Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 4 Differential Equations Ex 4.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.4

Question 1.

\(\frac { dy }{dx}\) – \(\frac { dy }{dx}\) = x

Solution:

The required solution is

Question 2.

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

Solution:

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

Here P = cos x; Q = sin x cos x

∫Pdx = ∫cos x dx = sin x

I.F = e^{∫pdx} = e^{sinx}

The required solution is

Y(I.F) = ∫Q (IF) dx + c

Y(e^{sinx}) = ∫Q (I-F) dx + c

y (e^{sinx}) = ∫sin x cos x e^{sinx} dx + c

Question 3.

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

Solution:

The given equation can be reduced to

\(\frac { dy }{dx}\) + \(\frac { 2y }{x}\) = x³

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

Here P = \(\frac { 2 }{x}\); Q = x³

∫pdx = ∫\(\frac { 2 }{x}\)dx = 2∫\(\frac { 1 }{x}\)dx = 2log x – log x²

I.F = e^{∫Pdx} = e^{logx²} = x²

The required solution is

y(I.F) = ∫Q (IF) dx + c

y(x²) = ∫x³ (x²) dx + c

x²y = ∫x^{5} dx + c

x²y = \(\frac { x^6 }{6}\) + c

Question 4.

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

Solution:

Question 5.

\(\frac { dy }{dx}\) + \(\frac { y }{x}\) = xe^{x}

Solution:

\(\frac { dy }{dx}\) + py = Q

Here P = \(\frac { 1 }{x}\); Q = xe^{x}

∫Pdx = ∫\(\frac { 1 }{x}\) dx = log x

I.F = e∫Pdx = e^{log} = x

The required solution is

y (I.F) = ∫Q (I.F) dx + c

Question 6.

\(\frac { dy }{dx}\) + y tan x = cos³ x

Solution:

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

Here P = tan x; Q = cos³ x

∫Pdx = ∫tan x dx = ∫\(\frac { sin x }{cos x}\) dx = -∫\(\frac { -sin x }{cos x}\) dx

= -log cos x = log sec x

I.F = e^{∫Pdx} = e^{log sec x} = sec x

The required solution is

y(I.F) = ∫Q(I.F) dx + c

y (sec x) = ∫cos³x (sec x) dx + c

y(sec x) = ∫cos³x \(\frac { 1 }{cos x}\) dx + c

y (sec x) = ∫cos²x dx + c

y (sec x)= ∫(\(\frac { 1+cos 2x }{2}\)) dx + c

y (sec x) = \(\frac { 1 }{2}\) ∫(1 + cos2x) dx + c

y (sec x) = \(\frac { 1 }{2}\) [x + \(\frac { sin2x }{2}\)] + c

Question 7.

If \(\frac { dy }{dx}\) + 2y tan x = sinx and if y = 0 when x = π/3 express y in terms of x

Solution:

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

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

Here P = 2tan x ; Q = sin x

∫Pdx = ∫2 tan x dx = 2∫tan xdx = 2 log sec x

log sec² x

I.F = e^{∫Pdx} = e^{log(sec²x)} = sec² x

The required solution is

y(I.F) = ∫Q(I.F) dx + c

y (sec² x) = ∫sin x (sec²x) dx + c

y(sec²x) = ∫sin x(\(\frac { 1 }{cos x}\)) sec x dx + c

y sec²x = ∫(\(\frac { sin x }{cos x}\)) sec x dx + c

y(sec²x) = ∫tan x sec x dx + c

⇒ y(sec²x) = sec x + c ………. (1)

If y = 0 when x = /3, then (1) ⇒

0(sec²(π/3)) = sec(π/3) + c

0 = 2 + c

⇒ c = -2

∴ Eqn (1) ⇒ y sec²x = sec x – 2

Question 8.

\(\frac { dy }{dx}\) + \(\frac { y }{x}\) = xe^{x}

Solution:

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

Here P = \(\frac { 1 }{x}\); Q = xe^{x}

∫Pdx = ∫\(\frac { 1 }{x}\) dx = log x

I.F = e^{∫Pdx} = e^{log x} = x

The required solution is

Question 9.

A bank pays interest by contionous compounding, that is by treating the interest rate as the instantaneous rate of change of principal. A man invests Rs 1,00,000 in the bank deposit which accures interest, 8% over year compounded continuously. How much will he get after 10 years?

Solution :

Let P(t) denotes the amount of money in the account at time t. Then the differential equation govemning the growth of money is

\(\frac { dp }{dt}\) = \(\frac { 8 }{100}\)p = 0.08 p

⇒ \(\frac { dp }{p}\) = 0.08 dt

Integrating on both sides

∫\(\frac { dp }{p}\) = ∫0.08 dt

log_{e} P = 0.08 t + c

P = e^{0.08 t} + c

P = e^{0.08 t}. e^{c}

P = C_{1} e^{0.08 t} ………. (1)

when t = 0, P = Rs 1,00,000

Eqn (1) ⇒ 1,00,000 = C_{1} e°

C_{1} = 1,00,000

∴ P = 100000 e^{0.08 t}

At t = 10

P= 1,00,000 . e^{0.08(10)}

= 1,00,000 e^{0.8} {∵ e^{0.8} = 2.2255}

= 100000 (2.2255)

p = Rs 2,25,550