Mathematics Formation of a Differential Equation whose General Solution is given Procedure to form a differential equation that will represent a given family of curves For CBSE-NCERT 3
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Class 12 Chapter 9 - DIFFERENTIAL EQUATIONS

Formation of a Differential Equation whose General Solution is given Procedure to form a differential equation that will represent a given family of curves For CBSE-NCERT 3

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`color {red} **` Formation of a Differential Equation whose General Solution is given
`color {red} **` Procedure to form a differential equation that will represent a given family of curves

Formation of a Differential Equation whose General Solution is given

● We know that the equation

`color{orange} {x^2 + y^2 + 2x – 4y + 4 = 0}` ... (1)

represents a circle having centre at (– 1, 2) and radius 1 unit.

● Differentiating equation (1) with respect to x, we get

`color{orange} {(dy)/(dx) = ( x+1)/(2 -y) \ \ \ \ ( y != 2)}` ........................(2)

which is a differential equation. You will find later on that this equation represents the family of circles and one member of the family is the circle given in equation (1).

● Now Let us consider the equation `color{orange} {x^2 + y^2 = r^2}` ......... (3)

By giving different values to r, we get different members of the family e.g.

`x^2 + y^2 = 1, x^2 + y^2 = 4, x^2 + y^2 = 9` etc. (see Fig).

Thus, equation (3) represents a family of concentric circles centered at the origin and having different radii.

`=>` To find a differential equation that is satisfied by each member of the family. The differential equation must be free from `r` because `r` is different for different members of the family. So the equation, obtained by differentiating equation (3) with respect to x, i.e.,

`color{orange} {2x + 2y (dy)/(dx) = 0}` or ` color{orange} {x + y (dy)/(dx) = 0}` ........ (4)

which represents the family of concentric circles given by equation (3).

`\color{green} ✍️` Again, let us consider the equation

`color{orange} {y = mx + c}` .................... (5)

By giving different values to the parameters m and c, we get different members of the family, e.g.,

`color {green} {y =x \ \ \ \ \ \ (m =1 , c =0) }`

`color {green} { y = sqrt3 x\ \ \ \ \ \ (m = 3 , c = 0)} `

`color {green} {y = x + 1 \ \ \ \ \ \ (m = 1, c = 1)}`

`color {green} {y = – x \ \ \ \ \ \ (m = – 1, c = 0)}`

`color {green} { y = – x – 1 \ \ \ \ \ \ (m = – 1, c = – 1) }` etc. ( see Fig 9.2).

Thus, equation (5) represents the family of straight lines, where m, c are parameters.

`\color{green} ✍️` To find differential equation that is satisfied by each member of the family. The equation must be free from `m` and `c` . This is obtained by differentiating equation (5) with respect to x, successively we get

`color {brown} {(dy)/(dx) = m ,` and ` color {brown} { (d^2y)/(dx^2) = 0`

The equation (6) represents the family of straight lines given by equation (5).

`\color{green} ✍️` Note that equations (3) and (5) are the general solutions of equations (4) and (6) respectively.

Procedure to form a differential equation that will represent a given family of curves

`=>`(a) If the given family `F_1` of curves depends on only one parameter then it is represented by an equation of the form

`color {red} {F_1 (x, y, a) = 0` ................... (1)

● For example, the family of parabolas `y^2 = ax` can be represented by an equation of the form f (x, y, a) :` y^2 = ax`.

Differentiating equation (1) with respect to x, we get an equation involving y′, y, x, and a, i.e.,

`color {red} {g (x, y, y′, a) = 0` ... (2)

● The required differential equation is then obtained by eliminating a from equations (1) and (2) as

`color {red} {F(x, y, y′) = 0` ... (3)

`=>` (b) If the given family F2 of curves depends on the parameters a, b (say) then it is represented by an equation of the from

`color {red} {F_2 (x, y, a, b) = 0` ... (4)

● Differentiating equation (4) with respect to x, we get an equation involving `y′, x, y, a, b`, i.e.,

`color {red} {g (x, y, y′, a, b) = 0` ... (5)

● But it is not possible to eliminate two parameters a and b from the two equations and so, we need a third equation. This equation is obtained by differentiating equation (5), with respect to x, to obtain a relation of the form

`color {red} {h (x, y, y′, y″, a, b) = 0` ... (6)

The required differential equation is then obtained by eliminating a and b from equations (4), (5) and (6) as

`color {red} {F (x, y, y′, y″) = 0` ... (7)


`"Key Concept : "` The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves.
Q 3176091876

Form the differential equation representing the family of curves y = mx, where, m is arbitrary constant.
Class 12 Chapter 9 Example 4
Solution:

We have

y = mx ... (1)

Differentiating both sides of equation (1) with respect to x, we get

`(dy)/(dx) = m`

Substituting the value of m in equation (1) we get `y = (dy)/(dx) * x`

or `x (dy)/(dx) - y = 0`

which is free from the parameter m and hence this is the required differential equation.
Q 3186091877

Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants.
Class 12 Chapter 9 Example 5
Solution:

We have

`y = a sin (x + b)` ... (1)

Differentiating both sides of equation (1) with respect to x, successively we get

`(dy)/(dx) = a cos (x +b)` ...................(2)

`(d^2y)/(dx^2) = - a sin (x + b)` .....................(3)

Eliminating a and b from equations (1), (2) and (3), we get

`(d^2y)/(dx^2) + y = 0` ...................(4)

which is free from the arbitrary constants a and b and hence this the required differential equation.
Q 3116091879

Form the differential equation representing the family of ellipses having foci on x-axis and centre at the origin.
Class 12 Chapter 9 Example 6
Solution:

We know that the equation of said family of ellipses (see Fig 9.3) is

`x^2/a^2 + y^2/b^2 = 1` ..................(1)

Differentiating equation (1) with respect to x, we get `(2x)/a^2 + (2y)/b^2(dy)/(dx) = 0`

or ` y/x ( (dy)/(dx)) = (-b^2)/a^2` ...................(2)

Differentiating both sides of equation (2) with respect to x, we get

`(y/x) ((d^2y)/(dx^2)) + ( (x (dy)/(dx) -y)/x^2 ) (dy)/(dx) = 0`

or `xy (d^2y)/(dx^2) + x ( (dy)/(dx) )^2 - y (dy)/(dx) = 0` ............................(3)

which is the required differential equation
Q 3156191974

Form the differential equation of the family of circles touching the x-axis at origin.
Class 12 Chapter 9 Example 7
Solution:

Let C denote the family of circles touching x-axis at origin. Let (0, a) be the coordinates of the centre of any member of the family (see Fig 9.4). Therefore, equation of family C is

`x^2 + (y – a)^2 = a^2` or `x^2 + y^2 = 2ay` ... (1)

where, a is an arbitrary constant. Differentiating both sides of equation (1) with respect to x,we get

`2x + 2 y (dy)/(dx) = 2a (dy)/(dx)`

or `x +y(dy)/(dx) = a (dy)/(dx) ` or ` a = (x +y (dy)/(dx) ) /( ( dy)/(dx))` ....................(2)

Substituting the value of a from equation (2) in equation (1), we get

`x^2 + y^2 = 2 y ([x +y (dy)/(dx) ] )/((dy)/(dx))`

or ` (dy)/(dx) ( x^2 + y^2) = 2 xy + 2 y^2 (dy)/(dx)`

or ` (dy)/(dx) = (2xy )/( x^2 -y^2)`

This is the required differential equation of the given family of circles.
Q 3106191978

Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Class 12 Chapter 9 Example 8
Solution:

Let P denote the family of above said parabolas (see Fig 9.5) and let (a, 0) be the focus of a member of the given family, where a is an arbitrary constant. Therefore, equation of family P is

`y^2 = 4ax` ..................... (1)

Differentiating both sides of equation (1) with respect to x, we get

`2y (dy)/(dx) = 4 a` .........................(2)

Substituting the value of 4a from equation (2) in equation (1), we get

`y^2 = (2y (dy)/(dx) ) (x)`

or ` y^2 - 2 xy (dy)/(dx) = 0`

which is the differential equation of the given family of parabolas.
 
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