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

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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.
(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)e^mx
∴ C.F = (Ax + B)e^kx
∴ The general solution is y = (Ax + B) e^kx

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

Question 6.
(4D² + 4D – 3) y = e^2x
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 = 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.
(D² – 3D + 2)y = e^3x 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 = Ae^2x + Be^x

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

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

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 = 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