Differential Differential Equations Equ ations
Overview : ●
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An equation involving derivative(s) of the dependent variable with respect to independent variable(s) is known as a differential equation. A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation and a differential equation involuting derivatives with respect to more than one independent variables in called partial differential equation. Order of a differential equation is the order of the highest order derivative occurring in the differential equation. For example - order of differential equation 4
dy + 3 y d 2 y = dx dx 2 ●
0 is 2
Degree of a differential equation is defined if it is a polynomial equation in its derivatives. Degree of the polynomial if defined is the highest power of the highest ordered derivative involved in the differential equation. Degree of a differential equation is a positive integer only. For example - degree of the differential equation d3y dx3
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3
d 2 y + x 2 = dx 2
0 is 1
A relation between the variables involved in the differential equation which satisfies the given differential equation is called its solution. Solution of differential equation
General Solution
— A solution which contains as many
Particular Solution
— A solution which is free from artitary
artitary co constants as as th the or order of of th the
constants is called is called a particular
differential equation is called the
solution.
general solution. ●
To form a differential equation from a given function, we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the artitrary constants.
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To from a differential equation representing family of curves given by y 2
= a(b2 − x 2 ) , we have to
differentiate the relation twice and then eliminate the artitrary constants a and b. ●
The order of a differential equation representing a family of curves is equal to the number of arbitrary constant(s) present in the equation representing the family of curves.
Types of Differential Equation There are three methods of solving a first order, first degree differential equation depending on its form. These are : (i) Differential equations with variables separable (ii) Homogeneous differential equations (iii) Linear differential equations Let us study each one of them in detail. ●
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‘Variable Separable Method’ is used to solve such an equation in which variables can be separated completely, i.e., terms containing x should remain with dx and terms containing y should remain with dy. A function f ( x , y) is said to be a homogeneous function of degree n is (i) f (λx, λy ) = λ n f ( x, y) for some non-zero constant
λ .
or
y or y n h x y x
n (ii) f ( x, y ) = x g
Note : A function f ( x , y) is said to be a homogeneous function of degree zero if (i) f (λx, λy ) = f ( x, y ) or
y or h x y x
(ii) f ( x , y ) = g ●
A homogeneous differential equation of degree zero can be expressed in the form dy dx
y x
= g
or
x = h dy y dx
To solve a homogeneous differential equation of the type dy dx
= f ( x, y ) ,
we make a substitution y = vx
and to solve a homogeneous differential equation of the type dx dy
= G( x , y), make a substitution x = vy.
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A differential equation of the form dy dx
+ Py = Q , where P and Q are either constants or functions of x is know as a first order linear
differential equation in y. Solution of such a differential equation is given by y × I.F =
∫ (Q × I.F)dx + C,
∫ where I.F (Integrating Factor) = e
Pdx
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A differential equation of the form dx dy
+ Px = Q, where p and Q are either constants or functions of y is known as first order linear
differential in x . Solution of such a differential equation is given by x × I.F =
∫ (Q × IF)dy + C
∫ where I.F (Integrating Factor) = e
Pdx
Question for Practice Very Short Answer Type Questions (1 Mark) 1.
Determine order and degree (if defined) of differential equation given below : (i)
d4y dx 4
+ sin( y ′′ ) = 0
order 4, degree not defined
3
(ii)
ds + 4s d 2 s = 0 dt dt 2
(iii) xy
d2y dx 2
order 2, degree 1
2
dy dy + x −y =0 dx dx 2
d 3 y (iv) x 3 dx
order 2, degree 1
4
dy + + y2 = 0 dx
order 3, degree 2
2
(v) y
=
dy dy x + 1 + dx dx
order 1, degree 2
2.
In each of the following verify that the given functions (explicit of implicit) is a solution of the corresponding differential equation : (i) y = cos x + k
dy
:
dx
+ sin x = 0
d2y
+ 9 y = 0
(ii) y = 4sin3x
:
(iii) x + y = tan −1 y
: y 2 y ′ + y 2 + 1 = 0
= A cos 2x − B sin 2x
(iv) y
(v) y = e
− x
+ ax + b
dx 2
d2y dx 2
+ 4 y = 0
d 2 y
x
e
dx 2
=1
Short and Long Answer Type 3.
Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. (i) y 2
xyy ′′ + x( y ′ ) y ′′′
(iii) y = ac3 x + be−2 x
y ′′ − y ′ − 6 y
(iv) ( y − b) 2
2 y ′′ + ( y ′)3
= 4( x − a)
a
= +b
y ′′ +
x
2
− yy′ = 0
=0
(ii) y = ax 2 + bx + c
(v) y 4.
= a(b2 − x 2 )
=0
=0
2
( y′) = 0 x
Form the differential equation of the family of circles touching the y-axis at origin.
2 2 xyy ′ + x
= y2
3
2 2 dy 2 2 d y 1 + dx = r dx 2
5.
Obtain the differential equation of all circles of radius r .
6.
Form the differential equation of the family of ellipses having foci on y-axis and centre at the origin 2
dy d2y dy x + xy y − =0 dx dx 2 dx
7.
Form the differential equation of the family of hyperboles having foci on x-axis and centre at origin xyy ′′ + x( y′ )
8.
2
− yy′ = 0
Find the general solution of the following differential equation 3 (i) sin x
dx dy
= sin y
(ii) sec 2 x tan ydx + sec 2 y tan x dy = 0
cos y −
3 4
cos x +
1 12
cos 3x = c
tan x tan y = c
1
(iii) (1 + y 2 )(1 + log x) dx + x dy = 0 dy
(iv)
dx
= (1 + x
2
2
(1 + log x )2 −1
)(1 + y )
x 2 (v) e 1 − y dx +
tan y = x +
2
y x
dy
=0
(vi) cos x(1 + cos y ) dx − sin y (1 + sin x) dy
x 3
+c
3
xe x − x = 1 − y 2
=0
+c
(1 + sin x )(1 + cos y ) = c
(vii) e x tan ydx + (1 − e x ) sec 2 y dy = 0 9.
= − tan 2 y + c
tan y = c(1 − e ) x
Solve the following initial value problem : 1− 1 x
(i) x( xdy − y dx ) = ydx , y(1) = 1 (ii) y ′ = y cot 2 x, y (π / 4) = 2 (iii) (1 + y 2 )(1 + log x) dx + x dy = 0 , given y = 1, when x = 1
, x ≠ 0.
y = xe
y = ± 2 sin 2x
1 2
(log x )2
+ log x + tan −1 y =
1 x dy = 3 y 3 given that y(0) = (iv) e dx 2
6 y 2 e − x −
π 4
1
= 1
3
10. Show that the given differential equation are homogeneous and solve each of them dy
2 (i) x
dx
= x 2 − 2 y 2 + xy
(ii) ( x3 + y 3 ) dy − x2 y dx = 0 (iii) x
dy
= y − x tan
dx
(v) (2 xe
(
− y)
x
(vi) 1 + e
y
dy dx
) dx + e
x + 2 y x − 2 y
− x 3 3 y 3
= 2 ye y y
+ log | y | = c
x
y x
=c
y − 1 x
cy = log
x
x
= log | x | + c
x sin
y dy − 2 x dy = 0 x
y
2 2
log
y
(iv) ydx + x log x
1
x
2e y
1 − x dy = 0 y
= c − log y x y
y e
+x=c
11. Solve the following differential equation and find the particular solution satisfying the given conditions (i)
dy dx
y ; y = 0 when x = 1 − + cosec y = 0 x
x
y = log | ex | x
cos
(ii) ( x
2
+y
2
3 − x 4
y =
) dx + xydy = 0, y(1) = 1
y − y dx + x dy = 0 x
2 x 2
y = log(cx ) x
2 (iii) x sin
cot
12. For each of the following differential equation find the general or particular solution as the case may be : (i) x
dy dx
+ 2y = x
2 (ii) cos x
dy dx
2
y =
log x
+ y = tan x
(iii) (1 + y ) dx = (tan 2
−1
x 2
16
−
(4 log x − 1) + cx 2
y = (tan x − 1) + ce
y − x) dy
x = (tan
−1
y − 1) + ce xy =
(iv) ydx + ( x − y ) dy = 0 3
2 (v) (1 + x )
dy dx
+ 2 xy =
1 1 + x
2
y (1 + x ) = tan 2
; y = 0 when x = 1
(vi) ye y dx = ( y3 + 2 xe y ) dy , y(0) = 1
x = y ( e 2
−1
− tan−1 y
y 4
4
−1
− tan x
+c
x −
π 4
− e− y ), y ≠ 0 x
(vii) ydx − ( x + 2 y 2 ) dy = 0
y
= 2 y + c
13. Solve the following differential equation :
y dy = y cos y + x x x dx
y = log x + c x
x cos
sin
14. Solve the following differential equation : dy dx
− y = cos x , given that if x = 0, y = 1
y
=
1 2
(sin x − cos x ) +
3 2
e x
15. Find the particular solution of the following differential equation, given that at x = 2, y = 1 : x
dy dx
+ 2 y = x 2 , ( x ≠ 0)
4 y = x 2
16. Find the particular solution of the differential equation : dy dx
+ y cot x = 2 x + x 2 cot x, x ≠ 0
given that y = 0, where x =
π
2 17. Find the particular solution of the differential equation : x
x
2 ye y dx + ( y − 2 x e y ) dy = 0 given that x = 0 when y = 1.
y = x
2
− x
2e y
π2 4
cosec x
+ log y = 2
18. Solve the following differential equation : dy dx
+ sec x.y = tan x , 0 ≤ x < π 2
y (sec x + tan x ) = sec x + tan x − x + c
19. Solve the following differential equation : 2 (1 + x )dy + 2 xydx = cot xdx( x ≠ 0)
y (1 + x ) = log sin x + c 2
20. Solve the following differential equation : x
dy dx
+ y − x + xy cot x = 0 , x ≠ 0
xy sin x = − x cos x + sin x + c
21. Find the particular solution of the differential equation : ( x 3 + x 2 + x + 1)
dy dx
= 2x 2 + x ; y = 1 when x = 0.
22. Solve the following differential equation :
y dx + ( y 2 − x2 log y dy = 0 x x
xy log
x
2
1 + 2log y + 4 y 2 (log y + c) = 0 x
23. Solve the following differential equation :
x cos y + y sin y y − y sin y − x cos y x dy = 0 x x x x dx 24. Solve :
e−2 x − x
tan
dx = 1, x ≠ 0 x dy
y
y x y
− log
= (2
y x
= 2 x + c
x + C ) l
−2
x
25. Solve the following differential equation : (1 + y )(1 + log x) dx + xdy = 0 2
log | x | +
(log x ) 2 2
= − tan −1 y + c
