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Ordinary Differential Equations (ODEs)
Differential Equation:
A differential equation is, in simpler terms, a statement of equality having a derivative or differentials.
An equation involving differentials or differential co-efficient is called a differential equation.
For examples, 2 0
2 =
x d
y
d and ydx+xdy=0 are two differential equations.
Ordinary Differential Equation:
If a differential equation contains one dependent variable and one independent variable, then the differential equation is called ordinary differential equation.
For example, (i) xSinx x
d
dy = ;
(ii) y x x
d y
d 6 tan
4 2
2 + = .
Partial Differential Equation:
If there are two or more independent variables, so that the derivatives are partial, then the differential equation is called partial differential equation.
For example, z
y y z x
x z =
∂ + ∂
∂
∂ .
Order:
By the order of a differential equation, we mean the order of the highest differential co- efficient which appears in it.
For example, 4 0
2
2 + =
dx xdy x
d y
d is a second order differential equation.
Degree:
By the degree of a differential equation, we mean the degree of the highest differential co- efficient after the equation has been put in the form free from radicals and fraction.
For example, 2 0
4 5 2
2 =
+
x d
y x d x
d y
d is a differential equation whose degree is 4.
The degree of 4 3
3 4 4
2
2 = −
+ x
x d
y x d x d
y
d is 9.
4
( )
133 1
4 4
2
2 = −
+ x
x d
y x d x d
y d
2
( )
43 3
4 4
2
2 = −
+ x
x d
y x d x d
y
d .
General Solution:
The solution of a differential equation in which the number of arbitrary constants is equal to the order of the differential equation is called the general solution.
For example, y = ax+b is the general solution of the differential equation 0
2
2 =
x d
y
d ,
where a and b are arbitrary constant.
Particular Solution:
If particular values are given to the arbitrary constants in the general solution, then the solution so obtained is called particular solution.
For example, putting a=2 and b=3 a particular solution of 2 0
2 =
x d
y
d is y = 2x+3.
How to solve 0
2
2 =
x d
y
d ?
Solution:
2 0
2
x = d
y
d , or =0,
dx dy dx
d or
∫
=a,dx d dy
or
∫
du=a where ,dx u=dy
or u=a,
or a,
dx dy=
or
∫
dy=a∫
dx+b, or y=ax+b.which geometrically represents a straight line.
Or one may solve it in the following way :
2 0
2
x = d
y
d , or =0;
dx dy dx
d
since the derivative of dx
dy is zero;
so =
dx
dy constant a (say)
or
∫
dy=a∫
dx+b or y=aqx+b.In fact 2 0
2
x = d
y
d is an ODE of order 2 and has as solution with 2 parameters a and b.
3 Formation of Ordinary Differential Equation:
Eliminating arbitrary constants, we can form ODE.
Form an ODE corresponding to y = ex
(
ACosx+BSinx)
Solution:
Given, y = ex
(
ACosx+BSinx)
Differentiating with respect to x we get
e
(
ACosx BSinx)
e(
ASinx BCosx)
x d
y
d x x
+
− + +
=
= y+ ex
(
−ASinx+BCosx)
Differentiating again with respect to x we get
e
(
ASinx BCosx)
e(
ACosx BSinx)
x d
y d x d
y
d x x
−
− + +
− +
2 =
2
e
(
ASinx BCosx)
e(
ACosx BSinx)
x d
y
d x x
+
− +
− +
=
y y x
d y d x
d y
d −
−
+
=
2 2 0
2
2 − + y=
x d
y d x
d y
d .
Derive an ODE corresponding to all circles lying in a plane.
Derivation:
The equation of all circles lying in a plane is
x2 +y2+2gx+2fy+c =0 … … … (i) Which contains three arbitrary constant g , f and c .
Differentiating (i) thrice successively, we have 2 +2 +2 +2 =0
x d
y f d x g
d y yd x
+ + + =0 x d
y f d x g
d y yd x
1 0
2 2 2
2 2
= +
+
+
x d
y f d x d
y yd x d
y d
1 2 +
(
+)
2 2 =0
+
x d
y f d x y
d y d
1 2
( )
2 2x d
y f d
x y d
y
d = − +
+ … … … … (ii)
4
2 22
( )
3 3 22x d
y d x d
y d x d
y f d y x
d y d x d
y
d = − + −
3 22
( )
3 3x d
y f d y x
d y d x d
y
d = − +
… … … (iii)
Dividing (ii) by (iii)
( )
( )
3 32 2
2 2 2
3 1
x d
y f d y
x d
y f d y
x d
y d x d
y d
x d
y d
+
− +
−
=
+
+
=
2
3 2 3
2 2
1
3 d x
y d x
d y d x
d y d x d
y d
Find the differential equation of the curves xy=Ae2x+B e−2x for different values of A and B. State order and degree of the derived equation.
Solution:
Given equation, xy=Ae2x+B e−2x Differentiating with respect to xwe get y Ae x B e x
dx y
xd + =2 2 −2 −2 Differentiating again with respect to xwe get
Ae x B e x
x d
y d x d
y d dx
y
xd 2 2
2 2
4
4 + −
= + +
dd xy
(
Ae x B e x)
dx y
xd 2 2
2 2
4
2 = + −
+
2 4 0
2 + + − xy=
x d
y d x d
y d dx
y xd
This is the required differential equation. The order of the above equation is 2 and the degree is 1.
Show that Ax2+By2=1is the solution of
dx ydy dx
dy dx
y yd
x =
+ 2
2 2
. Proof:
Given equation, Ax2+By2=1
Differentiating with respect to xwe get 2 +2 =0
dx y Byd Ax
+ =0 dx
y Byd
Ax … … … (i)
5
dx y d x y B
A= − … … … (ii)
Differentiating (i) again with respect to x we get
0
2 2
2 =
+
+ dx
y B d dx
y yd B A
2 2
2
−
−
= dx
y d dx
y yd B A
2 2
2
−
−
=
− dx
y d dx
y yd dx
y d x
y [Using (ii)]
dx ydy dx
dy dx
y yd
x =
+ 2
2 2
.
Find the differential equations of all circles passing through origin and having their centres on the X-axis.
Solution:
The equations of all circles passing through origin and having their centres on the X-axis is x2 +y2+2gx =0 … … … (i)
2gx = −
(
x2 +y2)
+
−
= x
y g x
2 2
2 … … … (ii)
Differentiating (i) with respect to xwe get 2 +2 +2g=0
x d
y yd x
2 2 0
2
2 =
+
−
+ x
y x x d
y yd
x .
