DIFFERENTIAL EQUATIONS
PAGE # 1 DIFFERENTIAL EQUATIONS
1.
An equation containi ng independent variable, dependent variable & diff erential coeff ecients of dependent variables w.r.t. independent variable is called differential equation.
2.
If all the differential coeffecients are present w.r.t. one independent variable then the differential equation is called ordinary differential equ. If they are present w.r.t. more than one independent variables then it is called partial differential equ. 3
d2 y
3.
dy x 2 dx dx
ODE
2 y 2 y 0 x 2 z 2
PDE.
Differential equation represents a family of curve which share some common characteristics
dy 5 dx
e.g.1 :
y = 5x + c
f am ily of lin es w i th s lope ‘5’
xy
e.g.2 :
dy 0 dx
x 2 y2 k 4.
5.
family of hyperbola with centre as (0, 0) & pair of asymptotes as y =
x
The family of curves obtained after intergrating the given differential equ., free from differential coeffecients & containing arbitrary constants is called general solution or solution or integral curves or primitive. If all the arbitrary constants present in general sol. are substituted with some specific values then the obtained equ. is called particular sol. O r d er & D e g r e e :
The order of highest order differential coeffecient present in the equation is called order of differential equation 2
e.g. :
8 d3 y 2 dy 3 y dx x 0 dx
order = 3.
The exponent of highest order differential coefficient present in the equation is called the degree of differential equation provided all the differential coefficients are present in polynomial form. Order of a differential equ. is always defined while degree need not to be defined. 7
e.g. : 1
order
d3 y d2 y 3 2 dy dx
25
y7 x 0
3
degree
degree
undefined
7
4
e.g.: 2
order
d5 y dy / dx x3 5 e dx
5 d4 y
e.g.: 3
order
4
4
e.g.: 4
order
e dx x 4
y" "
4
degree 3
1
3
7
y ' sin1 y 0 degree
DIFFERENTIAL EQUATIONS
PAGE # 2
2 1/ 3 y " " y ' x
e.g.: 5
or der - 4
degree
2
degree
1
sin1 y " x3 n x
(iv)
order
2
FORMATION OF DIFFERENTIAL EQUATION :
NOTE :
(A) If the equation of family of curves is given then (i) count the no. of independent arbitrary constants say ‘n’ The no. of independent arbitrary constants will be the order of differential equ. to be formed
sin3 x k1 k 2
e.g. 1: y =
e.g.2 : y =
1
kek 2 x k 4
order
3
A sin x B cos x
order
2
order
2
k1ek2 x k3 k 4
= e.g.3 : y = e.g.4 :
e.g.5 :
sin3 x k
order
y=
y A sin x c cos x D y
sin1 x cos1 x k1 k 2
k 3 k 4 sin x .
(ii) Differentiate the given equation ‘n’ times. Eliminate these ‘n’ arbitrary constants from these ‘n + 1’ equations, one given and ‘n’ obtained. (iii) The eliminant of above will be req. differential equ. (B) If the equations are given in terms of language then form the corresponding mathematical equation & hence differential equation
thus obtained is Reqd. D.E. Q.
Form the differential equ. for follow family of curves (i )
y k1ex k 2
y ' k1ex e x y ' k1
e x y ' e x y " 0 y” = y’. (ii) y =
k1e x k 2 e x
y ' k1e x k 2e x y " k1ex k 2 e x (iii)
y k1ex k 2 e x
y ' k1e x k 2e x y” =
k1ex k 2e x y .
y” = y..
DIFFERENTIAL EQUATIONS (iv)
y k1x3 5
y’ =
3k1x2
y’ =
3x 2
(v )
y eax b c
y=
keax c
y 5 x3
=
y 5 x
3
PAGE # 3
.
y ' ak eax e ax y ' ak
ae ax y ' e ax y " 0 y” = ay’ y”’ = ay”. (vi) Family of circles centred at origin
x 2 y2 k 2 x yy ' 0 (vii) Family of circles having centre at line x = 1 & radius = 2 2 2 x 1 y k 4
x 1 y k y ' 0 y k
x 1 y' 2
x 1 4. x 1 y ' 2 2
(viii) Family of parabolas axis // to x-axis & the vertex at (4, 5) 2 y 5 a x 4
2 y 5 y ' a 2 y 5 2 y 5 y ' x 4
y 5 2 x 4 y ' . (ix) Circle touching x-axis at origin 2
x2 y r r2 x 2 y2 2yr 0 x2 y 2r 0 y
DIFFERENTIAL EQUATIONS
2xy y ' x2
PAGE # 4
y' 0
y2
2xy x 2 y 2 y ' . (x)
y 2 2c x c
2yy ' 2c
y 2 2yy ' x c
y 2 2yy ' x yy '
y 2y ' x yy ' . SOLUTION OF DIFFERENTIAL EQUATION : 1.
A solu tion of a dif ferent ial equ. will sati sf y it ident ical ly e.g. : Find which one of the follow/s is/are sol. of differential equ.
(i) y =
So l .
y" y 3 0
a sin x b cos x
(ii)
y 3 asin x bcos x
(iii)
y 3 asinx bcos x
(iv)
y 4 sin x 3
Ch ec k i n g o p t io n (ii )
y " asin x bcos x
y " y 3
as bc as bc 3 3 0 So l .
Ch ec k in g op t i on (iv )
y " 4 sin x
4s 4s 3 3 0 . 2.
For a general curve (a) Tangent
Yy
dy X x dx
(b ) Normal
Yy
dx dy
X x
(c ) Intercepts made by tangent on both the axes are A :
3.
(a) R.P. = Length of tangent
Intercepted
y x y ' , 0 & B : 0, y xy ' .
portion of tangent b/w pt. & x-axis (b) P.S. = Intercepted portion of normal b/w pt. & x-axis Length of Normal. (c) RQ = subtangent = Projection of length of tangent on x-axis (d) QS = Projection of length of normal on x-axis Subnormal. Subnormal = QS = y tang = yy’
DIFFERENTIAL EQUATIONS
PAGE # 5
y 1 y '
Length of normal = PS =
subtangent =
y y'
Length of tangent =
4.
d =
2
1 y 1 y '
2
.
2 2 dx dy B
Length of curve = AB =
d
A
=
B
dx dy 2
2
2
1 y ' dx
A
METHODS OF SOLVING DIFFERENTIAL EQUATION : 1.
Var i ab l e s ep ar ab l e :
Take dy on one side of equation & dx on other side. If all the terms containing ‘y’ can be clubed with dy & the terms containing ‘x’ can be clubed with dx then it is variable separable form. Separate the terms & then intergrate.
sin x cos y
(i )
dy 0 dx
sin x dx cos y dy cos x sin y c . (ii)
2 cos x tan y 3
dy 0 dx
2 cos x dx 3cot y dy 2x sin x 3 n sin y c (iii)
2 sin 3x dy y2 1 dx 0
2 sin 3x dy y 2 1 dx dx dy 2 0 2 sin 3x y 1
dx 2 sin3x
+
y
dy 2
1
= 0
DIFFERENTIAL EQUATIONS
3x dx 2 3x 3x 2 2 tan2 2 tan 2 2
PAGE # 6
sec 2
+
tan 1 y c
.
dy e3x y ey sin x dx
(iv)
dy e3x .ey e y sin x dx
e y dy e3x sin x dx
e y e3x cos x c . 1 3
dy ax by dx
(v ) n
dy eax by dx
dy eax .e by dx eby dy eax dx
e by eax c. b a
( v i ) Find curve (which is not passing through origin) for which length of norm al at a point is equal to radius
vector ? Sol.
Length of normal = radius vector
y 2 1 y '
2
x
2
2
y 1 y ' x 2 y 2
y2
x
yy’ =
y2 x 2 k . (vii) Find curve for which segment of tangent contained b/w coordinate axes is bisected by the point & the curve passes through
(3, 4)
A :
y x y '
x A xB xp 2 x
y y'
y 0 y' x 2
x
DIFFERENTIAL EQUATIONS
PAGE # 7
dy dx y x n y
n x c
( (3, 4) lies on the curve)
xy = k = 12. Q. 8
Find the curve passing through (1, 2) for which mid pt. of (length of) tan gent is bisected at its pt. of ‘x’ with y-axis
x A xP 0 2 x
y x0 y'
2x
y y'
2dy dx y x 2 n y nx c y 2 kx (1, 2) k = 4
y 2 4x . Q. 9
Find curve for which area bounded by curve, the coordinate axes & a variable ordinate is equal to the length of corresponding arc in 1st quad. & given that curve passes through (0, 1) ?
Ar OAPQ PQ x
x
y dx =
0
2
1 y ' dx
0
x
f x dx
y = f(x)
0
1.y = 1 y ' y 2 1 y '
2
2
y ' y2 1 y=
1
y= 1
dy y2 1 n
y
dx
y2 1 x c
DIFFERENTIAL EQUATIONS n
y
PAGE # 8
y2 1 x
y y 2 1 e x y y2 1 e x
2y e x e x
y
ex e x 2
& y = 1.
(B ) Reducial to Variable separable :
dy f ax by c dx let ax + by + c = t a + by’ =
dt dx
a bf t
dt dx
dt dx . a f t
1.
dy dx
=
sin 2x 3y cos 2x 3y 4
2x + 3y = t 2 + 3y’ =
dt dx
2 + 3(sin (t) + cos(t) + 4) =
dt dx
dt dx . 3sin t 3cos t 14
2.
dy 2 x y dx
3.
dy x y 1 dx x y 2
4.
x y 1 dy x y 1 x y 2 dx x y 2
5.
dy x tan y x 1 dx
DIFFERENTIAL EQUATIONS 2.
x+y=t
dt dx
1 + y’ =
1+
dt dx
t2
dx =
dt
1 t
2
x + c = tan1 x y . 3.
x+y=t 1 + y’ =
1
dt dx
t 1 dt t 2 dx
t 2 t 1 dt t2 dx 2t 1 dt t 2 dx
t2
dx 2t 1 dt =
4.
.
x+y=t
dt dx
1 y '
1+
t 1 t 2 dt t 2 t 1 dx
t 2 t 2 t 2 t 2 dt dx t 2 t 1
2 t2 2
t 2 t 1 2dx dt
5.
t2 2
dt dx
xtant
dt dx
t 2 t 1
yxt y ' 1
dt dx
.
PAGE # 9
DIFFERENTIAL EQUATIONS
PAGE # 10
x dx cotdt =
x2 c n | sin y x | . 2 3.
Variable separable using Trigonometric substitution : (a)
y r sin
&x=
rcos
(b)
x 2 y 2 r 2
(c)
tan y / x
(d)
x dx ydy r dr
(e)
sec 2 d
x dy y dx x2
r 2 d xdy y dx . T-2:
(a)
x r sec rtan
y= (b)
x 2 y 2 r 2
(c)
y sin x
(d)
x dx y dy r dr
(e)
cos d
xdy ydx x2
x dy y dx r2 sec d .
1.
xdx ydy xdy ydx r dr r 2 sec d dr 2
1 r n
2.
r
1 x 2 y2 =
x 2 y2 1 r2
r2
sec d
1 r2 n sec tan c
x 2 xdx ydy 2y xdy ydx 0
r 2 cos2 r dr 2r sin r 2 d 0
cos2 .dr 2sin d 0 dr 2 tan sec d 0
.
DIFFERENTIAL EQUATIONS
PAGE # 11
r 2 sec c 0
xdx ydy 3.
x2 y 2
ydx xdy x2
4.
xdx ydy x xdy ydx
5.
x yy ' 1 x2 y2 xy ' y x2 y2
3.
r dr r 2 d 2 r r cos2
dr sec = –
r=
4.
r 2
d
tan c . r cos r 2 d
r dr =
1
2
dr cos d
1 r
c sin .
5.
1 x2 y2 x yy ' xy ' y x2 y 2
r cos r sin y ' 1 r 2 r cos y ' r sin r 2 x+
y
dy dx
dr 1 r 2
=
xdx y dy r dr 2 = xdy ydx r d
1 r 2 r
d .
a x b1y c1 dy 1 dx a2 x b2 y c 2
a2 xdy b2 ydy c 2 dy a1xdx b1ydx c1dx b2 ydy c 2 dy a1 xdx c1dx a2 xdy b1 ydx 0 If
a2 b1 0
a2 xdy b1ydx a2 xdy ydx
DIFFERENTIAL EQUATIONS
e.g.: 1
PAGE # 12
dy 3x 4y 5 dx 4x 8y 9
4xdy 8ydy 9dy 3xdx 4ydx 5dx 8ydy 9dy 3xdx 5dx 4 xdy ydx 0 . 4xy 4y 2 9y
e.g.2 :
3x 2 5x c 2
dy x 4y 9 dx 4x 16y 1
x + 4y = t. e.g.3 :
dy 2x 3y dx 3x 2y
Integrate term by term e.g.:4 :
3dy 9x 8y dx 8x 2y
e.g.: 5:
2
dy 3x 4y . dx 2x 9y 7
HOMOGENOUS D.Es :
The equation of the form
dy P x, y dx Q x, y
is called homogenous differential equation where P & Q are homogenous functions of
same degree.
In this case R.H.S ca also be written as
y f x
[By cancelling xn from numerator & D r of R.H.S. where n is the degree of P & Q.] To solve put y/x = t y = xt
dy dt tx dx dx f(t) = t +
x
dt dx
f(t) – t =
x
dt dx
1 dt dx x f t t
Q. 1
dy 3x 4y dx 4x 3y As it is homog. diff equ. using y/x = t
variable separable.
DIFFERENTIAL EQUATIONS
PAGE # 13
dx dt 3 4t x t 4 3t n x
4 3t dt 3 3t 2
dy y 2 y 2 x2 dx
2.
x
3.
dy x2 xy dx x 2 y2
4.
.
y y x sin dy y sin x dx . x x
5.
x 2 3xy y2 dx x 2dy
2.
2 2 dy y 2 y x dx x
dx dt x t 2 t2 1 t ln x
3.
1 2 n t t 1 c 2
dy x2 xy dx x 2 y2 dx dt 1 t x t 1 t2
=
1 t2 dt t 1 t 2
1 t 2 dt dx x 1 t3 1 t 2 dt n x 1 t t 2 t 1 + c.
4.
dy y sin y / x x dx x sin y / x
t = y/x.
DIFFERENTIAL EQUATIONS
PAGE # 14
dx dt t sin t 1 x t sint n x
5.
sin t dt .
dy x 2 3xy y 2 dx x2 dx dt x 1 3t t 2 t n x
6.
dt =
2
t 2t 1
dt
t 1
2
1 c t 1 .
Find the curve for which subnormal is equals to summation of coordinates
yy ' x y y’ =
xy y
dx dt t dt 1 t x = t 1 t t2 . t 7.
Find the curve for which ratio of subnormal to the sum of coordinates is equals to that of ordinate & absicca ?
yy ' y xy x y = 0 or y’ =
8.
.
If the area bounded by the curve b/w x = , x = x, y = 0 & the curve is = ratio of cube of ordinate to absicca x
y dx
y=
9.
xy x
y3 x
3y 2 y ' x y3 x2
x2 y y 3 3y 2 x
y' .
Find the curve for which angle formed by x-axis at any pt. is twice the tan =
y x
tan 2 y '
formed
by polar radius of pt. of tangency with x-axis
DIFFERENTIAL EQUATIONS
2y/x
y'
y 1 x
dx x
10 .
2
dt 2t t 1 t2
.
2
y y' 2xy ' y 0
y’ =
y’ =
2x 4x2 4y 2 2y
x x2 y2 y
dx dt x 1 1 t 2 t t
11.
.
1 e x / y dx e x / y 1 x / y dy 0 ex / y 1 x / y dx dy 1 ex / y dy dt t y e 1 t 1 e
12 .
ydx –
2
t
t .
xy x dy 0
dy y dx 2 xy x dx x
Q. 3
xy 2
dt t t 2 t 1
.
dy 1 x2 y2 x2 y2 . dx
REDUCIBLE TO HOMOGE NOUS DIFFERENTIAL EQUATION :
PAGE # 15
DIFFERENTIAL EQUATIONS
T-1.
PAGE # 16
a x b1y c1 dy 1 dx a2 x b2 y c 2 x=x+h y=y+k
a x b1y a1h b1k c1 dy 1 dx a2 x b2 y a2 h b2 k c 2
.
If h & k are chosen such that
a1h b1k c1 0 a 2 h b 2k c 2 0 . dy a1x b1y dx a2 x b2 y dx dt t. a1 b1t x a2 b2 t
Ex-1.
dy x 2y 5 dx 3x 4y 11 x=x+h y=y+k
dy x 2y dx 3x 4y h + 2k – 5 = 0 3h + 4k – 11 = 0 h = 1, k = 2
dx dt 3 4t dt = 2 1 2t x t 1 t 4t 3 4t T-2:
Q. 1
x
x – 1
y
If x & y is present as f(x) & g(y) then search f’(x) with dx & g’(y) with dy.
dy 3x3 4y 2 x 7x dx 4x 2 y 5y3 8y
=
2 2 x 3x 4y 7 y 4x2 5y2 8
y dy 3x 2 4y2 7 x dx 4x 2 5y2 8 dy2 dx2
3x2 4y 2 7 4x2 5y 2 8
y – 2.
DIFFERENTIAL EQUATIONS
PAGE # 17
dy 3x 4y 7 dx 4x 5y 8 4xy
2.
5 2 3 5 3 y 8y x2 7x c 4x 2 y 2 y 4 8y2 x4 7x2 c . 2 2 2 2
sec y 3cos x 4sin y dy dx cos ec x 4 cos x 5sin y
3.
x y2 dx 2xydy 0
4.
dy ex y ex e y dx
5.
yy 'sin x cos x sin x y 2
2.
cos y dy 3 cos x 4sin y sin x dx 4 cos x 5sin y
d sin y d cos x
.
3cos x 4 sin y 4cos x 5 sin y
y = sin y, x = –cos x
dy 3x 4y dx 4x 5y
4xy
3.
5y 2 2
=
3x 2 2
c .
x y2 dx 2xy dy 0 y2 y
x y dx x dy 0 dy y x dx x
4.
e y dy x
e dx
e x ey
e de d ey
.
x
x
e y Now ex x & ey y
dy xy . dx Reducible separable method.
DIFFERENTIAL EQUATIONS 5.
PAGE # 18
2yy ' sin x 2cos x sin x y2
sin x x
sinxdy2 2cos x sin x y dx
y2 y
dy 2 x y dx x
2y dy = dy. 2
6.
x 7 y 3 dy dx x 7 2 let X = X – 7 Y=Y+3
dY X Y dX X2
2
dX dt 2 X 1 t t
=
dt 1 t t2
where Y = tX
Ultimately we will put Y = Y + 3 & X = x – 7. LINEAR D.E. OF 1ST ORDER & FIRST DEGREE : 1.
Gen er al eq u at i o n :
dy p x . y Q x dx Define integrating factor I.F. = exp.
=
p x dx
e
px dx
multiply both sides by I.F.
e
px dx .
p x dx p x dx dy e .p x .y Q e dx
p dx p dx d y.e Q x e dx
y e
NOTE :
p dx
Q x .e
p dx
If an expression contains
dx dy dx
...(i)
linearly added with y(where coefficients could be functions of x) then it can be made perfect diff.
of (y. Integrating Factor). 2.
dx p y . x Q y . dy is linear differential equ. where y is dependent veriable and x is independent variable.
Q. 1
x 1 x2
y 1 x x n x . dy dx 2
2
Q. 2
x nx
dy y 2 n x . dx
DIFFERENTIAL EQUATIONS
Q. 3
Q. 5
Q. 7
PAGE # 19
dy 1 dx x cos y sin 2y
2
2
t 1 t dx x xt t
2
.
dt .
1 y x2 y dx x x3 dy 0 .
Q. 9
1 y2 dx tan1 y x dy .
Q.11
x y 1
1.
dy 1. dx
x n x dy 1 x2 y dx x 1 x 2 1 x2
I.F. =
=
=
=
1 x2 exp dx x 1 x2
1 1 x2 exp dx 1 x x
1 exp n x x x
1 x
.
Using (i)
x n x x2 1 . dx 1 x2 x
1 yx x
=
=
n
=
x n x 1 c .
x dx
Q. 4
Q. 6
dy y dx 2y n y y x
.
dy x 1 .y 2 dx 1 x 2x 1 x2
.
Q. 8
2xy cos x . 1 x dy dx
Q.10
2x 10y3 dy / dx y 0 .
Q.12
sin x
2
dy cos x 2 sin2 x y dx
.
DIFFERENTIAL EQUATIONS
2.
I.F. =
PAGE # 20
dx x n x
exp
exp n n x
=
= nx. From (i)
y nx
2
x
n x dx
y n x n2 x c . 3.
dx x cos y sin 2y dy dx
dy x cos y sin 2y xe
4.
sin y
sin 2y e sin y dy .
dx 2y n y y x dy y
dx 1 .x 2 n y 1 dy y I.F. = y
2y
xy =
9.
y y dy .
tan1 y dx x dy 1 y2 1 y2
x. e
5.
n
tan1 y
e
tan1 y
.tan1 y dy
1 y2
.
t 1 t 2 dx x xt 2 t 2 dt 2 dx x 1 t t2 dt t 1 t2 t 1 t2
dx 1 t x dt t 1 t2 I.F. =
1 1 1 dx exp. n t t t
exp.
x t 1 dt 2 t t 1 t
x tan1 t c . t
DIFFERENTIAL EQUATIONS
6.
I.F.
PAGE # 21
1 2x dx 2 2 1 x
exp.
=
1
y 1
7.
x 2 =
2x 1 x 2
2 dy 1 y x y dx x x3
1 dy 1 y dx x x 1 x2
I.F. =
8.
1 x2 dx
=
1 dx x x
exp
x dx 1 = tan x c x 1 x 2
xy =
xy =
tan1 x c .
dy 2x cos x .y dx 1 x2 1 x2 I.F. =
2x dx 1 x2 2 1 x
exp
y 1 x 2 cos x dx sin x c .
10 .
y dy dx 2x 10y3 dx 2 dx 2x 10y3 x 10y2 dy y dy y 2 2 x.y 10y 2 .y 2 dy xy2 2y5 c . y
11.
dx x y1 dy
=
exp. 1 x 2 1 x 2
1 1 dx . 2 x 1 x2
y 1 x2 1 dy dx x 1 x2 x 1 x2
1 exp. n 1 x 2 2
.
DIFFERENTIAL EQUATIONS
PAGE # 22
dx x y1 dy exp
I.F. =
dy e
y
e y .x e y y 1 dy .
12 .
dy cot x.y sin 2x dx I.F. =
exp
cot x dx exp
sin x
n sin x
sin x.y sin x.sin 2x dx . DIFFERENTIAL EQUATION REDUCIBLE TO LINEAR D.E. :
f ' y
dy p x f y Q x dx
f(y) = z
(diff w.r.t. ‘x’)
dz p x .z Q x dx Linear differential. equation in z. BERNOULI’S EQUATION :
1.
dy p x .y yn Q x dx 1 dy 1 Q x . p x yn dx yn 1 1 yn1
z.
e.g. :
1 y
2
x 2 y ' xy y2
y '
1 1 1 . 2 x y x
1/y = t
dt 1 1 t 2 dx x x I.F. 1/x
t dx 3 x x
2.
y '
xy 1 x2
.
x y.
DIFFERENTIAL EQUATIONS 2 y y n y n y . 2 x x
3.
y '
4.
y 2 y ' x y3 .
5.
sin y
6.
y ' xy y 2 ex
1 2.
PAGE # 23
dy cos x 2cos y sin2 x dx
y '
y
2
x y
/2
.
sin x
x
1 x2
y t
1 dy dt 2 y dx dx dt xt x 2 dx 2 1 x 2
I.F. =
1 2x dx 4 1 x 2
1 2 n 1 x 4
=
exp.
=
1 x 2
t 1 x2
3.
exp.
1/ 4
1/ 4
1/ 4
1 1 dy 1 1 . . dx y n y 2 x n y x2 1/ n y t dt 1 1 t 2 dx x x I.F. = 1/x
t 1 1 2 . dx . x x x
4.
x 1 x2 2
y2
dy x y3 dx
dx .
DIFFERENTIAL EQUATIONS
PAGE # 24
dy 3y3 3x dx
3y 2
y3 t dt 3t 3x dx exp 3 dx e3x
I.F. =
t e3x 3x e3x dx .
5.
dy cos x cos x sin2 x (2cosy) dx sin y sin y
6.
1 dy 1 x2 / 2 x e sin x y 2 dx y
.
1/ y t 2 dt x t ex / 2 sinx dx
exp
I.F. =
2
t e x
/2
x dx
ex
2
/2
=
e x
x2 x2 / 2 e 2
exp 2
/2
sin x dx
2
ex / 2 cos x c . y 7.
Find the curve such that the initial ordinate of any tangent is less than absicca of pt. of tangency by 2 units.
Y y y ' X x Initial ordinate =
y xy ', 0
put x = 0
x y xy ' 2 y ' 8.
1 2x y x x
.
Find the curve if the product of initial ordinate & absicca is constant ? IO=
y xy ', 0 = x y xy ' c .
FORMING PERFECT DIFFERENTIALS :
1.
d xy x dy y dx .
DIFFERENTIAL EQUATIONS
PAGE # 25
2.
d x2 y2 2 x dx y dy .
3.
y x dy y dx d x x2
.
4.
x y dx x dy d y y2
.
5.
x y dx x dy d tan1 y x2 y2
Q.
So l v e t h e fo l l o w i n g s :
1.
y2 x.dy y 3 dx y dx x dy
=
.
x x sec tan y y
x x x dy y dx 1 2 sec tan y dx x dy y y y d xy
y dx x dy y
2
x x sec y tan y
x x d xy sec tan d y y
xy = n
x x sec tan y y
x y
+ n
x cos c . y
2.
x3 dy x 2 y dx x dy y dx .
3.
x cos y dx x dy xy3 x dy y dx . y
4.
1 x
5.
y dx x dy xy dx
6.
x yy ' y4 x2 2y 2 2 y xy ' x
7.
xy2 x dx yx 2 y dy 0
x 2 y2 dx
y sin x cos2 xy 8.
cos
2
xy
x2 y2 1 y dy 0
dx
x dy cos2 xy
sin y dy 0
DIFFERENTIAL EQUATIONS 2.
x dy y dx
PAGE # 26
x dy y dx x2
d xy d y / x y x
xy =
3.
.
x y dx xdy cos xy x dy y dx y y2 x x cos d xy d xy y y 2
x xy sin c. 2 y
4.
dx x2 y2 x dx y dy y dy 0
1 2 x y 2 d x2 y2 y dy 0 2
dx x
1 2 2 x y2 2 3
y dx x dy 5.
y
2
3/2
y2 c. 2
x dx 1 d x dx y x y y
x xc. y
n
6.
x 2 y2 x dx y dy y dx x dy x2 x dx y dy
x
2
y2
2
1d x y 2 x2 y 2
2
2
x2
x
y 1 1 2 x 2 y2 x
2
x dy y dx
d y
2
.
DIFFERENTIAL EQUATIONS 7.
PAGE # 27
xy d xy x dx y dy 0
x 2 y 2 x2 y 2 k . y dx x dy 8.
cos2 xy
sin x dx sin y dy 0
tan xy cos x cos y c .
9.
x dy y dx
x dy y dx x2 y 2
x dy y dx
y2 x 1 x
x dx y dy x2 y2
x dx y dy
x
2
1
y tan1 x
3/ 2
3 / 2 1 2 x y2 d x 2 y2 2
2
10 .
y2
2
x y
2
c.
2 x y sin 2x dx 3y 2 cos 2x dy 0 2x dx 3y 2 dy cos 2x dy 2y sin 2x dx 0 2 3 d x d y d y cos 2x 0
x 2 y3 y cos 2x c .
11.
x dx y dy 1 x 2 y2 x dy y dx x2 y 2
1 x2 y2 1 2 2 d x y x dy y dx 2 x2 y 2
2 2 1d x y 2 x2 y 2
x dy y dx
y x 1 x 2
1 x2 y2
2
.
ORTHOGONAL TRAJECTORY : (O. T.) : 1.
O.T. of a family of curves is a family of curves each member of which intersect with each member of given family at 90°.
DIFFERENTIAL EQUATIONS
PAGE # 28
To find equ. of orthogonal trajectory 1st find differential equ. of given family, then replace integrate this DE to get req. O.T. Q. 1
Find orthogonal trajectory of follow :
x 2 y2 k 2
(i)
x yy' 0
1 For O.T. y’ =
y'
y 0 y'
x
xy ' y 0 x dy y dx 0 d xy 0 xy = c.
x 2 y 2 r 2
(ii)
x + yy’ = 0 For ORTHOGONAL TRAJECTORY y’
x
1 y'
y 0 y'
xy’ – y = 0 x dy – y dx = 0
dy dx 0 y x n(y)
– n(x) = c
y c x
n
y = cx. (iii)
x 2 / 3 y 2 / 3 a2 / 3
1 1/ 3
x
1 1/ 3
y
.y ' 0
For ORTHOGONAL TRAJECTORY y’
1 1/ 3
x
1 1/ 3
y
.
1 0 y'
y1/ 3 dy x1/ 3 dx 0
1 y'
dy dx
dx in this diff. equ. by
dy
& then
DIFFERENTIAL EQUATIONS
PAGE # 29
y4 / 3 x4 / 3 k . RATE CHANGE & FORMING CORRESPONDING D.E. : 1.
To analyse the problem consider a time interval ‘dt’ i.e. from t to (t + dt).
2.
In this time interval, a changing quantity can be assumed as constant.
3.
Form D.E. for corresponding physical phenomenon. Integrate the obt. differential equ. The given information in the ques. may be used in limits of integration or for indefinite
Q. 1
to evaluate arbitrary constant.
Let a spherical ball losses its volume which is directly proportional to its instantaneous area. Also at t = 0, r = 2m & at t = 3 months, r = 1 m. Find radius as function of time ?
dv 4r 2 dt dv k 4r 2 dt d 4 3 r k 4r 2 dt 3
4 dr 3r 2 k 4r 2 3 dt dr k dt r = kt + c t = 0, c = 2 t=3 3k + 2 = 1
Q. 2
r=1
1
k=
r=
r=2
3
1 3
t2.
Let a cone is filled with water & water evaporates from it at the rate directly proportional to suface area expossed to environment proportionality constant being 2 m/s. Find the height of water column as funct. of time given at t = 0, h = 100 m & the semivertical angle is 60°.
v
1 2 2 r h htan60 h 3 3
= h 3
dv 2r 2 6h2 dt
6h
d h3 dt
2
DIFFERENTIAL EQUATIONS
PAGE # 30
dh 6h2 dt
3 h2
dh 2 dt h = –2t + k h = 100 h = –2t + 100. Q. 3
t=0
k = 100
Let a right circular cylinder of cross-sectional area A is provided with a circular opening at bottom of area ‘a’, which is covered with a diaphrogm z is opening at a constant rate & is completely opened at the time, t = T. The ht. of circular vessel is H & is comp. filled column as a funct. of time ? (i) t < T (ii) t > T
a t kt a t a when t = T a T
k=
at
a t. T
vreduced v out
Adh a t v.dt
(v is the velocity of water moving out)
Adh at 2gh dt a t dt 2gh T
dh
A
A
h
2g H A 2g 2A 2g
a t dt h T0
2 h
t
dh
h H
a t2 T 2
h H
2gat 2 h H 4TA
Q. 4
0
a t2 T 2
,
A dh t>T
t
2gh
t
adt .
Let a container contains ‘v 0 ’ L of fresh water. At time t = 0 a salted water is run into it at the rate ‘v 1’ L/min, each litre of which contains ‘a’ kg of salt. At t = 0 water also starts flowing out at the rate of ‘v2’ L/min then find quantity of salt present in container as a function of time, all the time mixture is kept uniform by stirring. am = In – out
DIFFERENTIAL EQUATIONS
PAGE # 31
m v 2 dt v0 v1 v2 t
av1 dt
=
Volume of water =
v 0 v1t v 2 t
m v 2 dt v 0 v1 v2 dt
dm av1dt
v2 dm m av1 dt v 0 v1 v2 t
v2 dt v 0 v1 v2 t
exp.
If =
v2 ln v 0 v1 v 2 t v1 v 2
=
exp.
=
v 0 v1 v2 t
m v 0 v1 v 2 t
v2 v1 v 2
v2 v1 v 2 =
m v0 v1 v2 t
v2 v1 v 2 =
av v 1
0
v1 v2 t
av1 v0 v1 v2 t
v2 v1 v 2
dt
v2 1 v1 v 2
v v1 v 2 v 2v 1 1 2
+ c .
STQ:
1.
f’(x) + p(x).f(x)
Linear Diff. Expression.
2.
Let for a given diff. equation
dy f x, y dx
If g(x) is one of the solution then
dg f x, g x . dx
Q. 1
Let y = f(x) be a solution of differential equ. curve at a given point ?
dy 1 3 2 dx 1 T = y + 4 = –2(x – 1) N
1
y 4 2 x 1 .
dy x 2 3x dx y 5
& it passes through (1, – 4) then find equ. of tangent & normal to this
DIFFERENTIAL EQUATIONS Q. 2
Let p(1) = 0 and
d p x dx
PAGE # 32
p x , x 1
x 1
T.P.T. p(x) > 0
dp p 0 dx
d p x e x 0 dx
f’(x) > 0
ing
f(x)
then if x > 1 f(x) > f(1) p(x)
e x p 1 e1
p(x)
e x 0
p(x) > 0. Al t er n at i v e :
d p x .e x 0 dx
Integrating both sides within limits 1 to x
p x e x
x
1 0
p x e x p 1 e1 0 p(x) > 0.
Q. 3
Let f(x, y, c) = 0
be the integral curves of the differential equ.
dy g x, y dx k x, y
where g & k ar e homogenous exp equ. of s ame degree T.P.T. tangent s
drawn to all the integral curves at their pt. of intersection with a straight line passing through origin, will be parallel.
dy g x, y dx k x, y dy y f x dx
Q. 4
MT1
dy f m dx x1, y1
MT2
dy f m . dx x , y 2 2
Let two curves are given y = f(x) & y = g(x) satisfy the follow two properties (i) tangents drawn at the pts. of equal absicca intersect on y-axis. (ii) normals drawn at the point of equal absical intersect on x-axis then find the curves.
DIFFERENTIAL EQUATIONS
T1 Y f
df X x dx
T2 Y g
dg X x dx
PAGE # 33
As both pass es throu gh A
f xf ' g xg' N1 Y f
dx
N2 Y g
df
X x
dx dg
X x
As both pas ses throu gh B
x f f ' x gg'
...(ii)
f 2 g2 c
f g f g c
...(ii)
f g x f ' g'
from (i)
f g x
d f g dx
d f g dx f g x
f g kx
...(iv)
from (iii) & (iv) kx(f + g) = c f+g=
c c 1 kx x
f – g = kx
f=
c1 1 kx 2 x
g=
1 c1 kx 2 x
.
Q. 5
Let two curves y = f(x) & y = g(x) are given where g(x) =
f x dx
& also tangents drawn at pts. with equal absicca to both
curves
intersect
1 0, 4 x
f g x f' g'
on
x-axis
find
the
As th ey in ters ect on x-axi s
curves
if
f(x)
passes
through
(0,
1)
&
g(x)
passes
through
DIFFERENTIAL EQUATIONS
f ' g' f g
PAGE # 34
f = kg(x)
x
f x dx
g(x) =
g’(x) = f(x)
g’(x) = kg(x)
g' k g
Q. 6
n g
kx c
g(x) =
ekx c k1 ekx
f(x) =
kk1 ekx
1 0, 4
k1 = 1/4
(0, 1)
1 = k.
f(x) =
e4x
g(x) =
1 4x e 4
Let
1 ae
two solutions of
Q
y
1
dx
dy2 py 2 Q dx y 2 y1z y1
dy dz z 1 py1z Q dx dx
y1
dz zQ Q dx
y1
dz Q 1 z dx
Q dz dx z 1 y1
Q
y dx c 1
z 1 k e
Q
y
1
dx
dy p x y dx
= Q(x) & also
where a is arbitrary constant
dy1 py1 Q dx
n z 1
k = 4
.
y1 & y 2 are
T.P.T. z =
1 4
y 2 y1z
