Consider a family of exponential curves (y = Aex), where A is an arbitrary constant for different values of A, we get different members of the family. Differentiating the relation (y = Aex) w.r.t.x, we get
dy/dx = Aex
Eliminating the arbitrary constant between y = Aex and dy/dx = Aex, we get dy/dx = y. This is the differential equation of the family of curves represented by y = Aex
Thus, by eliminating one arbitrary constant, a differential equation of first order is obtained
Now consider the family of curves given by
y = A cos 2x + B sin 2x … (1)
Where A and B are arbitrary consists.
Differentiating (1) w.r.t, x we get
dy/ dx = – 2Asin 2x + 2Bcos 2x … (2))
Differentiating (2) w.r.t. x we get
d²y/ dx² = – 4Acos ax – 4Bsin 2x … (3))
Eliminating A and B from equations (1) and (2) (3), we get
d²y/ dx² = – 4y ⇒ d²y/ dx² + 4y = 0
Here we note that by eliminating two arbitrary consists, a differential equation of second order is obtained.
Step I: write the given equation involving independent variable x (say), dependent variable y (say) and the arbitrary constants.
Step II: obtained the numbers of a arbitrary constants in step in step I. let there be n arbitrary consists.
Step III: differentiate the relation in step in times with respect to x.
Step IV: eliminate arbitrary constants with the help of n equations involving differential coefficient obtained in step III and an equation in step I.
The equation so obtained is the desired differential equation.
Example: Show that the differential equation that represented all parabolas having their axis symmetry coincident with the axis of x is yy₂ + y₁² = 0
Solution: The equation that represents a family of parabolas having their axis of symmetry coincident with the axis of x is
y² = 4a(x – h) … (1)
This equation contains two arbitrary constants, so we shall differentiate twice to obtain second order differential equation.
Differentiating (1) w.r.t x we get
2y dy/dx = 4a ⇒ y dy/ dx = 2a … (2))
Differentiating (2) w.r.t, x we get
y d²y/ dx² + (dy/ dx)² = 0 ⇒ yy₂ + y₁² = 0
Which is the required differential equation.
Example: Find the differential equation of all non-horizontal lines in a plane.
Solution: The equation of the family of all non-horizontal line in a plane is given by
Ax + by = 1 … (1)
Where a, b are arbitrary constants such that (a ≠ 0)
Differentiating (1) w.r.t, x we get
a dx/dy + b = 0
Differentiating this w.r.t y we get
ad²x/ dy² = 0
⇒ d²x/ dy² = 0
Hence, the differential equation of all non-horizontal lines in a plane is d²x/ dy² = 0.
Example: Find the differential equation of all non- vertical lines in a plane.
Solution: The general equation of all non-vertical lines in a plane is (ax + by = 1) where (b ≠ 0)
ax + by = 1
a + b dy/dx = 0 [differentiating w.r.t.x]
b d²y/ dx = 0 [differentiating w.r.t.x]
d²y/ dx² = 0 [∵ b ≠ 0]
Hence, the differential equation is d²y/ dx² = 0
Solution of a differential equation: The solution of a differential equation is a relation between the variable involved which satisfies the differential equation. Such a relation and the derivates obtained therefore when substituted in the differential equation, makes left hand right hand sides identically equal.
General solution: The solution which contains as many as arbitrary constants as the order of the differential equations is called the general solution of the differential equation.
For example, y = Acos x + Bsin x is the general solutions one arbitrary constant.
Particular solution: Solution obtained by giving particular values to the arbitrary constant in the general solution of a differential equation is called a particular solution.
Example: Show that xy = aex + be– x + x² is a solution of the differential equation
x d²y/ dx² + 2 dy/dx – xy + x² – 2 = 0
Solution: We are given that
xy = aex + be– x + x² … (1)
Differentiating w.r.t.x, we get
x dy/ dx + y = aex – be– x + 2x
Differentiating again w.r.t.x, we get
xd²y/ dx² + dy/dx + dy/dx = aex + be– x + 2
xd²y/ dx² + 2dy/dx = aex + be– x + 2 … (2)
Now x d²y/ dx² + 2dy/dx – xy + x2 – 2
= [aex + be– x + 2] – [aex + be– x + x²] + x² – 2
= 0 [using (1) and (2)]
Thus, (xy = aex + be– x + x²) is a solution of the given differential equation.