Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.4

For each of the differential equations in questions 1 to 10, find the general solution:

Question 1.

\(\frac{dy}{dx}\) = \(\frac{1-cosx}{1+cosx}\)

Solution:

∴ y = 2 tan \(\frac{x}{2}\) – x + C

This is the required solution.

Question 2.

\(\frac{dy}{dx}\) = \(\sqrt{4-y^{2}}\) (- 2 < y < 2)

Solution:

\(\frac{dy}{dx}\) = \(\sqrt{4-y^{2}}\) or \(\frac{d y}{\sqrt{4-y^{2}}}\) = dx

Integrating both sides, we get

∫ \(\frac{d y}{\sqrt{4-y^{2}}}\) = ∫dx

⇒ sin^{-1}\(\frac{y}{2}\) = x + C or y = 2 sin(x + C).

which is the required solution.

Question 3.

\(\frac{dy}{dx}\) + y = 1 (y ≠ 1)

Solution:

Given differential equation is \(\frac{dy}{dx}\) + y = 1

or \(\frac{dy}{dx}\) = 1 – y or \(\frac{dy}{1-y}\) = dx

Integrating both sides, we get

∫ – log (1 – y) = x + log C

∴ x = – log C – log (1 – y) = – log C(1 – y)

∴ C(1 – y) = e^{-x}

or C(1 – y)e^{x} = 1

⇒ 1 – y = \(\frac{1}{C}\)e^{-x} ⇒ y = 1 – \(\frac{1}{C}\)e^{-x}.

Put – \(\frac{1}{C}\) = A; y = 1 + Ae^{-x} is the required solution.

Question 4.

sec^{2}x tan y dx + sec^{2}y tan x dy = 0

Solution:

We have:

sec^{2}x tan y dx + sec^{2}y tan x dy = 0

⇒ \(\frac{\sec ^{2} x}{\tan x}\) dx + \(\frac{\sec ^{2} x}{\tan y}\)dy = 0

Integrating both sides, we get

∫ \(\frac{\sec ^{2} x}{\tan x}\) dx + ∫ \(\frac{\sec ^{2} y}{\tan y}\)dy = 0

⇒ log|tan x| + log|tan y| = log C

⇒ log|tan x tan y| = log C

⇒ tan x tan y = C

which is the required solution, where x ≠ odd multiple of and x ∈ R.

Question 5.

(e^{x} + e^{-x})dy – (e^{x} – e^{-x})dx = 0

Solution:

We have:

(e^{x} + e^{-x})dy = (e^{x} – e^{-x})dx

⇒ dy = \(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)dx

Integrating both sides, we get

∫ dy = ∫ \(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)dx

⇒ ∫ dy = ∫\(\frac{dt}{t}\),

where e^{x} + e^{-x} = t so that (e^{x} – e^{-x}) dx = dt.

⇒ y = log |t| + C

⇒ y – C = log|e^{x} + e^{-x}| (x ∈ R)

or y = log (e^{x} + e^{-x}) + C (∵ e^{x}, e^{-x} > 0)

which is the required solution.

Question 6.

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

Solution:

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

or \(\frac{d y}{1+y^{2}}\) = (1 + x^{2}) dx

Integrating both sides, we get

∫ \(\frac{d y}{1+y^{2}}\) = ∫

⇒ tan^{-1}y = x + \(\frac{x^{3}}{3}\) + C,

which is the required equation.

Question 7.

y log y dx – x dy = 0

Solution:

y log y dx – x dy = 0

or y log y dx = x dy or \(\frac{dy}{ylogy}\) = \(\frac{dx}{x}\)

Integrating both sides, we get

∫ \(\frac{dy}{ylogy}\) = ∫\(\frac{dx}{x}\)

Putting log y = t, \(\frac{1}{y}\)dy = dt

So, ∫ \(\frac{dt}{t}\) = log x + log C

∴ log(log y) = log Cx

∴ log y = Cx

or y = e^{Cx} is the required Solution.

Question 8.

x^{5}\(\frac{dy}{dx}\) = – y^{5}

Solution:

Putting – 4C = C. This is the required solution.

Question 9.

\(\frac{dy}{dx}\) = sin^{-1}x

Solution:

\(\frac{dy}{dx}\) = sin^{-1}x

or dy = sin^{-1}x

Integrating both sides, we get

∫ dy = ∫sin^{-1} x dx + C

or y = ∫(sin^{-1} x). 1 dx + C

Integrating by parts, taking sin^{-1}x as first function, we get

Putting this value of I_{1} in (1), we get

y = x sin^{-1}x + \(\sqrt{1-x^{2}}\) + C

∴ The Solution is y = x sin^{-1}x + \(\sqrt{1-x^{2}}\) + C.

Question 10.

e^{x}tan y dx + (1 – e^{x}) sec^{2}y dy = 0

Solution:

e^{x} tan y dx + (1 – e^{x}) sec^{2}y dy = – e^{x} tan y dx = 0

or (1 – e^{x}) sec^{2}y dy = – e^{x} tan y dx

Dividing by (1 – e^{x}) tan, we get

\(\frac{\sec ^{2} y}{\tan y}\) dy = + \(\frac{-e^{x}}{1-e^{x}}\)dx

Integrating both sides, we get

∫ \(\frac{\sec ^{2} y}{\tan y}\)dy = ∫\(\frac{-e^{x}}{1-e^{x}}\)dx

∴ log tan y = log (1 – e^{-x}) + log C

= log C(1 – e^{-x})

∴ tan y = C(1 – e^{-x}) is the required solution.

For each of the differential equations in questions from 11 to 14, find a particular solution satisfying the given condition:

Question 11.

(x^{3} + x^{2} + x + 1)\(\frac{dy}{dx}\) = 2x^{2} + x; y = 1, when x = 0

Solution:

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

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

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

Integrating both sides, we get

⇒ 2x^{2} + x = A(x^{2} + 1) + (Bx + C)(x + 1)

= A(x^{2} + 1) + B(x^{2} + x) + C(x + 1) ………….. (2)

Put x = – 1 in (2), we get

2 – 1 = A(1 + 1) ⇒ A = \(\frac{1}{2}\)

Comparing the coefficients of x^{2} and x, we get

2 = A + B and 1 = B + C.

∴ B = 2 – A = 2 – \(\frac{1}{2}\) = \(\frac{3}{2}\).

C = 1 – B = 1 – \(\frac{3}{2}\) = – \(\frac{1}{2}\).

Therefore, from (1), we get

We have y = 1, when x = 0.

Putting these values in (3), we get

1 = \(\frac{1}{2}\) log 1 + \(\frac{3}{4}\) log 1 – \(\frac{1}{2}\)tan^{-1}0 + C ……………. (3)

[Note : for integrating ∫ \(\frac{2 x}{x^{2}+1}\)dx, put x^{2} + 1 = t

We have y = 1, when x = 0.

Putting these values in (3), we get

1 = \(\frac{1}{2}\) log 1 + \(\frac{3}{4}\) log 1 – \(\frac{1}{2}\)tan^{-1}0 + C

= 0 + C ⇒ C = 1.

Thus, the solution is

This is the required particular solution.

Question 12.

x(x^{2} – 1)\(\frac{dy}{dx}\) = 1; y = 0, when x = 2

Solution:

Question 13.

cos(\(\frac{dy}{dx}\)) = a; (a ∈ R), y = 2, when x = 0

Solution:

Question 14.

\(\frac{dy}{dx}\) = y tan x; y = 1, when x = 0

Solution:

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

Integrating, we get

∫ \(\frac{dy}{y}\) = ∫ tan x dx

⇒ log y = – log cos x + log C

or log y + log cos x = log C

∴ y cos x = C

Putting y = 1, x = 0 and ⇒ C = 1.

∴ y cos x = C

Putting y = 1, x = 0 ⇒ C = 1.

∴ y cos x = 1

or y = sec x is the required particular solution.

Question 15.

Find the equation of the curve passing through the point (0, 0) and whose differential equation is y’ = e^{x} sin x.

Solution:

y’ = e^{x}sin x or \(\frac{dy}{dx}\) = e^{x}sin x

∴ dy = e^{x} sin x dx

Integrating both sides, we get

∫ dy = ∫e^{x} sin x dx

Integrating by parts, taking e^{x} as the first function, we get

y = e^{x}(- cos x) – ∫e^{x} sin x dx

= – e^{x}cos x + ∫e^{x}cos x dx

Again integrating by parts, taking e^{x} as first function, we get

y = – e^{x} cos x + e^{x} sin x – ∫e^{x} sin x dx

⇒ y = – e^{x} cos x + e^{x} sin x – ∫dy

⇒ y = – e^{x} cos x + e^{x} sin x + C

∴ y = \(\frac{e^{x}}{2}\)[- cos x + sin x] + C

Put x = 0, y = 0.

0 = – \(\frac{1}{2}\) + C ∴ C = \(\frac{1}{2}\).

∴ Solution is y = \(\frac{e^{x}}{2}\)(sin x – cos x) + \(\frac{1}{2}\)

Which is the required equation of the curve.

Question 16.

If for the differential equation xy\(\frac{dy}{dx}\) = (x + 2)(y + 2), find the solution curve passing through the point (1, – 1).

Solution:

The differential equation is xy\(\frac{dy}{dx}\) = (x + 2)(y + 2)

or xy dy = (x + 2)(y + 2)dx

Dividing by x(y + 2), we get

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

Integrating, we get

⇒ y – 2 log (y + 2) = x + 2 log x + C

The curve passes through (1, – 1)

∴ – 1 – 2 log 1 = 1 + 2 log 1 + C

⇒ – 1 = 1 + C [log 1 = 0]

∴ C = – 2.

Putting C = – 2, we get

y – 2 log (y + 2) = x + 2 log x – 2

or y – x = 2[log (y + 2) + log x] – 2

= 2 log x (y + 2) – 2

∴ Solution curve is y = x + 2 log x (y + 2) – 2.

Question 17.

Find the equation of the curve passing through the point (0, – 2) given that at any point (x, y) on the curve,

the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Solution:

Slope of tangent to the curve at (x, y) = \(\frac{dy}{dx}\).

We are given: y(\(\frac{dy}{dx}\)) = x

∴ ydy = xdx

Integrating, we get

\(\frac{y^{2}}{2}\) = \(\frac{x^{2}}{2}\) + C

or y^{2} = x^{2} + 2C.

The curve passes through (0, – 2).

∴ 4 = 0 + 2C.

∴ 2C = 4.

∴ Equation of the curve is

y^{2} = x^{2} + 4 or y^{2} – x^{2} = 4.

Question 18.

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (- 4, – 3).

Find the equation of the curve given that it passes through (- 2, 1).

Solution:

Slope of the tangent to the curve = \(\frac{dy}{dx}\)

Slope of the line joining (x, y) and (- 4, – 3) = \(\frac{y+3}{x+4}\)

We are given: \(\frac{dy}{dx}\) = 2(\(\frac{y+3}{x+4}\))

∴ dy = \(\frac{2(y+3)}{x+4}\) dx

Dividing by y + 3, we get

∴ \(\frac{dy}{y+3}\) = \(\frac{2}{x+4}\)dx

Integrating, we get

∫ \(\frac{dy}{y+3}\) = 2∫\(\frac{dx}{x+4}\)

or log(y + 3) = 2 log (x + 4) + log C

or log (y + 3) – log (x + 4)^{2} = log C

i.e; log \(\frac{y+3}{(x+4)^{2}}\) = C ∴ \(\frac{y+3}{(x+4)^{2}}\) = C.

The curve passes through (- 2, 1).

∴ \(\frac{1+3}{(-2+4)^{2}}\) = C = \(\frac{4}{4}\) = 1.

∴ Equation of the curve is \(\frac{y+3}{(x+4)^{2}}\) = 1.

or y + 3 = (x + 4)^{2} or y = (x + 4)^{2} – 3.

Question 19.

The volume of a spherical balloon, being inIated, changes at a constant rate.

If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of balloon after t seconds.

Solution:

Let V be the volume of the balloon.

Putting values of k and C in (1), we get

Question 20.

In a bank, principal increases at the rate of r% per year. Find the value of r, if ₹100 double itself in 10 years (log 2 = 0.6931).

Solution:

Let p be the principal an rate of interest = r%.

∴ r = 10 × 0.6931 = 6.931.

Thus, the value of r = 6.931.

Question 21.

In a bank, principal increases at the rate of 5% per year.

An amount of ₹1000 is deposited with this bank. How much will it worth after 10 years (e^{0.5} = 1.648)?

Solution:

Let p be the principal and Rate of interest is 5%.

∴ \(\frac{dp}{dt}\) = \(\frac{5}{100}\)p

∴ \(\frac{dp}{p}\) = 0.05 dt

Integrating, we get log p = 0.05 t + log C

or log p – log C = 0.05 t or log \(\frac{p}{C}\) = 0.05 t.

\(\frac{p}{C}\) = e^{0.005t} ∴ p = Ce^{0.005t} …………… (1)

Initially p = ₹1000, t = 0

∴ 1000 = C e^{0} = C

∴ C = 1000.

Putting this value in (1), we get

p = 1000 e^{0.005t}

When t = 10, p = 1000 e^{0.005×10} = 1000 e^{0.5}

p = 1000 × 1.648 [∵ e^{0.5} = 1.648]

= 1648

After 10 years, ₹1000 will amount to ₹1648.

Question 22.

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours.

In how many hours, will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solution:

Let y denote the number of bacteria at any instant t.

Then, according to the problem,

\(\frac{dy}{dt}\) ∝ y

⇒ \(\frac{dy}{y}\) = k dt ……………… (1)

where k is the constant of proportionality, taken to be positive. On integrating (1), we get

log y = kt + C ………………. (2)

where C is a parameter.

Let y_{0} be the initial number of bacteria, i.e., at t = 0.

Using this in (2), we get C = log y_{0}.

So, we have: log y = kt + log y_{0}

⇒ log \(\frac{y}{y_{0}}\) = kt

According to the problem,

y = (y_{0} + \(\frac{10}{100}\)y_{0}) = \(\frac{11 y_{0}}{10}\), when t = 2

So, from (3) we get,

Le the number of bacteria becomes from 1,00,000 to 2,00,000 in t_{1} hours i.e; y = 2y_{0}, when t = t_{1} hours.

Using (5) we get,

Question 23.

The general solution of the differential equation \(\frac{dy}{dx}\) = e^{x+y} is

(A) e^{x} + e^{-y} = c

(B) e^{x} + e^{y} = c

(C) e^{-x} + e^{y} = c

(D) e^{-x} + e^{-y} = c

Solution:

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

dy = e^{x}.e^{y} dx

Dividing by e^{y}, we get

e^{-y}dy = e^{x}dx

Integrating, we get

∫ e^{-y} dy = ∫ e^{x} dx

or – e^{-y} = e^{-x} – c = e^{x} + e^{-y} = c.

∴ Part (A) is the correct answer.