### In each of the exercises 1 to 5, from a differential equation representing the given family of curves by eliminating arbitrary constraints a and b.

### Question 1.

**Solution:**

Given:

We can also write

bx + ay = ab

On differentiating we get

b + ay’ = 0

y’ = -b/a

Again differentiating we get

y” = 0

### Question 2.

**Solution:**

Given:

On differentiating we get

2y.y’=-2ax

Again differentiating we get

xyy” + x(y’)

^{2 }– yy’ = 0

### Question 3. y = ae^{3x}+be^{-2x}

**Solution:**

y = ae

^{3x }+ be^{-2x}-(1)On differentiating we get

y’=3ae

^{3x}-2be^{-2x}-(2)Again differentiating we get

y”=9ae

^{3x}+4be^{2x}Now on multiply eq(1) by 6

6y = 6ae

^{3x }+ 6be^{-2x}And add with eq(2)

6y + y’ = 6ae

^{3x }+ 6be^{-2x }+ 3ae^{3x }– 3be^{-2x}6y + y’ = 9ae

^{3x }+ 4be^{-2z }= y”y” – y’ – 6y = 0

### Question 4. y = e^{2x}(a + bx)

**Solution:**

Given: y = e

^{2x}(a + bx) -(1)On differentiating we get

y’ = e

^{2x}(b) + (a + bx).2e^{2x}y’ = e

^{2x}(b + 2a + 2bx) -(2)Now on multiply eq(1) by 2

2y = e

^{2x}(2a + 2bx)And add with eq(2)

y’ – 2y = e

^{2x}(b + 2a + 2bx) – e^{2x}(2a + 2bx)y’ – 2y = be

^{2x}-(3)Again differentiating we get

y” – 2y’ = 2be

^{2x}Now put the value of be

^{2x}from eq(3)y” – 2y’ = 2(y’ – 2y)

y” – 2y’ = 2y’ – 4y

y” – 2y’ – 2y’ + 4y = 0

y” – 4y’ + 4y = 0

### Question 5.

**Solution:**

Given: y = e

^{x}(a cos x + b sin x) -(1)On differentiating we get

y’ = e

^{x}[a cos x + b sin x – a sin x + b cos x]y’ = y + e

^{x}[b cos x – a sin x] -(2)Again differentiating we get

y’ ‘ =y’ + e

^{x}[b cos x – a sin x – b sin x – a cos x]y” = y’ + e

^{x}[b cos x – a sin x] – e^{x}[a cos x + b sin x]From eq(1) and (2), we get

y” = y’ + [y’ – y] – y

y” – 2y’ + 2y = 0

### Question 6. Form the differential equation of the family of circles touching the y-axis at the origin.

**Solution:**

Given that the family of circles touching the y-axis at the origin.

So, the center of the circle is (a, 0) and radius a

Let the equation of a circle is

(x – a)

^{2 }+ y^{2}= a^{2}= x

^{2 }+ y^{2}= 2ax -(1)On differentiating we get

2x + 2yy’ = 2a

x + yy’ = a

Now substitute the value of a in eq(1), we get

x

^{2 }+ y^{2}= 2(x + yy’)xx

^{2 }+ y^{2}= 2x^{2}+ 2xyy’x

^{2 }+ y^{2}– 2x^{2}– 2xyy’y

^{2}= x^{2}+ 2xyy’

### Question 7. Form the differential equation of the family of parabolas having a vertex at origin and axis along positive y-axis.

**Solution:**

Given that the family of parabolas having a vertex at origin and axis along positive y-axis.

So the equation of parabola is:

x

^{2 }= 4ay -(1)On differentiating we get

2x = 4ay’ -(2)

Now divide eq(2) by (1), we have

2x/ x

^{2 }= 4ay’ /4ay2/x

^{ }= y’ /yy’x = 2y

^{ }y’x – 2y = 0

### Question 8. Form the differential equation of the family of ellipses having foci on y-axis and center at the origin.

**Solution:**

Given that the family of ellipses having facii or y-axis & center at the origin.

So the equation of parabola is

On differentiating we get

Again differentiating we get

xyy” + x(y’)

^{2 }– yy’ = 0

### Question 9. Form the differential equation of the family of hyperbolas having foci on the x-axis and center at the origin.

**Solution:**

Given that the family of hyperbolas having foci on the x-axis and center at the origin.

So the equation of hyperbola is

On differentiating we get

Again differentiating we get

### Question 10. Form the differential equation of the family of circles having a center on y-axis and radius 3 units.

**Solution:**

Given that the of circles having a center on y-axis and radius 3 units.

So the center be (0, k)

General equation of the circle is,

x

^{2 }+ (y – k)^{2 }= 3^{2}(y – k)

^{2 }= 9 – x^{2}y – k =

k =

On differentiating we get

Squaring on both sides we get

### Question 11. Which of the following differential equations has y = c_{1}e^{x }+ c_{2}e^{-x} as the general solution?

### (A)

### (B)

### (C)

### (D)

**Solution:**

y = c

_{1}e^{x }+ c_{2}e^{-x}On differentiating we get

y’ = c

_{1}e^{x }– c_{2}e^{-x}Again differentiating we get

y” = c

_{1}e^{x }+ c_{2}e^{-x}y” = y

y” – y = 0

Hence, the correct option is B

### Question 12. Which of the following differential equations has y = x as one of its particular solution?

### (A)

### (B)

### (C)

### (D)

**Solution:**

y = x

On differentiating we get

y’ = 1

Again differentiating we get

y” = 0

Now substitute the value of y, y’ and y” in each option to check for correct option

(A)

= 0 – x

^{2}(1) + x.x = x0 ≠ x

(B)

= 0 + x(1) + x.x = x

= x + x

^{2}≠ x(C)

= 0 – x

^{2}(1) + x.x = 0= 0 = 0

Hence, the correct option is C.