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

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
Aem1x + Bem2x
C.F = Ae2x + Be4x
∴ The general solution is y = Ae2x + Be4x

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)emx
∴ C.F = (Ax + B)e2x
∴ The general solution is y = (Ax + B) e2x

Question 3.
(D2 + 2D + 3) y = 0
Solution:
The auxiliary equations A.E is m2 + 2m + 3 = 0
⇒ m2 + 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)emx
∴ C.F = (Ax + B)ekx
∴ The general solution is y = (Ax + B) ekx

Question 5.
(D2 – 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 m2 – 2m – 15 = 0
(m – 5)(m + 3) = 0
m = 5, -3
C.F = Ae5x + Be-3x
The general solution is y = Ae5x + Be-3x …….. (1)

Question 6.
(4D2 + 4D – 3) y = e2x
Solution:
The auxiliary equations is 4m2 + 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 = eαx [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.
(D2 – 3D + 2)y = e3x which shall vanish for x = 0 and for x = log 2
Solution:
A.E is m2 – 3m + 2 = 0
⇒ (m – 2) (m – 1) = 0
⇒ m = 2, 1
CF = Ae2x + Bex

Question 9.
(D2 + D – 6)y = e3x + e-3x
Solution:
A.E is m2 + m – 6 = 0
(m + 3) (m – 2) = 0

Question 10.
(D2 – 10D + 25)y = 4e5x + 5
Solution:
A.E is m2 – 10m + 25 = 0
⇒ (m – 5)2 = 0
⇒ m = 5, 5
C.F = (Ax + B) e5x

Question 11.
(4D2 + 16D + 15) y = 4$$e^{\frac{-3}{2} x}$$
Solution:
A.E is 4m2 + 16m + 15 = 0
(2m + 3) (2m + 5) = 0

Question 12.
(3D2 + D – 14)y = 13e2x
Solution:

Question 13.
Suppose that the quantity demanded Qd = 13 – 6p + 2$$\frac{d p}{d t}+\frac{d_{2} p}{d t^{2}}$$ and quantity supplied Qs = -3 + 2p where p is the price. Find the equilibrium price for market clearance.
Solution:
For market clearance, the required condition is Qd = Qs