ( ) x y = ( ( )( ) ) x y = (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 y) = ∫ ( x x̅̅̅̅̅̅̅̅) x = ∫ x = x = x ( x y) = ∫ x ( x x̅̅̅̅̅̅̅̅) x = ∫ x ( x ) x
= x ( ) ( ) = sin ( x y x̅̅̅̅̅̅̅̅) ( ̅̅̅̅̅̅̅̅) = sin (2x + y) + ( x y) Approach 2 P.I = ( )( ) ( x y) = ( ) . ∫ ( x x̅̅̅̅̅̅̅̅) x = ( ) . ∫ ( x) x = ( ) . ( ) = ( ) . ( ) = ( ) . (y x) ∫ ( x̅̅̅̅̅̅̅̅ x) x = ∫ x = (y x) Approach 3 P.I. = ( )( ) ( x y) = ( x y) = x ( x y) = x ( x y) = x ( ) ( ) = D ( x y) = ( x y) x = ( 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.
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
