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

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems

Question 1.

Suppose that Qd = 30 – 5P + 2\(\frac { dp }{dt}\) + \(\frac { d^2p }{dt^2}\) and Qs = 6 + 3P Find the equilibrium price for market clearance.

Solution :

Qd = 30 – 5P + 2\(\frac { dp }{dt}\) + \(\frac { d^2p }{dt^2}\) and

Qs = 6 + 3P

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

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

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

The general solution is y = C.F + P.I

y = Ae^{-ut} + Be^{2t} + 3

Question 2.

Form the differential equation having for its general solution y = ax² + bx

Solution:

y = ax² + bx ……….. (1)

Since we have two arbitary constants, differentiative twice.

Question 3.

Solve yx²dx + e^{-x} dy = 0

Solution:

yx²dx + e^{-x} dy = 0

e^{-x} dy = -yx²dx

\(\frac { 1 }{y}\) dy = -x² e^{x} dx

Integrating on both sides

Question 4.

Solve (x² + y²) dx + 2xy dy = 0

Solution:

(x² + y²) dx + 2xy dy = 0

2xy dy = – (x² + y²) dx

\(\frac { dy }{dx}\) = \(\frac { -(x^2+y^2) }{2xy}\) ………. (1)

This is a homogeneous differential equation

\(\frac { 1 }{3}\) log(3v² + 1) = – logx + logc

log (3v² + 1)^{1/3} + log x = log c

logx (3v² + 1)^{1/3} = log c

⇒ x (3v² + 1)^{1/3} = c

⇒ x[\(\frac { 3y^2 }{x^2}\) + 1]^{1/3} = c

Question 5.

Solve x \(\frac { dy }{dx}\) + 2y = x^{4}

Solution:

This solution is

y(I.F) = ∫Qx (I.F) dx + c

y(x²) = ∫(x³ × x²) dx + c

yx² = ∫x^{5} dx + c

⇒ yx² = \(\frac { x^6 }{6}\) + c

Question 6.

A manufacturing company has found that the cost C of operating and maintaining the equipment is related to the length’m’ of intervals between overhauls by the equation m² \(\frac { dc }{dm}\) + 2mc = 2 and c = 4 and whem m = 2. Find the relationship between C and m.

Solution:

m² \(\frac { dc }{dm}\) + 2mc = 2

÷ each term by m²

\(\frac { dc }{dm}\) + \(\frac { 2mc }{m^2}\) = \(\frac { 2 }{m^2}\)

\(\frac { dc }{dm}\) + \(\frac { 2c }{m}\) = \(\frac { 2 }{m^2}\)

This is a first order linear differential equation of the form

\(\frac { dc }{dm}\) + Pc = Q where P = \(\frac { 2 }{m}\) and Q = \(\frac { 2 }{m^2}\)

∫Pdm = 2 ∫\(\frac { 1 }{m}\)dm = 2 log m = log m²

I.F = e^{∫Pdm} = e^{logm²} = m²

General solution is

C (I.F) = ∫Q × (IF) dm + k

C(m²) = ∫\(\frac { 2 }{m^2}\) × (m²) dm + k

C(m²) = ∫2dm + k

Cm² = 2m + k ……….. (1)

when C = 4 and m = 2, we have

(4) (2)² = 2(2) + k

16 = 4 + k = 12

Equation (1)

Cm² = 2m + 12

Cm² = 2(m + 6)

Question 7.

Solve (D² – 3D + 2) y = e^{4x}

Solution:

(D² – 3D + 2) y = e^{4x}

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

(m – 1) (m – 2) = 0

m = 1, 2

The roots are real and different

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

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

The general solution is y = C.F + P.I

y = Ae^{x} + Be^{2x} + \(\frac { e^{4x} }{6}\) ………. (1)

When x = 0; y = 0

Question 8.

Solve \(\frac { dy }{dx}\) + y cos x = 2 cos x

Solution:

\(\frac { dy }{dx}\) + y cos x = 2 cos x

This is of the form \(\frac { dy }{dx}\) + Py = Q

Here P = cos x and Q = 2 cos x

∫Pdx = ∫cos x dx = sin x

I.F = e^{∫pdx} = e^{sin x}

The solution is

y (I.F) = ∫Q (I.F) dx + c

ye^{sin x} = ∫(2 cos x) e^{sin x} dx

Question 9.

Solve x²ydx – (x³ + y³) dy = 0

Solution:

x²ydx = (x³ + y³) dy = 0

x²ydx = (x³ + y³) dy

\(\frac { dy }{dx}\) = \(\frac { x^2y }{(x^3+y^3)}\) ……… (1)

This is a homogeneous differential equation, same degree in x and y

Question 10.

Solve \(\frac { dy }{dx}\) = xy + x + y + 1

Solution: