GSEB Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.4

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.
sec²x tan y dx + sec²y tan x dy = 0
Solution:
We have:
sec²x tan y dx + sec²y 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 + y²)
Solution:
$\frac{dy}{dx}$ = (1 + x)²(1 + y²)
or $\frac{d y}{1+y^{2}}$ = (1 + x²) dx
Integrating both sides, we get
∫ $\frac{d y}{1+y^{2}}$ = ∫
⇒ tan^-1y = 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.
x⁵$\frac{dy}{dx}$ = – y⁵
Solution:

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

Question 9.
$\frac{dy}{dx}$ = sin^-1x
Solution:
$\frac{dy}{dx}$ = sin^-1x
or dy = sin^-1x
Integrating both sides, we get
∫ dy = ∫sin^-1 x dx + C
or y = ∫(sin^-1 x). 1 dx + C
Integrating by parts, taking sin^-1x as first function, we get

Putting this value of I₁ in (1), we get
y = x sin^-1x + $\sqrt{1-x^{2}}$ + C
∴ The Solution is y = x sin^-1x + $\sqrt{1-x^{2}}$ + C.

Question 10.
e^xtan y dx + (1 – e^x) sec²y dy = 0
Solution:
e^x tan y dx + (1 – e^x) sec²y dy = – e^x tan y dx = 0
or (1 – e^x) sec²y 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.
(x³ + x² + x + 1)$\frac{dy}{dx}$ = 2x² + x; y = 1, when x = 0
Solution:
(x³ + x² + x + 1)$\frac{dy}{dx}$ = 2x² + x
or (x³ + x² + x + 1)dy = (2x² + x)dx
or dy = $\frac{2 x^{2}+x}{x^{3}+x^{2}+x+1}$ dx
Integrating both sides, we get

⇒ 2x² + x = A(x² + 1) + (Bx + C)(x + 1)
= A(x² + 1) + B(x² + 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 x² 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^-10 + C ……………. (3)
[Note : for integrating ∫ $\frac{2 x}{x^{2}+1}$dx, put x² + 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^-10 + C
= 0 + C ⇒ C = 1.
Thus, the solution is

This is the required particular solution.

Question 12.
x(x² – 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^xsin x or $\frac{dy}{dx}$ = e^xsin 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^xcos x + ∫e^xcos 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 y² = x² + 2C.
The curve passes through (0, – 2).
∴ 4 = 0 + 2C.
∴ 2C = 4.
∴ Equation of the curve is
y² = x² + 4 or y² – x² = 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 e⁰ = 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 y₀ be the initial number of bacteria, i.e., at t = 0.
Using this in (2), we get C = log y₀.
So, we have: log y = kt + log y₀
⇒ log $\frac{y}{y_{0}}$ = kt
According to the problem,
y = (y₀ + $\frac{10}{100}$y₀) = $\frac{11 y_{0}}{10}$, when t = 2
So, from (3) we get,

Le the number of bacteria becomes from 1,00,000 to 2,00,000 in t₁ hours i.e; y = 2y₀, when t = t₁ 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^-ydy = e^xdx
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.