# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.2 | Set 2

### Question 11. Assume that a raindrop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the raindrop.

**Solution:**

Let us considered ‘r’ be the radius of rain drop, volume of the drop be ‘V’ and area of the drop be ‘A’

(dV/dt) proportional to A

(dV/dt) – kA -(V decreases with increasing in t so negative sing)

Here, k is proportionality constant,

= -k(4π r

^{2})4πr

^{2}(dr/dt) = -k(4πr^{2})(dr/dt) = -k

### Question 12. Find the differential equation of all the parabolas with latus rectum 4a’ and whose axes are parallel to the x-axis.

**Solution:**

Equation of parabola whose area is parallel to x-axis and vertices at (h, k).

(y – k)

^{2 }= 4a(x – h) -(1)On differentiating w.r.t x,

2(y – k)(dy/dx) = 4a

(y – k)(dy/dx) = 2a

-(2)

Again, differentiating w.r.t x,

d

^{2}y/dx^{2}(y – k) + (dy/dx)(dy/dx) = 02a(d

^{2}y/dx^{2}) + (dy/dx)^{3 }= 0

### Question 13. Show that the differential equation of which is a solution, is (dy/dx) + 2xy = 4x^{3}

**Solution:**

-(1)

On differentiating w.r.t x,

On adding 2xy in R.H.S and L.H.S,

On putting the value of y in above equation,

=

(dy/dx) + 2xy = 4x

^{3}

### Question 14. From the differential equation having y = (sin^{-1}x)^{2} + A cos^{-1}x + B, where A and B are arbitrary constants, as its general solution.

**Solution:**

y = (sin

^{-1}x)^{2}+ A cos^{-1}x + BOn differentiating w.r.t x,

Again, on differentiating w.r.t x,

### Question 15. Form the differential equation of the family of curves represented by the equation (a being the parameter)

### (i) (2x + a)^{2 }+ y^{2 }= a^{2}

**Solution:**

(2x + a)

^{2 }+ y^{2 }= a^{2}-(1)On differentiating w.r.t x,

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

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

a = -2x – y(dy/dx) -(2)

On putting the value of ‘a’ in eq(1), we have

y

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

^{2 }– 4x^{2 }– 4xy(dy/dx) = 0

### (ii) (2x – a)^{2 }– y^{2 }= a^{2}

**Solution:**

(2x – a)

^{2 }– y^{2 }= a^{2}4x

^{2 }– 4ax + a^{2 }– y^{2 }= a^{2 }4ax = 4x

^{2 }– y^{2}a = (4x

^{2 }– y^{2})/4xOn differentiating w.r.t x,

4x

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

### (iii) (x – a)^{2 }+ 2y^{2 }= a^{2 }

**Solution:**

(x – a)

^{2 }+ 2y^{2 }= a^{2}-(1)On differentiating w.r.t x,

2(x – a) + 4y(dy/dx) = 0

(x – a) + 2y(dy/dx) = 0

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

On putting the value of a in eq(1)

2y

^{2 }– 4xy(dy/dx) – x^{2 }= 0

### Question 16. Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

### (i) x^{2 }+ y^{2 }= a^{2}

**Solution:**

x

^{2 }+ y^{2 }= a^{2}On differentiating w.r.t x,

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

x + y(dy/dx) = 0

### (ii) x^{2 }– y^{2 }= a^{2}

**Solution:**

x

^{2 }– y^{2 }= a^{2}On differentiating w.r.t x,

2x – 2y(dy/dx) = 0

x – y(dy/dx) = 0

### (iii) y^{2 }= 4ax

**Solution:**

y

^{2 }= 4ax(y2/x) = 4a

On differentiating w.r.t x,

2xy(dy/dx) – y

^{2 }= 02x(dy/dx) – y = 0

### (iv) x^{2 }+ (y – b)^{2 }= 1

**Solution:**

x

^{2 }+ (y – b)^{2 }= 1 -(1)On differentiating w.r.t x,

2x + 2(y – b)(dy/dx) = 0

On putting the value of (y – b) in eq(1)

x

^{2}(dy/dx)^{2 }+ x^{2 }= (dy/dx)^{2}x

^{2}[(dy/dx)^{2 }+ 1] = (dy/dx)^{2}

### (v) (x – a)^{2 }– y^{2 }= 1

**Solution:**

(x – a)

^{2 }– y^{2 }= 1 -(1)On differentiating w.r.t x,

2(x – a) – 2y(dy/dx) = 0

(x – a) – y(dy/dx) = 0

(x – a) = y(dy/dx)

On putting the value of (y – b) in eq(i), we get

y

^{2}(dy/dx)^{2 }– y^{2 }= 1y

^{2}[(dy/dx)2 – 1] = 1

### (vi)

**Solution:**

We have,

-(1)

{(bx)

^{2 }– (ay)^{2}} = (ab)^{2}-(2)On differentiating w.r.t x,

2xb

^{2 }– 2a^{2}y(dy/dx) = 0xb

^{2 }– a^{2}y(dy/dx) = 0 -(3)Again, differentiating w.r.t x,

On putting the value of b

^{2}in equation(3), we getxb

^{2 }– a^{2}y(dy/dx) = 0

### (vii) y^{2 }= 4a(x – b)

**Solution:**

We have,

y

^{2 }= 4a(x – b)On differentiating w.r.t x,

2y(dy/dx) = 4a

Again differentiating w.r.t x,

[(dy/dx)

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

### (viii) y = ax^{3}

**Solution:**

We have,

y = ax

^{3}-(1)On differentiating w.r.t x,

(dy/dx) = 3ax

^{2}From eq(1),

a=(y/x

^{3}-(1)On putting the value of a in eq(1)

dy/dx = 3(y/x

^{3}) × x^{2}x(dy/dx) = 3y

### (ix) x^{2 }+ y^{2 }= ax^{3}

**Solution:**

We have,

x

^{2 }+ y^{2 }= ax^{3}a = (x

^{2 }+ y^{2})/(x^{3})On differentiating w.r.t x,

2x

^{3}y(dy/dx) = x^{4 }+ 3x^{2}y^{2}2x

^{3}y(dy/dx) = x^{2}(x^{2 }+ 3y^{2})2xy(dy/dx) = (x

^{2 }+ 3y^{2})

### (x) y = e^{ax}

**Solution:**

We have,

y = e

^{ax}-(1)On differentiating w.r.t x,

dy/dx = ae

^{ax}dy/dx = ay -(2)

y = e

^{ax }On taking log both side, we get

logy = ax

a = (logy/x)

Now, put the value of ‘a’ in eq(2)

(dy/dx) = logy/x) × y

x(dy/dx) = ylogy

### Question 17. Form the differential equation representing the family of ellipses having foci on the x-axis and the centre at the origin.

**Solution:**

We have,

Equation of ellipse having foci on the x-axis,

-(where a > b)

(bx)

^{2 }+ (ay)^{2 }= (ab)^{2}-(1)On differentiating above equation w.r.t x,

2b

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

^{2}x + a^{2}y(dy/dx) = 0 -(2)Again, differentiating w.r.t x,

On putting the value of b

^{2}in eq(2),xb

^{2 }+ a^{2}y(dy/dx) = 0x[y(d

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

### Question 18. Form the differential equation of the family of hyperbolas having foci on the X-axis and centre at the origin

**Solution:**

We have,

Equation of a hyperbola having a Centre at the origin and foci along x-axis

-(1)

(bx)

^{2}-(ay)^{2}=(ab)^{2}-(2)On differentiating above equation w.r.t x,

2xb

^{2}-2a^{2}y(dy/dx)=0xb

^{2}-a^{2}y(dy/dx)=0 -(3)Again, differentiating above equation w.r.t x,

Putting the value of b

^{2}in equation(3),xy(d

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

^{2}y/dx^{2}) +(dy/dx)^{2}] = y(dy/dx)This is required differential equation.

### Question 19. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axis.

**Solution:**

We have,

Let (-a, a) be the coordinates of the centre of circle

So, the equation of circle is given by,

(x + a)

^{2 }+ (y – b)^{2 }= a^{2}-(1)x

^{2 }+ 2ax + a^{2 }+ y^{2 }– 2ay + a^{2 }= 0 -(2)On differentiating above equation w.r.t x,

2x + 2a + 2y(dy/dx) – 2a(dy/dx) = 0

x + a + y(dy/dx) – a(dy/dx) = 0

On substituting the value of ‘a’ in eq(2)

Let, (dy/dx) = p

[xp – x + x + yp]

^{2 }+ [yp – y – x – yp]^{2 }= [x + yp]^{2}(x + y)

^{2}p^{2 }+ (x + y)^{2 }= (x + yp)^{2}(x + y)

^{2}[p^{2 }+ 1] = (x + yp)^{2}-(where (dy/dx) = p)