Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Students can download 12th Business Maths Chapter 4 Differential Equations Additional Problems 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 Additional Problems

I. One Mark Questions

Choose the correct answer.

Question 1.
The differential equation of straight lines passing through the origin is ______
(a) \(\frac{x d y}{d x}=y\)
(b) \(\frac{d y}{d x}=\frac{x}{y}\)
(c) \(\frac{d y}{d x}=0\)
(d) \(\frac{x d y}{d x}=\frac{1}{y}\)
Answer:
(a) \(\frac{x d y}{d x}=y\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q1

Question 2.
The solution of x dy + y dx = 0 is _______
(a) x + y = c
(b) x2 + y2 = c
(c) xy = c
(d) y = cx
Answer:
(c) xy = c
Hint:
x dy + y dx = 0
d(xy) = 0
xy = c

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 3.
The solution of x dx + y dy = 0 is _____
(a) x2 + y2 = c
(b) \(\frac{x}{y}\) = c
(c) x2 – y2 = c
(d) xy = c
Answer:
(a) x2 + y2 = c
Hint:
x dx = -y dy
\(\frac{x^{2}}{2}=\frac{-y^{2}}{2}+c_{1}\)
x2 + y2 = c

Question 4.
The solution of \(\frac{d y}{d x}\) = ex-y is ______
(a) ey ex = c
(b) y = log cex
(c) y = log(ex + c)
(d) ex+y = c
Answer:
(c) y = log(ex + c)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q4

Question 5.
The solution of \(\frac{d p}{d t}\) = ke-t (k is a constant) is ________
(a) \(c-\frac{k}{e^{t}}=p\)
(b) p = ket + c
(c) t = log\(\frac{c-p}{k}\)
(d) t = logc p
Answer:
(a) \(c-\frac{k}{e^{t}}=p\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q5

Question 6.
The integrating factor of (1 + x2) \(\frac{d y}{d x}\) + xy = (1 + x2)3 is _______
(a) \(\sqrt{1+x^{2}}\)
(b) log(1 + x2)
(c) \(e^{\tan ^{-1} x}\)
(d) \(\log ^{\left(\tan ^{-1} x\right)}\)
Answer:
(a) \(\sqrt{1+x^{2}}\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems I Q6

Question 7.
The complementary function of the differential equation (D2 – D) y = ex is _____
(a) A + B ex
(b) (Ax + B) ex
(c) A + B e-x
(d) (A + Bx) e-x
Answer:
(a) A + B ex
Hint:
m2 – m = 0
m(m – 1) = 0
CF = Ae0x + Bex = A + B ex

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 8.
Match the following
(a) \(\frac{d^{4} y}{d x^{4}}\) + sin y = 0 – (i) order 1, degree 1
(b) y’ + y = ex – (ii) order 3, degree 2
(c) y'” + 2y” + y’ = 0 – (iii) order 4, degree 1
(d) (y”’)2 + y’ + y5 = 0 – (iv) order 3, degree 1
Answer:
(a) – (iii)
(b) – (i)
(c) – (iv)
(d) – (ii)

Question 9.
Fill in the blanks
(a) The general solution of the equation \(\frac{d y}{d x}+\frac{y}{x}=1\) is ______
(b) Integrating factor of \(\frac{x d y}{d x}\) – y = sin x is ______
(c) The differential equation of y = A sin x + B cos x is _______
(d) The D.E \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}\) is a ________ differential equation.
Answer:
(a) \(y=\frac{x}{2}+\frac{c}{x}\)
(b) \(\frac{1}{x}\)
(c) \(\frac{d^{2} y}{d x^{2}}+y=0\)
(d) linear

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 10.
State true or false
(a) y = 3 sin x + 4 cos x is a particular solution of the differential equation \(\frac{d^{2} y}{d x^{2}}\) + y = 0
(b) The solution of \(\frac{d y}{d x}=\frac{x+2 y}{x}\) is x + y = kx2
(c) y = 13ex + 4e-x is a solution of \(\frac{d^{2} y}{d x^{2}}\) – y = 0
Answer:
(a) True
(b) True
(c) True

II. 2 Marks Questions

Question 1.
Form the D.E of the family of curves y = ae3x + bex where a, b are parameters
Solution:
y = ae3x + bex
\(\frac{d y}{d x}\) = 3 ae3x + bex
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q1

Question 2.
Find the D.E of a family of curves y = a cos (mx + b), a and b are constants.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q2

Question 3.
Find the D.E by eliminating the constants a and b from y = a tan x + b sec x
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q3

Question 4.
Solve: \(\frac{d y}{d x}=e^{7 x+y}\)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q4

Question 5.
Solve: (x2 – ay) dx = (ax – y2) dy
Solution:
Writing the equation as
x2 dx + y2 dy = a (x dy + y dx)
x2 dx + y2 dy = a d(xy)
∫x2 dx + ∫y2 dy = a ∫d(xy) + c
\(\frac{x^{3}}{3}+\frac{y^{3}}{3}\) = axy + c
Hence the general solution is x3 + y3 = 3axy + c

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 6.
Solve (sin x + cos x) dy + (cos x – sin x) dx = 0
Solution:
The given equation can be written as
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems II Q6

III. 3 and 5 Marks Questions

Question 1.
Solve \(\frac{x d y}{d x}\) + cos y = 0, given y = \(\frac{\pi}{4}\) when x = √2
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q1

Question 2.
The slope of a curve at any point is the reciprocal of twice the ordinate of the point. The curve also passes through the point (4, 3). Find the equation of the curve.
Solution:
Slope at any point (x, y) is the slope of the tangent at (x, y)
\(\frac{d y}{d x}=\frac{1}{2 y}\)
⇒ 2y dy = dx
⇒ ∫2y dy = ∫dx + c
⇒ y2 = x + c
Since the curve passes through (4, 3)
we have 9 = 4 + c ⇒ c = 5
Equation of the curve is y2 = x + 5

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 3.
The net profit P and quantity x satisfy the differential equation \(\frac{d p}{d x}=\frac{2 p^{3}-x^{3}}{3 x p^{2}}\). Find the relationship between the net profit and demand given that p = 20 when x = 10
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q3
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q3.1

Question 4.
Solve: \(\frac{d y}{d x}\) + ay = ex (a ≠ -1)
Solution:
The given equation is of the form \(\frac{d y}{d x}\) + Py = Q
Here P = a, Q = ex
The general solution is y (I.F) = ∫Q (I.F) dx + c
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q4

Question 5.
Solve: \(\frac{d y}{d x}\) + y cos x = \(\frac{1}{2}\) sin 2x
Solution:
Here P = cos x, Q = \(\frac{1}{2}\) sin 2x
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q5

Question 6.
Solve (D2 – 6D + 25) y = 0
Solution:
The auxiliary equations is m2 – 6m + 25 = 0
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q6
The Roots are complex and of the form,
α ± β with α = 3 and β = 4
The complementary function = e3x (A cos 4x + B sin 4x)
The general solution is y = e3x (A cos 4x + B sin 4x)

Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

Question 7.
Solve (D2 + 10D + 25) y = \(\frac{5}{2}\) + e-5x
Solution:
The auxiliary equations is m2 + 10m + 25 = 0
(m + 5)2 = 0
m = -5, -5
The complementary function = (Ax + B) e-5x
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q7

Question 8.
Suppose that the quantity demanded Qd = 40 – 4p – 4\(\frac{d p}{d t}+\frac{d^{2} p}{d t^{2}}\) and quantity supplied Qs = -6 + 8p where p is the price. Find the equilibrium price for market clearance.
Solution:
For market clearance, the required condition is Qd = Qs
Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems III Q8