The availability of step-by-step ML Aggarwal Class 12 Solutions ISC Chapter 9 Differential Equations Ex 9.8 can make challenging problems more manageable.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.8

Question 1.
The surface area of a balloon being inflated changes at a constant rate. If initially, its radius is 3 units and after 2 seconds, it is 5 units, find the radius after t seconds.
Solution:
Let r be the radius of bal loon.
Then surface area of balloon = 4πr²
according to given condition, we have
$$\frac{d}{d t}$$ (4πr²) = K (say)
⇒ r dr = $$\frac{K}{8 \pi}$$ dt ;
on integrating
⇒ $$\frac{r^2}{2}=\frac{K}{8 \pi} t+\mathrm{C}$$ ………………….(1)
initially t = 0 ; r = 3 units
∴ from (1) ;
$$\frac{9}{2}$$ = C
Thus eqn. (1) gives ;
$$\frac{r^2}{2}=\frac{K}{8 \pi} t+\frac{9}{2}$$
⇒ r2 = $$\frac{K}{4 \pi}$$ + 9 ………………..(2)
Also when t = 2 seconds, r = 5 units
∴ from (2) ; we have
25 = $$\frac{K}{4 \pi}$$ × 2 + 9
⇒ 16 = $$\frac{K}{2 \pi}$$
⇒ K = 32 π
∴ from (2) ;
r2 = $$\frac{32 \pi}{4 \pi}$$ t + 9
= 8t + 9
⇒ r = $$\sqrt{8 t+9}$$
which is required radius after time t.

Question 2.
The surface area of a balloon being inflated changes at a rate proportional to time. If initially its radius is 1 unit and after I second ¡t is 3 units, find its radius after time t.
Solution:
Let r be the radius after time t
Then $$\frac{d}{d t}$$ (4πr²) ∝ t
⇒ $$\frac{d}{d t}$$ (4πr²) = Kt
where K be the constant of proportionality
8πr $$\frac{d r}{d t}$$ = Kt
⇒ r dr = $$\frac{K}{8 \pi}$$ t dt
On integrating ; we have
$$\frac{r^2}{2}=\frac{K}{8 \pi} \times \frac{t^2}{2}+\mathrm{C}$$ …………….(1)
initially t = 0, r = 1 units
∴ from (1) ;
$$\frac{1}{2}=\frac{\mathrm{K}}{16 \pi}$$ × 0 + C
⇒ C = $$\frac{1}{2}$$
∴ from (1) ; we have
$$\frac{r^2}{2}=\frac{K}{16 \pi} t^2+\frac{1}{2}$$ …………………….(2)
$$r^2=\frac{\mathrm{K}}{8 \pi} t^2+1$$
When t = 1 second ;
r = 3 units
∴ from (2) ;
9 = $$\frac{K}{8 \pi}$$ + 1
K = 64 π
∴ from (2) ;
r2 = $$\frac{64 \pi}{8 \pi}$$ t2 + 1
r = $$\sqrt{8 t^2+1}$$ units
which is the required radius after time t.

Question 3.
A population grows at the rate of 2 % per year. How long does it take for the population to double itself?
Solution:
Let P0 be the population initially i.e. at t = 0
and P be the population after t years.
According to given condition, we have
$$\frac{d P}{d t}$$ = 2% of P
$$\frac{d P}{d t}$$ = $$\frac{2}{100}$$ P
= $$\frac{t}{50}$$ P
On integrating ; we have
log P = $$\frac{t}{50}$$ + C ………………(1)
When t = 0; P = P0
∴ from (1) ;
log P0 = C
Thus eqn. (1) gives;
log P = $$\frac{t}{50}$$ + log P0
⇒ log $$\frac{\mathrm{P}}{\mathrm{P}_0}=\frac{t}{50}$$ ……………..(2)
Let t = t1
When P = 2P0
∴ from (2) ;
log $$\frac{2 \mathrm{P}_0}{\mathrm{P}_0}=\frac{t_1}{50}$$
⇒ t1 = 50 log 2
Hence it takes 50 log 2 years for the population to grow 2 times.

Question 4.
It is given that the rate at which some bacteria multiply is proportional to the instantaneous number present. If the original number of bacteria doubles in 2 hours, in how many hours will it be five times ?
Solution:
Let P0 be the number of bacteria present initially
i.e. t = 0 and P be the bacteria present after t years.
Then $$\frac{d P}{d t}$$ ∝ P
⇒ $$\frac{d P}{d t}$$ = KP ;
where K = constant of proportionality
on integrating ; we have
∫ $$\frac{d P}{P}$$ = K ∫ dt
⇒ log P = Kt + C …………….(1)
When t = 0 ; P = P0
∴ from (1) ;
log P0 = C
Thus eqn. (1) gives ;
log $$\frac{\mathrm{P}}{\mathrm{P}_0}$$ = Kt ……………..(2)
also P = 2P0 ; t = 2 hours
∴ from (2);
log $$\frac{2 \mathrm{P}_0}{\mathrm{P}_0}$$ = 2K
⇒ K = $$\frac{1}{2}$$ log 2
Thus eqn. (2) gives;
log $$\frac{\mathrm{P}}{\mathrm{P}_0}=\left(\frac{1}{2} \log 2\right) t$$ ………………(3)
Let t = t1 when P = 5P0
∴ eqn. (3) gives ;
log $$\frac{5 \mathrm{P}_0}{\mathrm{P}_0}$$ = ($$\frac{1}{2}$$ log 2) t1
⇒ t1 = $$\frac{2 \log 5}{\log 2}$$ hours
Hence $$\frac{2 \log 5}{\log 2}$$ hours, bacteria will be five times the no. of bacteria present initially.

Question 5.
Assume that a spherical rain drop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the rain drop at any time.
Solution:
Let r be the radius of rain drop after timer.
Then according to given condition,
$$\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)$$ ∝ 4πr²
⇒ $$\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)$$ = – K 4πr²
where K be the constant of proportionality
$$\frac{4 \pi}{3} \times 3 r^2 \frac{d r}{d t}$$ = – K 4πr² ;
⇒ $$\frac{d r}{d t}$$ = – K
on integrating
⇒ dr = – K ∫ dt + C
⇒ r = – Kt + C …………..(1)
When t = 0 : r = 3
∴ from (1) ;
3 = C
∴ eqn. (1) gives;
r = – Kt + 3 ………………(2)
further when t= 1 hour, r = 2 mm
∴ from (2) ;
2 = – K + 3
⇒ K = 1
putting the value of K in eqn. (2); we get
r = – t + 3 ; 0 < t < 3
Which gives the required radius of rain drop atany time t.

Question 5 (old).
A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the mind loses half its moisture the first hour, when will it have lost 90% moisture, whether conditions remaining the same ?
Solution:
Let M0 be the moisture content initially at t = 0
and M be the moisture constant after time t hours.
Then according to given condition, we have
$$\frac{d \mathrm{M}}{d t}$$ ∝ M
$$\frac{d \mathrm{M}}{d t}$$ = – KM
where K = constant of proportionality
$$\frac{d \mathrm{M}}{\mathrm{M}}$$ = – K dt [after variable separation]
∫ $$\frac{d \mathrm{M}}{\mathrm{M}}$$ = ∫ – K dt + C [on integrating]
⇒ log M = – Kt + C …………………..(1)
When t = 0 ;
M = M0
∴ from (1) ;
log M0 = C
∴ eqn. (1) gives ;
log M = – Kt + log M0
⇒ log $$\frac{\mathrm{M}}{\mathrm{M}_0}$$ = – Kt ………………..(2)
When t = 1, M = $$\frac{1}{2}$$ M0
∴ from (1) ;
log $$\left(\frac{\frac{1}{2} M_0}{M_0}\right)$$ = – K × 1
⇒ – K = log $$\frac{1}{2}$$
⇒ – K = log 1 – log 2
⇒ – K = – log 2
⇒ K = log 2
putting the value of K in eqn. (2) ; we have
log $$\left(\frac{M}{M_0}\right)$$ = – (log 2) t ………………….(3)
When the sheet loses 90% of the moisture.
i.e. When M = M0 – $$\frac{90}{100}$$ M0
= $$\frac{\mathrm{M}_0}{10}$$ t = t1
∴ from (3) ;
log $$\left(\frac{\mathrm{M}_0}{10 \mathrm{M}_0}\right)$$ = – (log 2) t1
⇒ t1 = $$\frac{1}{-\log 2} \log \frac{1}{10}$$
= $$\frac{\log 1-\log 10}{-\log 2}$$
= $$\frac{\log 10}{\log 2}$$

Question 6.
The acceleration of a particle moving along the line OX at any time t seconds is given by a = 2 (t – 5). If at t = 0, the velocity and displacement of the particle are 9 m/sec and 18 m respectively, find its velocity v and displacement x at any time t seconds. What is the velocity at t = 1 sec?
Solution:
Given acceleration of particle at any time t is given by
a = $$\frac{d^2 x}{d t^2}$$
= 2 (t – 5)
= $$\frac{d v}{d t}$$ ……………….(1)
where v be the velocity of particle at any time t.
On integrating eqn. (1) ; we have
v = 2 $$\left(\frac{t^2}{2}-5 t\right)$$ + C ………………(2)
given at t = 0 ; v = 9 m/sec
∴ from (2) ;
9 = 0 + C
⇒ C = 9
Substituting the value of C in eqn. (2) ; we have
$$\frac{d x}{d t}$$ = v = t2 – 10t + 9 ………………(3)
where x be the displacement of particle at any time t.
On integrating both sides of eqn. (3) ; we have
x = $$\frac{t^3}{3}-10 \frac{t^2}{2}+9 t+C_1$$ ………………(4)
Further at t = 0; x = 18 m
∴ from (4) ;
18 = C1
Thus, eqn. (4) gives;
x = $$\frac{t^3}{3}$$ – 5t2 + 9t + 18 ………………(5)
From (3) ;
(v)t = 1 = 1 – 10 + 9 = 0 m/sec.

Question 7.
Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve at any point (x, y) is $$\frac{y-1}{x^2+x}$$. (NCERT Exemplar)
Solution:
Given slope of tangent to the curve at any point (x, y) = $$\frac{y-1}{x^2+x}$$
Also, slope of tangent to curve at any point (x, y) = $$\frac{d y}{d x}$$

which gives the family of curves satisfied by diff. eqn. (1).
Now curve given by eqn. (2) passes through (1, 0).
i.e. when x = 1, y = 0
∴ from (2) ;
– 1 = $$\frac{C}{2}$$
⇒ C = – 2
putting the value of C in eqn. (2); we have
(y – 1) (x + 1) + 2x = 0, which is the required eqn. of curve.

Question 8.
Find the equation of a curve passing through the point (2, 1) if the slope of the tangent to the curve at any point (x, y) is $$\frac{x^2+y^2}{2 x y}$$.
Solution:
We know that, slope of tangent at any point (x, y) to the curve = $$\frac{d y}{d x}$$
Also, slope of tangent to curve at any point (x, y) = $$\frac{x^2+y^2}{2 x y}$$
∴ $$\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}$$ ………………..(1)
which is homogeneous differential equation
Put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$

which represents the family of curves satisfied by given differential equation (1).
Since the particular member of the family of curve passes through the point (2, 1);
we have
12 – 22 = C × 2
⇒ 2C = – 3
⇒ C = – $$\frac{3}{2}$$
putting the value of C in eqn. (2); we get
(y2 – x2) = – $$\frac{3 x}{2}$$
⇒ 2 (x2 – y2) = 3x,
which is the required equation of curve.

Question 8 (old).
Find the equation of a curve passing through the point (1, 1) whose differential equation ¡s x dy = (2x2 + 1) dx, x ≠ 0. (NCERT)
Solution:
Given differential eqn. can be written as
$$\frac{d y}{d x}=\frac{2 x^2+1}{x}$$ ;
on integrating
y = ∫ $$\frac{2 x^2+1}{x}$$ dx + C
⇒ y = $$\frac{2 x^2}{2}$$ + log |x| + C …………….(1)
Now (1) represents family of curves.
Now the curve passes through (I, 1)
∴ from (1) ;
1 = 1 + C
⇒ C = 0
∴ eqn.(1) gives ;
y = x2 + log |x| istherequired particular solution.

Question 9.
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point. (NCERT)
Solution:
We know that slope of tangent at any poiont (x, y) of the curve = $$\frac{d y}{d x}$$
also according to given condition, we have
$$\frac{d y}{d x}$$ = x + y
⇒ $$\frac{d y}{d x}$$ – y = x
which is linear differential equation of the form
$$\frac{d y}{d x}$$ + Py = Q ;
where P = – 1 and Q = x
∴ I.F. = e∫ P dx
= e∫ – 1 dx
= e– x
and solution is given by
ye∫ P dx = ∫ Q . e∫ P dx dx + c
⇒ ye– x = ∫ x e– x dx + c
⇒ y e– x = [- x e– x – e– x] + c
⇒ y = – (x + 1) + cex ……………………(1)
it is given that curve given by eqn (1) passes through (0, 0)
i.e. when x = 0 ; y = 0
∴ from (1) ; we have
0 = – (0 + 1) + c
⇒ c = 1
Thus from eqn. (1) ; we have
y = – x – 1 + ex
⇒ x + y = ex – 1
which gives the required equation of curve.

Question 10.
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point (x, y) on the curve exceeds the slope of the tangent to the curve at the point by 5. (NCERT)
Solution:
We know that,
slope of tangent to curve at any point (x,y) = $$\frac{d y}{d x}$$
according to given condition, we have
$$\frac{d y}{d x}$$ + 5 = x + y
⇒ $$\frac{d y}{d x}$$ – y = x – 5
which is linear duff. eqn. in y and is of the form $$\frac{d y}{d x}$$ + Py = Q
where P = – 1 and Q = x – 5
⇒ IF. = e∫ P dx
= e∫ – dx
= e– x
and solution is given by
y . e∫ P dx = ∫ Q . e∫ P dx dx + C
⇒ ye-x = ∫ (x – 5) e– x dx + C
⇒ ye– x = (x – 5) e– x – e– x + C
⇒ ye– x = – (x – 4) e– x + C
⇒ y = – (x – 4) + C e– x …………….(1)
Since the curve passing through the point (0, 2)
∴ eqn. (1) passes through (0, 2).
i.e. when x = 0 ; y = 2
∴ from (1) ;
2 = – (0 – 4) + C
⇒ C = – 2
Thus eqn. (1) gives ;
y = – (x – 4) – 2 ex
which gives the required eqn. of curve.

Question 10 (old).
Find the equation of a curve passing through the point (- 2, 3) given that the slope of the tangent to the curve at any point (x, y) is $$\frac{2 x}{y^2}$$. (NCERT)
Solution:
Given slope of tangent to curve at any point (x, y) = $$\frac{2 x}{y^2}$$
Also, slope of tangent to curve at any point (x, y) = $$\frac{d y}{d x}$$ ……………….(1)
∴ $$\frac{d y}{d x}=\frac{2 x}{y^2}$$
⇒ ∫ y2 dy = ∫ 2x dx + C
On integrating; we have
∫ y2 dy = ∫ 2x dx + c
⇒ $$\frac{y^3}{3}$$ = x2 + C ………………….(2)
So eqn. (2) represents the family of curves satisfied by diff eqn. (1).
Now given curve (2) passes through (- 2, 3),
i.e., when x = – 2, y = 3
∴ from (2) ;
9 = 4 + C
⇒ C = 5
Thus, from (2) gives ;
⇒ $$\frac{y^3}{3}$$ = x2 + 5
⇒ y3 = 3x2 + 15, which is the required equation of curve.