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

### Question 1. Form the differential equation of the family of curves represented by y^{2 }= (x – c)^{3}

**Solution:**

y

^{2 }= (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

y

^{2}= [(2y/3)(dy/dx)]^{3/2}On squaring both side, we get

y

^{4 }= [(2y/3)(dy/dx)]^{3}y

^{4}= (8y^{3}/27)(dy/dx)^{3}27y

^{4 }= 8y^{3}(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) y^{2 }= 4ax

**Solution:**

y

^{2 }= 4ax^{ }-(1)y

^{2}/4x = aOn 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(y

^{2}/4x)2y(dy/dx) = y

^{2}/x2x(dy/dx) = y

### (ii) y = cx + 2c^{2 }+ c^{3}

**Solution:**

y = cx + 2c

^{2 }+ c^{3 }-(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 = a^{2 }

**Solution:**

xy = a

^{2 }-(1)On differentiating the given equation w.r.t x,

x(dy/dx) + y = 0

### (iv) y = ax^{2 }+ bx + c

**Solution:**

y = ax

^{2 }+ 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,

d

^{2}y/dx^{2 }= 2a^{ }-(3)Again, differentiating the given equation w.r.t x, we get

d

^{3}y/dx^{3 }= 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,

d

^{2}y/dx^{2 }= 4Ae^{2x }+ 4Be^{-2x}d

^{2}y/dx^{2 }= 4(Ae^{2x }+ Be^{-2x})d

^{2}y/dx^{2 }= 4y

### Question 5. Find the differential equation of the family of curves, x = Acosnt + 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,

d

^{2}y/dx^{2 }= -An^{2}cosnt – Bn^{2}sinntd

^{2}y/dx^{2 }= -n^{2}(Acosnt + Bsinnt)d

^{2}y/dx^{2 }+ n^{2}x = 0

### Question 6. Form the differential equation corresponding to y^{2} = a(b – x^{2}) by eliminating a and b.

**Solution:**

y

^{2 }= a(b – x^{2})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 y^{2 }– 2ay + x^{2 }= a^{2 }by eliminating a.

**Solution:**

y

^{2 }– 2ay + x^{2 }= a^{2 }-(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

(x

^{2 }– 2y^{2})y’^{2 }– 4xyy’ – x^{2 }= 0(x

^{2 }– 2y^{2})(dy/dx)^{2 }– 4xy(dy/dx) – x^{2 }= 0

### Question 8. Form the differential equation corresponding to (x – a)^{2 }+ (y – b)^{2 }= r^{2 }by eliminating a and b.

**Solution:**

(x – a)

^{2 }+ (y – b)^{2 }= r^{2 }-(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)(d

^{2}y/dx^{2}) + (dy/dx)(dy/dx) = 0

^{ }-(3)On putting the value of (y – b) in eq(2),

(x – a)(d

^{2}y/dx^{2}) – (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 d

^{2}y/dx^{2}= y”y’

^{2}(1 + y’^{2})^{2 }+ (1 + y’^{2})^{2 }= r^{2}y”^{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 }= r^{2 }Here, a and b are the centre of the circle.

(x – a)

^{2 }+ (y – b)^{2 }= r^{2 }-(1)When the centre lies on the y-axis, so a = 0

x

^{2 }+ (y – b)^{2 }= r^{2 }-(2)So, when the circle is passing through origin, so the equation is

0

^{2 }+ b^{2 }= r^{2 }-(3)x

^{2 }+ (y – b)^{2 }= r^{2}On squaring both side, we have

x

^{2 }+ y^{2 }– 2yr + r^{2 }= r^{2}-(Since r^{2 }= b^{2})2yr = x

^{2 }+ y^{2}r = (x

^{2 }+ y^{2})(2y)On differentiating the equation(1) w.r.t x, we get

0 = 4xy + 4y

^{2}(dy/dx) – 2x^{2}(dy/dx) – 2y^{2}(dy/dx)0 = y

^{2}(dy/dx) – x^{2}(dy/dx) + 2xy(x

^{2 }– y^{2})(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 }= r^{2}Here, a and b are the centre of the circle.

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

(x – a)

^{2 }+ y^{2 }= r^{2 }-(1)When the circle is passing through origin, so the equation is

a

^{2 }+ 0^{2 }= r^{2}^{ }-(2)(x – a)

^{2 }+ y^{2 }= r^{2}On squaring both side, we get,

x

^{2 }– 2ax + a^{2 }+ y^{2 }= r^{2}x

^{2 }+ y^{2 }– 2xr + r^{2 }= r^{2}-(Since r^{2 }= a^{2})2xr = x

^{2 }+ y^{2 }r = (x

^{2 }+ y^{2})(2x)^{ }-(3)On differentiating the equation w.r.t x, we get

0 = 2x

^{2 }+ 2xy(dy/dx) – x^{2 }– y^{2}(x

^{2 }– y^{2}) + 2xy(dy/dx) = 0