Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 4 Differential Equations Ex 4.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5

Question 1.

\(\frac { d^2y }{dx^2}\) – 6\(\frac { dy }{dx}\) + 8y = 0

Solution:

Given (D^{2} – 6D + 8) y = 0, D = \(\frac{d}{d x}\)

The auxiliary equations is

m^{2} – 6m + 8 = 0

(m – 4)(m – 2) = 0

m = 4, 2

Roots are real and different

The complementary function (C.F) is (Ae^{4x} + Be^{2x})

The general solution is y = Ae^{4x} + Be^{2x}

Question 2.

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

Solution:

The auxiliary equations A.E is m^{2} – 4m + 4 = 0

(m – 2)^{2} = 0

m = 2, 2

Roots are real and equal

The complementary function (C.F) is (Ax + B) e^{2x}

The general solution is y = (Ax + B) e^{2x}

Question 3.

(D² + 2D + 3) y = 0

Solution:

The auxiliary equation is m² + 2m + 3 = 0

Here a = 1, b = 2, c = 3

The complementary function is

e^{ax} (Acosßx + Bsinßx)

∴ C.F = e^{-x} [Acos√2x + Bsin √2x]

∴ The general solution is

y = e^{-x} (Acos√2x + Bsin√2x)

Question 4.

\(\frac { d^2y }{dx^2}\) – 2k\(\frac { dy }{dx}\) + k²y = 0

Solution:

Given (D^{2} – 2kD + k^{2})y = 0, D = \(\frac{d}{d x}\)

The auxiliary equations is m^{2} – 2km + k = 0

⇒ (m – k)^{2} = 0

⇒ m = k, k

Roots are real and equal

The complementary function (C.F) is (Ax + B) e^{kx}

The general solution is y = (Ax + B) e^{kx}

Question 5.

(D² – 2D – 15) y = 0

Solution:

The auxiliary equation is

m² – 2m + 15 = 0

m² + 3m – 5m – 15 = 0

m (m + 3) – 5 (m + 3) = 0

(m + 3) (m – 5) = 0

m = -3, 5

Roots are real and different

∴ The complementary function is

Ae^{m1x} + Be^{m2x}

C.F = Ae^{-3x} + Be^{5x}

∴ The general solution is

y = (Ae^{-3x} + Be^{5x}) ………… (1)

\(\frac { dy }{dx}\) = Ae^{-3x} (-3) + Be^{5x} (5)

\(\frac { dy }{dx}\) = -3Ae^{-3x} + 5Be^{5x} ………… (2)

\(\frac { d^2y }{dx^2}\) = 9Ae^{-3x} + 25Be^{5x} ……….. (3)

when x = 0; \(\frac { dy }{dx}\) = 0

-3 Ae° + 5Be° = 0

-3A + 5B = 0 ………. (4)

when x = 0; \(\frac { d^2y }{dx^2}\) = 2

Eqn (3) ⇒ 9Ae° + 25Be° = 2

9A + 25B = 2 ……… (5)

Solving equation (4) & (5)

Question 6.

(4D² + 4D – 3) y = e^{2x}

Solution:

The auxiliary equation is

4m² + 4m – 3 = 0

4m² + 6m – 2m – 3 = 0

2m (2m + 3) – 1 (2m + 3) = 0

(2m + 3) (2m – 1) = 0

2m = -3; 2m = 1

m = -3/2, 1/2

Roots are real and different

The complementary function is

Ae^{m1x} + Be^{m2x}

Question 7.

\(\frac { d^2y }{dx^2}\) + 16y = 0

Solution:

Given (D^{2} + 16) y =0

The auxiliary equation is m^{2} + 16 = 0

⇒ m^{2} = -16

⇒ m = ± 4i

It is of the form α ± iβ, α = 0, β = 4

The complementary function (C.F) is e^{0x} [A cos 4x + B sin 4x]

The general solution is y = [A cos 4x + B sin 4x]

Question 8.

(D² – 3D + 2) y = e^{3x} which shall vanish for x = 0 and for x = log 2

Solution:

(D² – 3D + 2) y = e^{3x}

The auxiliary equation is

m² – 3m + 2 =0

(m – 1) (m – 2) = 0

m = 1, 2

Roots are real and different

The complementary function is

C.F = Ae^{m1x} + Be^{m2x}

C.F = A^{x} + Be^{2x}

when x = log 2; y = 0

Ae^{log 2} + Be^{2log 2} + \(\frac { e^{3xlog2} }{2}\) = 0

Ae^{log 2} + Be^{log (2)²} + \(\frac { e^{log2³} }{2}\) = 0

2A + 4B + \(\frac { 8 }{2}\) = 0

2A + 4B + 4 = 0

2A + 4B = -4 ……… (3)

Solving equation (2) & (3)

Eqn (2) × 2 ⇒ 2A + 2B = -1

Question 9.

(D² + D – 2) y = e^{3x} + e^{-3x}

Solution:

The auxiliary equation is

m² + m – 6 = 0

(m + 3) (m – 2) = 0

Roots are real and different

The complementary function is

C.F = Ae^{m1x} + Be^{m2x}

C.F = Ae^{-3x} + Be^{2x}

Question 10.

(D² – 10D + 25) y = 4e^{5x} + 5

Solution:

The auxiliary equation is

m² – 10m + 25 = 0

(m – 5) (m – 5) = 0

m = 5, 5

Roots are real and equal

C.F = (Ax + B) e^{mx}

C.F = (Ax + B) e^{5x}

P.I_{(1)} = x. \(\frac { 4 }{2D-10}\) e^{5x}

Replace D by 5, 2D – 10 = 0 when D = 5

Question 11.

(4D² + 16D +15) y = 4e^{\(\frac { -3 }{2}\)}x

Solution:

The auxiliary equation is 4m² + 16m + 15 = 0

4m² + 16m + 10m + 15 = 0

2m (2m + 3) + 5 (2m + 3) = 0

(2m + 3) (2m + 5) = 0

2m = -3, -5

∴ m = -3/2, -5/2

Roots are real and different

C.F = (Ax + B) e^{m1x} + Be^{m2x}

C.F = Ae^{-3/2 x} + Be^{-5/2 x}

Question 12.

(3D² + D – 14) y – 13 e^{2x}

Solution:

The auxiliary equation is 3m² + m – 14 = 0

3m² – 6m + 7m – 14 = 0

3m (m – 2) + 7 (m – 2) = 0

(m – 2) (3m + 7) = 0

m = 2; 3m = -7

m = 2, -7/3

Roots are real and different

C.F = (Ax + B) e^{m1x} + Be^{m2x}

C.F = Ae^{2x} + Be^{-7/3 x}

Question 13.

Suppose that the quantity demanded Qd = 13 – 6p + 2\(\frac { dp }{dt}\) + \(\frac { d^2p }{dt^2}\) = and quantity supplied Qd = -3 + 2p where is the price. Find the equilibrium price for market clearence.

Solution:

For market clearance, the required condition is Qd = Qs

The auxiliary equation is

m² + 2m – 8 = 0

(m + 4) (m – 2) = 0

m = -4, 2

Roots are real and different