This set of Class 12 Maths Chapter 9 Multiple Choice Questions & Answers (MCQs) focuses on “Linear First Order Differential Equations – 1”.

1. What is the differential equation whose solution represents the family c(y + c)^{2} = x^{3}?

a) [2x/3 *(dy/dx) – y][x/3 *dy/dx]^{2} = x^{3}

b) [x/3 *(dy/dx) – y][x/3 *dy/dx]^{2} = x^{3}

c) [2x/3 *(dy/dx) – y][ 2x/3 *dy/dx]^{2} = x^{3}

d) [x/3 *(dy/dx) – y][ 2x/3 *dy/dx]^{2} = x^{3}

View Answer

Explanation: The given family is c(y + c)

^{2}= x

^{3}

Differentiating once, we get

c[2(y + c)]dy/dx = 3x

^{2}

=> 2x

^{3}(y + c)/(y + c)

^{2}* dy/dx = 3x

^{2}

=> 2x

^{3}/(y + c) * dy/dx = 3x

^{2}

Or, 2x (y + c)/(y + c)

^{2}* dy/dx = 3

=> 2x/3 *[dy/dx] = (y + c)

=> c = 2x/3 *[dy/dx] – y

Substituting c back into equation (1), we get

[2x/3 *(dy/dx) – y][ 2x/3 *dy/dx]

^{2}= x

^{3}

which is the required differential equation

2. What is the value of the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?

a) -c/3

b) 3c/2

c) 2c/3

d) -2c/3

View Answer

Explanation: dy/dx = (6x + 9y – 7)/(2x + 3y – 6)

So, dy/dx = (3(2x + 3y) – 7)/(2x + 3x – 6) ……….(1)

Now, we put, 2x + 3y = z

Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]

Or, dy/dx = 1/3(dz/dx – 2)

Therefore, from (1) we get,

1/3(dz/dx – 2) = (3z – 7)/(z – 6)

Or, dz/dx = 2 + (3(3z – 7))/(z – 6)

= 11(z – 3)/(z – 6)

Or, (z – 6)/(z – 3) dz = 11 dx

Or, ∫(z – 6)/(z – 3) dz = ∫11 dx

Or, ∫(1 – 3/(z – 3)) dz = 11x + c

Or, z – log |z – 3| = 11x + c

Or, 2x + 3y – 11x – 3log|2x + 3y -3| = c

Or, 3y – 9x – 3log|2x + 3y – 3| = c

Or, 3x – y + log|2x + 3y – 3| = -c/3

3. What is the order of the differential equation of the family of circles with one diameter along the line y = x axis?

a) 3

b) 2

c) 1

d) 0

View Answer

Explanation: Centre of the circles can be taken as (a, a) and radius as r for some real numbers a and r.

Thus, the family is a two parameter family.

Hence, order of the corresponding differential equation is 2.

4. What will be the general solution of differential equation for 2^{x – y} dx + 2^{y – x} dy = 0?

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

b) 2(2^{2x} + 2^{2y}) = c

c) 3(2^{2x} + 2^{2y}) = c

d) 4(2^{2x} + 2^{2y}) = c

View Answer

Explanation: We have, 2

^{x – y}dx + 2

^{y – x}dy = 0

Or, (2

^{x}/2

^{y})dx + (2

^{y}/2

^{x})dy = 0

Integrating both sides,

∫2

^{2x}dx + ∫2

^{2y}dy = k

Taking k = c/2log2 as, k, c and 2log2 are constant.

Or, 2

^{2x}/2log2 + 2

^{2y}/2log2 = c/2log2

Or, 2

^{2x}+ 2

^{2y}= c

5. What will be the value of dy/dx = (x + 2y + 3)/(2x + 3y + 4)?

a) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) + (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]

b) [(2 + √3)/2√3 * (log (√3(y + 2) + (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) + (x – 1)))]

c) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]

d) [(2 + √3)/2√3 * (log (√3(y + 2) + (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]

View Answer

Explanation: Put x = X + h, Y = Y + k,

We have, dY/dX = (X + 2Y +(h + 2k + 3))/ 2X + 3Y + (2h + 3k + 4)

So, (a – b)x = (a – b)

To determine h and k we set,

2h + 3k + 4 = 0 and h + 2k + 3 = 0

=> h = 1 and k = – 2

Therefore, dY/dX = (X + 2Y) / (2X + 3Y)

Putting Y = VX, we get,

V + X dV/dX = (1 + 2V)/(2 + 3V)

= (1 + 2V)/(3V

^{2}– 1)*dV = -dX/X

=> [(2 + √3)/(2(√3V – 1)) – (2 – √3)/(2(√3V – 1))] dV = -dX/X

Simplifying it further, we get;

[(2 + √3)/2√3 * (log (√3Y – X)) – (2 – √3)/2√3 * (log (√3Y – X))]

Where, X = x – 1 and Y = y + 2

6. What will be the value of the differential equation dy/dx = (x + y)^{2}/(x + 2)(y – 2)?

a) ke^{Y/X} where, X = x + 2 and Y = y – 2

b) ke^{2Y/X} where, X = x + 2 and Y = y – 2

c) ke^{2Y/X} where, X = x and Y = 2y

d) ke^{2X/Y} where, X = x – 2 and Y = y + 2

View Answer

Explanation: Let, X = x + 2 and Y = y – 2

Then, dY/dX = (X + Y)

^{2}/XY

So, let, Y = vX

dY/dX = v + x dv/dX

=>v + Xdv/dX = (v + 1)

^{2}/v

=>v/1 + 2v dv = dX/X

=> (1 – 1/1 + 2v)dv = 2dx/x

=> v – ½ log (1 + 2v) = 2 log X + c

This means, X

^{4}(1 + 2Y/X)

= ke

^{2Y/X}where, X = x +2 and Y = y – 2

7. What will be the value of x(x – 1)dy/dx – y = x^{2}(x – 1)^{2}?

a) xy = (x + 1)(x^{3}/3 + c)

b) depends on x

c) depends on y

d) xy = (x – 1)(x^{3}/3 + c)

View Answer

Explanation: x(x – 1)dy/dx – y = x

^{2}(x – 1)

^{2}

or, dy/dx – 1/(x(x – 1))*y = x(x – 1) ……(1)

we have, ∫- 1/(x(x – 1)) dx = – ∫ [1/(x – 1) – 1/x] dx

= – [log(x – 1) – log x]

= log x – log(x – 1)

= log(x/(x – 1))

Thus, integrating factor = e

^{∫- 1/(x(x – 1)) dx}= x/(x – 1)

Thus, multiplying both sides of (1) by x/(x – 1), we get,

x/(x – 1)*dy/dx – 1/(x – 1)

^{2}* y = x

^{2}

or, d/dx[x/(x – 1) * y] = x

^{2}…….(2)

integrating both sides of (2) we get,

x/(x – 1) * y = ∫x

^{2}dx = x

^{3}/3 + c

or xy = (x – 1)(x

^{3}/3 + c)

8. What is the solution of the equation(y + x(√xy)(x + y))dx – (y + y(√xy)(x + y))dy = 0?

a) x^{2} + y^{2} = 2tan^{-1}(√(y/x)) + c

b) x^{2} + y^{2} = 4tan^{-1}(√(y/x)) + c

c) x^{2} + y^{2} = tan^{-1}(√(y/x)) + c

d) x^{2} + y^{2} = 2tan^{-1}(√(x/y)) + c

View Answer

Explanation: The given equation can be written as on simplifying,

xdx + ydy + (ydx – xdy)/(√xy)(x + y) = 0

the above equation can be written as,

or, 1/2d(x

^{2}+ y

^{2}) = (xdy – ydx)/(x

^{2}(√y/x) (1 + y/x))

= 2/(1 + y/x)* d(√y/x)

Thus, x

^{2}+ y

^{2}= 4tan

^{-1}(√(y/x)) + c

9. A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant

length k. Then what is the differential equation describing such curve?

a) y(dy/dx) = ± √(k^{2} + x^{2})

b) y(dy/dx) = ± √(k^{2} – 2y^{2})

c) y(dy/dx) = ± √(k^{2} – y^{2})

d) Depends on the value of k

View Answer

Explanation: Equation of the normal at a point P(x, y) is given by

Y-y = -1/(dy/dx)(X – x) ….(1)

Let the point Q at the x-axis be (x

_{1}, 0).

From (1), we get

y(dy/dx) = x

_{1}– x ….(2)

Now, giving that PQ

^{2}= k

^{2}

Or, x

_{1}– x + y

^{2}= k

^{2}

=> y(dy/dx) = ± √(k

^{2}– y

^{2})

10. A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant

length k. What is the equation of such curve passing through (0, k)?

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

b) 2x^{2} + y^{2} = k^{2}

c) x^{2} – y^{2} = k^{2}

d) x^{2} + y^{2} = k^{2}

View Answer

Explanation: Equation of the normal at a point P(x, y) is given by

Y-y = -1/(dy/dx)(X – x) ….(1)

Let the point Q at the x-axis be (x

_{1}, 0).

From (1), we get

y(dy/dx) = x

_{1}– x ….(2)

Now, giving that PQ

^{2}= k

^{2}

Or, x

_{1}– x + y

^{2}= k

^{2}

=> y(dy/dx) = ± √(k

^{2}– y

^{2}) ….(3)

(3) is the required differential equation for such curves,

Now solving (3) we get,

∫-dy/√(k

^{2}– y

^{2}) = ∫-dx

Or, x

^{2}+ y

^{2}= k

^{2}passes through (0, k)

**Sanfoundry Global Education & Learning Series – Mathematics – Class 12**.

To practice all chapters and topics of class 12 Mathematics, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

**Related Posts:**

- Practice Class 12 - Physics MCQs
- Check Class 12 - Mathematics Books
- Practice Class 12 - Chemistry MCQs
- Practice Class 11 - Mathematics MCQs
- Practice Class 12 - Biology MCQs