 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
Click for Only Video

### Topic Covered

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. 