Linear First Order Differential Equations - Class 12 Maths MCQ

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 = x³?
a) [2x/3 *(dy/dx) – y][x/3 *dy/dx]² = x³
b) [x/3 *(dy/dx) – y][x/3 *dy/dx]² = x³
c) [2x/3 *(dy/dx) – y][ 2x/3 *dy/dx]² = x³
d) [x/3 *(dy/dx) – y][ 2x/3 *dy/dx]² = x³
View Answer

Answer: c
Explanation: The given family is c(y + c)² = x³
Differentiating once, we get
c[2(y + c)]dy/dx = 3x²
=> 2x³ (y + c)/(y + c)² * dy/dx = 3x²
=> 2x³/(y + c) * dy/dx = 3x²
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 = x³
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

Answer: a
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

Answer: b
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

Answer: a
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,
∫2^2xdx + ∫2^2ydy = 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

Answer: c
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)/(3V² – 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

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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

Answer: b
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, X⁴(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 = x²(x – 1)²?
a) xy = (x + 1)(x³/3 + c)
b) depends on x
c) depends on y
d) xy = (x – 1)(x³/3 + c)
View Answer

Answer: d
Explanation: x(x – 1)dy/dx – y = x²(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 = x²
or, d/dx[x/(x – 1) * y] = x² …….(2)
integrating both sides of (2) we get,
x/(x – 1) * y = ∫x²dx = x³/3 + c
or xy = (x – 1)(x³/3 + c)

8. What is the solution of the equation(y + x(√xy)(x + y))dx – (y + y(√xy)(x + y))dy = 0?
a) x² + y² = 2tan^-1(√(y/x)) + c
b) x² + y² = 4tan^-1(√(y/x)) + c
c) x² + y² = tan^-1(√(y/x)) + c
d) x² + y² = 2tan^-1(√(x/y)) + c
View Answer

Answer: b
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(x² + y²) = (xdy – ydx)/(x² (√y/x) (1 + y/x))
= 2/(1 + y/x)* d(√y/x)
Thus, x² + y² = 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) = ± √(k² + x²)
b) y(dy/dx) = ± √(k² – 2y²)
c) y(dy/dx) = ± √(k² – y²)
d) Depends on the value of k
View Answer

Answer: c
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 (x₁ , 0).
From (1), we get
y(dy/dx) = x₁ – x ….(2)
Now, giving that PQ² = k²
Or, x₁ – x + y² = k²
=> y(dy/dx) = ± √(k² – y²)

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) x² + 2y² = k²
b) 2x² + y² = k²
c) x² – y² = k²
d) x² + y² = k²
View Answer

Answer: d
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 (x₁ , 0).
From (1), we get
y(dy/dx) = x₁ – x ….(2)
Now, giving that PQ² = k²
Or, x₁ – x + y² = k²
=> y(dy/dx) = ± √(k² – y²) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k² – y²) = ∫-dx
Or, x² + y² = k² passes through (0, k)

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