Students can download 12th Business Maths Chapter 4 Differential Equations Ex 4.5 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

Solve the following differential equations:

Question 1.

\(\frac{d^{2} y}{d x^{2}}-6 \frac{d y}{d x}+8 y=0\)

Solution:

The auxiliary equation is

m² – 6m + 8 = 0

m² – 4m – 2m + 8 = 0

m (m – 4) – 2 (m – 4) = 0

(m – 2) (m – 4) = 0

m = 2, 4

Roots are real and different

∴ The complementary function is

Ae^{m1x} + Be^{m2x}

C.F = Ae^{2x} + Be^{4x}

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

Question 2.

\(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+4 y=0\)

Solution:

The auxiliary equation is

m² – 4m + 4 = 0

m² – 2m – 2m + 4 = 0

m (m – 2) – 2 (m – 2) = 0

(m – 2) (m – 2) = 0

m = 2, 2

Roots are real and equal

∴ The complementary function is

(Ax + B)e^{mx}

∴ C.F = (Ax + B)e^{2x}

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

Question 3.

(D^{2} + 2D + 3) y = 0

Solution:

The auxiliary equations A.E is m^{2} + 2m + 3 = 0

⇒ m^{2} + 2m + 1 + 2 = 0

⇒ (m + 1)^{2} = -2

⇒ m + 1 = ± √2i

⇒ m = – 1 ± √2i

It is of the form α ± iβ

The complementary function (C.F) = e^{-x} [A cos √2 x + B sin √2 x]

The general solution is y = e^{-x} [A cos √2 x + B sin √2 x]

Question 4.

\(\frac{d^{2} y}{d x^{2}}-2 k \frac{d y}{d x}+k^{2} y=0\)

Solution:

The auxiliary equation is

m² – 2km + k² = 0

m² – km – km + k² = 0

m (m – k) – k (m – k) = 0

(m – k) (m – k) = 0

m = k, k

Roots are real and equal

∴ The complementary function is

(Ax + B)e^{mx}

∴ C.F = (Ax + B)e^{kx}

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

Question 5.

(D^{2} – 2D – 15) y = 0 given that \(\frac{d y}{d x}\) = 0 and \(\frac{d^{2} y}{d x^{2}}\) = 2 when x = 0

Solution:

A.E is m^{2} – 2m – 15 = 0

(m – 5)(m + 3) = 0

m = 5, -3

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

The general solution is y = Ae^{5x} + Be^{-3x} …….. (1)

Question 6.

(4D^{2} + 4D – 3) y = e^{2x}

Solution:

The auxiliary equations is 4m^{2} + 4m – 3 = 0

Question 7.

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

Solution:

The auxiliary equation is m² + 16 = 0

m² = -16

m = ±\(\sqrt { -16}\)

m = ± 4i

It is of the form of α ± ß

Here α = 0 ß = 4

C.F = e^{αx} [A cosß x + B sinß x]

= e0 [A cos4x + B sin4x]

= A cos4x + B sin4x

∴ The general solution is

y = A cos4x + B sin4x

Question 8.

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

Solution:

A.E is m^{2} – 3m + 2 = 0

⇒ (m – 2) (m – 1) = 0

⇒ m = 2, 1

CF = Ae^{2x} + Be^{x}

Question 9.

(D^{2} + D – 6)y = e^{3x} + e^{-3x}

Solution:

A.E is m^{2} + m – 6 = 0

(m + 3) (m – 2) = 0

Question 10.

(D^{2} – 10D + 25)y = 4e^{5x} + 5

Solution:

A.E is m^{2} – 10m + 25 = 0

⇒ (m – 5)^{2} = 0

⇒ m = 5, 5

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

Question 11.

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

Solution:

A.E is 4m^{2} + 16m + 15 = 0

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

Question 12.

(3D^{2} + D – 14)y = 13e^{2x}

Solution:

Question 13.

Suppose that the quantity demanded Q_{d} = 13 – 6p + 2\(\frac{d p}{d t}+\frac{d_{2} p}{d t^{2}}\) and quantity supplied Q_{s} = -3 + 2p where p is the price. Find the equilibrium price for market clearance.

Solution:

For market clearance, the required condition is Q_{d} = Q_{s}