Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 4 Differential Equations Ex 4.6 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6

Choose the most suitable answer from the given four alternatives:

Question 1.

The degree of the differential equation

\(\frac { d^2y }{dx^4}\) – (\(\frac { d^2y }{dx^2}\)) + \(\frac { dy }{dx}\) = 3

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

(a) 1

Hint:

Since the power of \(\frac{d^{4} y}{d x^{4}}\) is 1

Question 2.

The order and degree of the differential equation \(\sqrt{\frac { d^2y }{dx^2}}\) = \(\sqrt{\frac { dy }{dx}+5}\) are respectively.

(a) 2 and 2

(b) 3 and 2

(c) 2 and 1

(d) 2 and 3

Solution:

(a) 1

Hint:

Squaring on both sides

\(\frac { d^2y }{dx^2}\) = \(\frac { dy }{dx}\) + 5

Highest order derivative is \(\frac { d^2y }{dx^2}\)

∴ order = 2

Power of the highest order derivative \(\frac { d^2y }{dx^2}\) = 1

∴ degree = 1

Question 3.

The order and degree of the differential equation

(\(\frac { d^2y }{dx^2}\))^{3/2} – \(\sqrt{(\frac { dy }{dx})}\) – 4 = 0

(a) 2 and 6

(b) 3 and 6

(c) 1 and 4

(d) 2 and 4

Solution:

(a) 2 and 6

Hint:

Highest order derivative is \(\frac { d^2y }{dx^2}\)

∴ Order = 2

Power of the highest order derivative \(\frac { d^2y }{dx^2}\) is

∴ degree = 6

Question 4.

The differential equation (\(\frac { dx }{dy}\))³ + 2y^{1/2} = x

(a) of order 2 and degree 1

(b) of order 1 and degree 3

(c) of order 1 and degree 6

(d) of order 1 and degree 2

Solution:

(b) of order 1 and degree 3

Hint:

Highest order derivative is \(\frac { dy }{dx}\)

∴ order = 1

Power of the highest order derivative \(\frac { dy }{dx}\) is 3

∴ degree = 3

Question 5.

The differential equation formed by eliminating a and b from y = ae^{x} + be^{-x}

(a) \(\frac { d^2y }{dx^2}\) – y = 0

(b) \(\frac { d^2y }{dx^2}\) – \(\frac { dy }{dx}\)y = 0

(c) \(\frac { d^2y }{dx^2}\) = 0

(d) \(\frac { d^2y }{dx^2}\) – x = 0

Solution:

(a) \(\frac { d^2y }{dx^2}\) – y = 0

Hint:

Question 6.

If y = ex + c – c³ then its differential equation is

(a) y = x\(\frac { dy }{dx}\) + \(\frac { dy }{dx}\) – (\(\frac { dy }{dx}\))³

(b) y + (\(\frac { dy }{dx}\))³ = x \(\frac { dy }{dx}\) – \(\frac { dy }{dx}\)

(c) \(\frac { dy }{dx}\) + (\(\frac { dy }{dx}\))³ – x\(\frac { dy }{dx}\)

(d) \(\frac { d^3y }{dx^3}\) = 0

Solution:

(a) y = x\(\frac { dy }{dx}\) + \(\frac { dy }{dx}\) – (\(\frac { dy }{dx}\))³

Hint:

y = cx + c – c³ ……… (1)

\(\frac { dy }{dx}\) = c

(1) ⇒ y = x\(\frac { dy }{dx}\) + \(\frac { dy }{dx}\) – (\(\frac { dy }{dx}\))³

Question 7.

The integrating factor of the differential equation \(\frac { dy }{dx}\) + Px = Q is

(a) e^{∫pdx}

(b) e^{∫}Pdx

(c) e^{∫}Pdy

(d) e^{∫pdy}

Solution:

(d) e^{∫pdy}

Question 8.

The complementary function of (D² + 4) y = e^{2x} is

(a) (Ax+B)e^{2x}

(b) (Ax+B)e^{-2x}

(c) A cos2x + B sin2x

(d) Ae^{-2x} + Be^{2x}

Solution:

(c) A cos2x + B sin2x

Hint:

A.E = m^{2} + 4 = 0 ⇒ m = ±2i

C.F = e^{0x} (A cos 2x + B sin 2x)

Question 9.

The differential equation of y = mx + c is (m and c are arbitrary constants)

(a) \(\frac { d^2y }{dx^2}\) = 0

(b) y = x\(\frac { dy }{dx}\) + o

(c) xdy + ydx = 0

(c) ydx – xdy = 0

Solution:

(a) \(\frac { d^2y }{dx^2}\) = 0

Hint:

y = mx + c

\(\frac { dy }{dx}\) = m ⇒ \(\frac { d^2y }{dx^2}\) = 0

Question 10.

The particular intergral of the differential equation \(\frac { d^2y }{dx^2}\) – 8\(\frac { dy }{dx}\) + 16y = 2e^{4x}

(a) \(\frac { x^2e^{4x} }{2!}\)

(b) y = x\(\frac { e^{4x} }{2!}\)

(c) x²e^{4x}

(d) xe^{4x}

Solution:

(c) x²e^{4x}

Hint:

Question 11.

Solution of \(\frac { dx }{dy}\) + Px = 0

(a) x = ce^{py}

(b) x = ce^{-py}

(c) x = py + c

(d) x = cy

Solution:

(b) x = ce^{-py}

Question 12.

If sec^{2x} x isa na intergranting factor of the differential equation \(\frac { dx }{dy}\) + Px = Q then P =

(a) 2 tan x

(b) sec x

(c) cos 2 x

(d) tan 2 x

Solution:

(a) 2 tan x

Hint:

I.F = sec² x

e^{∫pdx} = sec²x

∫pdx = log sec² x

∫pdx = 2 log sec x

∫pdx = 2∫tan x dx ⇒ p = 2 tan x

Question 13.

The integrating factor of the differential equation is x \(\frac { dy }{dx}\) – y = x²

(a) \(\frac { -1 }{x}\)

(b) \(\frac { 1 }{x}\)

(c) log x

(c) x

Solution:

(b) \(\frac { 1 }{x}\)

Hint:

Question 14.

The solution of the differential equation where P and Q are the function of x is

(a) y = ∫Q e^{∫pdx} dx + c

(b) y = ∫Q e^{-∫pdx} dx + c

(c) ye^{∫pdx} = ∫Q e^{∫pdx} dx + c

(c) ye^{∫pdx} = ∫Q e^{-∫pdx} dx + c

Solution:

(c) ye^{∫pdx} = ∫Q e^{∫pdx} dx + c

Question 15.

The differential equation formed by eliminating A and B from y = e^{-2x} (A cos x + B sin x) is

(a) y_{2} – 4y_{1} + 5 = 0

(b) y_{2} + 4y – 5 = 0

(c) y_{2} – 4y_{1} + 5 = 0

(d) y_{2} + 4y_{1} – 5 = 0

Solution:

(d) y_{2} + 4y_{1} – 5 = 0

Hint:

y = e^{-2x} (A cosx + B sinx)

y e^{2x} = A cosx + B sinx ………. (1)

y(e^{2x}) (2) + e^{2x} \(\frac { dy }{dx}\) = A(-sin x) + B cos x ………. (2)

Differentiating w.r.to x

Question 16.

The particular integral of the differential equation f (D) y = e^{ax} where f(D) = (D – a)²

(a) \(\frac { x^2 }{2}\) e^{ax}

(b) xe^{ax}

(c) \(\frac { x }{2}\) e^{ax}

(d) x² e^{ax}

Solution:

(a) \(\frac { x^2 }{2}\) e^{ax}

Question 17.

The differential equation of x² + y² = a²

(a) xdy + ydx = 0

(b) ydx – xdy = 0

(c) xdx – ydx = 0

(d) xdx + ydy = 0

Solution:

(d) xdx + ydy = 0

Hint:

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

⇒ 2x + 2y \(\frac{d y}{d x}\) = 0

⇒ x dx + y dy = 0

Question 18.

The complementary function of \(\frac { d^y }{dx^2}\) – \(\frac { dy }{dx}\) = 0 is

(a) A + Be^{x}

(b) (A + B)e^{x}

(c) (Ax + B)e^{x}

(d) Ae^{x} + B

Solution:

(a) A + Be^{x}

Hint:

A.E is m^{2} – m = 0

⇒ m(m – 1) = 0

⇒ m = 0, 1

CF is Ae^{0x} + Be^{x} = A + Be^{x}

Question 19.

The P.I of (3D² + D – 14) y = 13e^{2x} is

(a) \(\frac { 1 }{2}\) e^{x}

(b) xe^{2x}

(c) \(\frac { x^2 }{2}\) e^{2x}

(d) Ae^{x} + B

Solution:

(b) xe^{2x}

Hint:

(3D² + D – 14) y = 13e^{2x}

Question 20.

The general solution of the differential equation \(\frac { dy }{dx}\) = cos x is

(a) y = sinx + 1

(b) y = sinx – 2

(c) y = cosx + C, C is an arbitary constant

(d) y = sinx + C, C is an arbitary constant

Solution:

(d) y = sinx + C, C is an arbitary constant

Hint:

\(\frac { dy }{dx}\) = cos x

dy = cos x dx

∫dy = ∫cos x dx ⇒ y = sin x + c

Question 21.

A homogeneous differential equation of the form \(\frac { dy }{dx}\) = f(\(\frac { y }{x}\)) can be solved by making substitution.

(a) y = v x

(b) y = y x

(c) x = v y

(d) x = v

Solution:

(a) y = v x

Question 22.

A homogeneous differential equation of the form \(\frac { dy }{dx}\) = f(\(\frac { x }{y}\)) can be solved by making substitution.

(a) x = v y

(b) y = v x

(c) y = v

(d) x = v

Solution:

(a) y = v x

Question 23.

The variable separable form of \(\frac { dy }{dx}\) = \(\frac { y(x-y) }{x(x+y)}\) by taking y = v x and \(\frac { dy }{dx}\) = v + x \(\frac { dy }{dx}\)

(a) \(\frac { 2v^2 }{1+v}\) dv = \(\frac { dx }{x}\)

(b) \(\frac { 2v^2 }{1+v}\) dv = –\(\frac { dx }{x}\)

(c) \(\frac { 2v^2 }{1-v}\) dv = \(\frac { dx }{x}\)

(d) \(\frac { 1+v }{2v^2}\) dv = –\(\frac { dx }{x}\)

Solution:

(d) \(\frac { 1+v }{2v^2}\) dv = –\(\frac { dx }{x}\)

Hint:

Question 24.

Which of the following is the homogeneous differential equation?

(a) (3x – 5) dx = (4y – 1) dy

(b) xy dx – (x³ + y³) dy = 0

(c) y²dx + (x² – xy – y²) dy = 0

(d) (x² + y) dx (y² + x) dy

Solution:

(c) y²dx + (x² – xy – y²) dy = 0

Question 25.

The solution of the differential equation \(\frac { dy }{dx}\) = \(\frac { y }{x}\) + \(\frac { f(\frac { y }{x}) }{ f(\frac { y }{x}) }\) is

(a) f\(\frac { y }{x}\) = k x

(b) x f\(\frac { y }{x}\) = k

(c) f\(\frac { y }{x}\) = k y

(d) x f\(\frac { y }{x}\) = k

Solution:

(a) f\(\frac { y }{x}\) = k x