Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.2 | Set 1

Differential equations are mathematical tools that describe relationships involving rates of change. They are central in many fields such as physics, engineering, economics, and biology, because they effectively model dynamic systems. A differential equation involves a function and its derivatives, expressing how the function changes over time or space.

Question 1. Form the differential equation of the family of curves represented by y² = (x - c)³

Solution:

y² = (x - c)³

On differentiating the given equation w.r.t x,

2y(dy/dx) = 3(x - c)²

(x - c)² = (2y/3)(dy/dx)

(x - c) = [(2y/3)(dy/dx)]^1/2          -(1)

On putting the value of (x - c) in equation (1), we get

y² = [(2y/3)(dy/dx)]^3/2

On squaring both side, we get

y⁴ = [(2y/3)(dy/dx)]³

y⁴ = (8y³/27)(dy/dx)³

27y⁴ = 8y³(dy/dx)³

27y = 8(dy/dx)³

Question 2. Form the differential equation corresponding to y = e^mx by eliminating m.

Solution:

y = e^mx          -(1)

On differentiating the given equation w.r.t x,

dy/dx = me^mx          -(2)

From eq(1), we get

y = e^mx

logy = mx

m = (logy/x)

Now, put the value of m in equation(2), we get

x(dy/dx) = ylogy

Question 3. Form the differential equations from the following primitives where constants are arbitrary.

(i) y² = 4ax

Solution:

y² = 4ax         -(1)

y²/4x = a

On differentiating the given equation w.r.t x,

2y(dy/dx) = 4a         -(2)

Now, put the value of a in equation(2), we get

2y(dy/dx) = 4(y²/4x)  

2y(dy/dx) = y²/x

2x(dy/dx) = y

(ii) y = cx + 2c² + c³

Solution:

y = cx + 2c² + c³          -(1)

On differentiating the given equation w.r.t x,

dy/dx = c         -(2)

Now, put the value of c in equation(1), we get

y = x(dy/dx) + 2(dy/dx)² + (dy/dx)³

(iii) xy = a²  

Solution:

xy = a²          -(1)

On differentiating the given equation w.r.t x,

x(dy/dx) + y = 0

(iv) y = ax² + bx + c   

Solution:

y = ax² + bx + c         -(1)

On differentiating the given equation w.r.t x,

dy/dx = 2ax + b         -(2)

Again differentiating the given equation w.r.t x,

d²y/dx² = 2a         -(3)

Again, differentiating the given equation w.r.t x, we get

d³y/dx³ = 0  

Question 4. Find the differential equation of the family of curves y = Ae^2x + Be^-2x where A and B are arbitrary constants.

Solution:

y = Ae^2x + Be^-2x          -(1)

On differentiating the given equation w.r.t x,

(dy/dx) = 2Ae^2x - 2Be^-2x          -(2)

Again, differentiating the given equation w.r.t x,

d²y/dx² = 4Ae^2x + 4Be^-2x

d²y/dx² = 4(Ae^2x + Be^-2x)

d²y/dx² = 4y

Question 5. Find the differential equation of the family of curves, x = Aconst + Bsinnt where A and B are arbitrary constants.

Solution:

x = Acosnt + Bsinnt         -(1)

On differentiating the given equation w.r.t x,

dy/dx = -Ansinnt + Bncosnt         -(2)

Again, differentiating the given equation w.r.t x,

d²y/dx² = -An²cosnt - Bn²sinnt 

d²y/dx² = -n²(Acosnt + Bsinnt) 

d²y/dx² + n²x = 0

Question 6. Form the differential equation corresponding to y² = a(b - x²) by eliminating a and b.

Solution:

y² = a(b - x²) 

On differentiating the given equation w.r.t x,

2y(dy/dx) = a(0 - 2x)

Again, differentiating the given equation w.r.t x,

x[

Question 7. Form the differential equation corresponding to y² - 2ay + x² = a² by eliminating a.

Solution:

y² - 2ay + x² = a²          -(1)

On differentiating the given equation w.r.t x,

2y(dy/dx) - 2a(dy/dx) + 2x = 0

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

    -(2)

Let us considered dy/dx = y' 

On putting the value of 'a' in the eq(1), we get 

On solving this equation, we get

(x² - 2y²)y'² - 4xyy' - x² = 0

(x² - 2y²)(dy/dx)² - 4xy(dy/dx) - x² = 0

Question 8. Form the differential equation corresponding to (x - a)² + (y - b)² = r² by eliminating a and b.

Solution:

(x - a)² + (y - b)² = r²          -(1)

On differentiating the given equation w.r.t x,

2(x - a) + 2(y - b)(dy/dx) = 0   

(x - a) + (y - b)(dy/dx) = 0         -(2)

Again, differentiating the given equation w.r.t x,

1 + (y - b)(d²y/dx²) + (dy/dx)(dy/dx) = 0    

        -(3)

On putting the value of (y - b) in eq(2),

(x - a)(d²y/dx²) - (dy/dx)³ - (dy/dx) = 0  

     -(4)

On putting the value of (x - a) and (y - b) in eq(1)

Put (dy/dx) = y' and d²y/dx² = y'' 

y'²(1 + y'²)² + (1 + y'²)² = r²y''²

Question 9. Find the differential equation of all the circles which pass through the origin and whose centres lie on-axis.

Solution:

Equation of the circle is (x - a)² + (y - b)² = r² 

Here, a and b are the centre of the circle.

(x - a)² + (y - b)² = r²          -(1)

When the centre lies on the y-axis, so a = 0 

x² + (y - b)² = r²          -(2)

So, when the circle is passing through origin, so the equation is

0² + b² = r²          -(3)

x² + (y - b)² = r²

On squaring both side, we have

x² + y² - 2yr + r² = r²          -(Since  r² = b²)

2yr = x² + y²

r = (x² + y²)(2y)

On differentiating the equation(1) w.r.t x, we get

0 = 4xy + 4y²(dy/dx) - 2x²(dy/dx) - 2y²(dy/dx)

0 = y²(dy/dx) - x²(dy/dx) + 2xy

(x² - y²)(dy/dx) = 2xy 

Question 10. Find the differential equation of all the circles which pass through the origin and whose centres lie on the x-axis.

Solution:

Equation of the circle is (x - a)² + (y - b)² = r²

Here, a and b are the centre of the circle.

When the centre lies on the x-axis, so b = 0 

(x - a)² + y² = r²          -(1)

When the circle is passing through origin, so the equation is

a² + 0² = r²         -(2)

(x - a)² + y² = r²

On squaring both side, we get,

x² - 2ax + a² + y² = r²

x² + y² - 2xr + r² = r²           -(Since  r² = a²)

2xr = x² + y² 

r = (x² + y²)(2x)         -(3)

On differentiating the equation w.r.t x, we get

0 = 2x² + 2xy(dy/dx) - x² - y²

(x² - y²) + 2xy(dy/dx) = 0

Practice Questions

1).Form the differential equation of all circles with center at the origin.

2).Find the differential equation of all parabolas having their axis parallel to the y-axis.

3).Form the differential equation of all curves for which the slope of the tangent at any point is equal to the square of the x-coordinate of that point.

4).Derive the differential equation of all curves whose subtangent is constant and equal to 'a'.

5).Form the differential equation of all curves in which the sum of the intercepts on the coordinate axes is always equal to 'a'.

6).Find the differential equation of the family of curves y = ax² + bx + c, where a, b, and c are arbitrary constants.

7).Form the differential equation of all straight lines passing through the point (1, 2).

8).Derive the differential equation of all curves for which the perpendicular from the origin to the tangent at any point is equal to the abscissa of that point.

9).Find the differential equation of all circles passing through the points (0, 0) and (a, 0).

10).Form the differential equation of all ellipses with center at the origin and eccentricity e.

Summary

Chapter 22 of RD Sharma's Class 12 mathematics textbook focuses on Differential Equations, with Exercise 22.2 specifically dealing with the formation and solution of differential equations. This exercise helps students understand how to derive differential equations from given conditions or equations, and how to solve first-order differential equations using various methods such as separation of variables, integrating factors, and substitution. The problems in this exercise range from simple equation formations to more complex applications in physics and geometry, providing a comprehensive practice for students to develop their skills in handling differential equations.

Q: What is a differential equation?

A: A differential equation is an equation that involves derivatives of a function.

Q: What is the order of a differential equation?

A: The order of a differential equation is the highest order of derivative present in the equation.

Q: What is meant by the formation of a differential equation?

A: Formation of a differential equation involves deriving an equation containing derivatives from given conditions or a family of functions.

Q: What is the degree of a differential equation?

A: The degree of a differential equation is the power of the highest order derivative after the equation has been made free from fractions and radicals in derivatives.

Q: What is a general solution of a differential equation?

A: A general solution is a solution that involves arbitrary constants and represents a family of functions satisfying the differential equation.