# Formation of differential equations

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

⇒ 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)

Now,

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.