# RD Sharma Class 12 Ex 22.1 Solutions Chapter 22 Differential Equations

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## RD Sharma Class 12 Ex 22.1 Solutions Chapter 22 Differential Equations

### Question 1. The surface area of a balloon being inflated changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.

Solution:

Let us considered radius = r

and the surface area of the balloon at a particular time ‘t’ = S

Surface area is given by,

S = 4πr2   …..(1)

We have,

(dS/dt)∝ t

(dS/dt) = kt (where k is proportional constant)

On differentiating eq(1), we get

d/dt(4πr2) = kt

8πr(dr/dt) = kt

ktdt = 8πrdr

On integrating both sides, we get

∫ktdt = ∫8πrdr

kt2/2 = 8π(r2/2) + c    …..(2)

At t = 0, r = 1 unit and at t = 3sec, r = 2units

0 = 4π + c

c = -4π

And,

k(3)2/2 = 8π(22/2) – 4π

(9/2)k = 12π

k = (8π/3)

On putting the values of k in equation (2)

(8π/3)(t2/2) = 8π(r2/2) – 4π

4t2/3 = 4r2-4

r= 1 + (t2/3)

Hence, the radius after time t = ### Question 2. A population grows at the rate of 5% per year. How long does it take for the population to double?

Solution:

Let us considered the initial population = P0

and the population at a particular time ‘t’ = P’

We have,

dP/dt = 5%P

dP/dt = 5P/100

dP/P = 0.05dt

On integrating both sides, we get

∫(dP/P) = ∫0.05dt

Log|P| = 0.05t + c

At t = 0, P = P0

log|P0| = c

Log|P| = 0.05t + log|P0|

Log|P/P0| = 0.05t

Now we find the time population becomes double,

P = 2P0

Log|2P0/P0| = 0.05t

Log|2| = 0.05t

t = 20Log|2| years

### Question 3. The rate of growth of a population is proportional to the number present population of a city doubled in the past 25 years, and the present population 100000, when will the city have a population of 500000?

Solution:

Let us considered

The initial population = P0

the population at a particular time ‘t’ = P

and the growth of population = g’

We have,

dP/dt = gP

dP/P = gdt

On integrating both sides, we get

∫(dP/P) = g∫dt

Log|P| = gt + c

At t = 0, P = P0

log|P0| = c

Log|P| = gt + log|P0|

Log|P/P0| = gt

Population of city is doubled in 25 years.

At t = 25, P = 2P0

Log|2P0/P0| = 25g

g = Log|2|/25

g = 0.0277

Log|P/P0| = 0.0277t

For P = 500000 and P= 100000

Log|500000/100000| = 0.0277t

t = Log|5|/0.0277

t = 58.08 year

t = 58 year

### Question 4. In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number of present?

Solution:

Let us assume the number of bacteria count at a particular time ‘t’ = P

We have,

dP/dt ∝ P

(dP/dt) = kP (where k is proportional constant)

(dP/P) = kdt

On integrating both sides, we get

∫(dP/P) = ∫kdt

Log|P| = kt + c

At t = 0, P = 100000

Log|100000| = c

Log|P| = kt + Log|100000|

After t = 2 hours number is increases by 10%.

Therefore, P = 100000 + (100000)(5/100)

P = 110000

Log|110000| = 2k + Log|100000|

k = (1/2)Log|11/10|

Log|P| = (t/2)Log|11/10| + Log|100000|      …(i)

Putting the value of k in equation (i)

Let at t = T P = 200000

Log|200000| = (T/2)Log|11/10| + Log|100000|

Log|2| = (T/2)Log|11/10|

### Question 5. If the interest is compounded continuously at 6% per annum, how much worth RS 1000 will be after 10 years? How long will it take to double RS. 1000?

Solution:

Let us assume be the initial amount = P0

and the amount at a particular time ‘t’ = P

dP/dt = 6%P

dP/dt = 6P/100

dP/P = 0.06dt

On integrating both sides, we get

∫(dP/P) = ∫0.06dt

Log|P| = 0.06t + c

At t = 0, P = P0

log|P0| = c

Log|P| = 0.06t + log|P0|

Log|P/P0| = 0.06t

At t = 10 years find the amount

Log|P/P0| = 0.06 × 10

Log|P/P0| = 0.6

P/P= e0.6

P = P0 × 1.8221

P = 1000 × 1.8221

P = 1822

At what time amount becomes double,

P = 2P0

Log|2P0/P0| = 0.06t

Log|2| = 0.06t

t = 16.66Loge|2|

t = 11.55 years

### Question 6. The rate of increase in the number of bacteria in a certain bacteria culture proportional to the number present, Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also, find the time necessary for the number of bacteria to be 10 times the number of initial present.

Solution:

Let us considered

The initial count of bacteria = P0

the count of bacteria at a particular time ‘t’ = P

and the growth of bacteria = g times.

We have,

dP/dt ∝ P

dP/dt = gP

dP/P = gdt

On integrating both sides, we get

∫(dP/P) = g∫dt

Log|P| = gt + c

At t = 0, P = P0

log|P0| = c

Log|P| = gt + log|P0|

Log|P/P0| = gt

At t = 5 hours, P = 3P0

Log|3P0/P0| = 5g

g = Loge|3|/5

g = 0.219722

Log|P/P0| = 0.219722t

At t = 10 hours find the numbers of bacteria.

Log|P/P0| = 0.219722 × 10

|P/P0| = e2.19722

|P/P0| = 9

P = 9P0

At ‘T’ time, a number of bacteria become 10 times.

At t = T, P = 10P0

Log|10P0/P0| = Loge|3|/5T

Log|10| = T(Loge|3|/5)

### Question 7. The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000. What will be the population in 2010?

Solution:

Let us considered

The initial population = P0

the population at a particular time ‘t’ = P

and the growth of population = g times.

We have,

dP/dt ∝ P

dP/dt = gP

dP/P = gdt

On integrating both sides, we get

∫(dP/P) = g∫dt

Log|P| = gt + c                …(i)

At t = 1990, P = 200000 and at t = 2000, P = 250000

Log|200000| = 1990g + c          …(ii)

Log|250000| = 2000g + c          …(iii)

On subtracting eq (iii) from (ii)

10g = Log|250000/200000|

g = (1/10)Log|5/4|

On putting the value of ‘g’ in equation (i)

Log|200000| = 1990 × (1/10)Log|5/4| + c

c = Log|200000| – 199 × Log|5/4|

Population in 2010,

Log|P| = (1/10)Log|5/4| × 2010 + Log|200000| – 199 × Log|5/4|

Log|P| = 201Log|5/4| – 199Log|5/4| + Log|200000|

Log|P| = Log|5/4|201 – Log|5/4|199 + Log|200000|

Log|P| = Log|(5/4)201(4/5)199| + log|200000|

Log|P| = Log|5/4|+ log|200000|

Log|P| = Log|(25/16)200000|

Log|P| = Log|312500|

P = 312500

### Question 8. If the marginal cost of manufacturing a certain item is given by C'(x) = (dC/dx) = 2 + 0.15x. Find the total cost function C(x), given that C(0) = 100

Solution:

We have,

dC/dx = 2 + 0.15x

dC = (2 + 0.15x)dx

On integrating both sides, we get

∫dC = ∫(2 + 0.15x)dx

C(x) = 2x + (0.15/2)x+ c1

At C(0) = 100, we have

100 = 2(0) + (0.15/2)(0)+ c1

c= 100

C(x) = 0.075x+ 2x + 100

### Question 9. A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.

Solution:

Let us considered

The initial population = P0

and the population at a particular time ‘t’ = P

We have,

dP/dt = (8/100)P

dP/dt = (2/25)P

dP/P = (2/25)dt

On integrating both sides, we get

∫(dP/P) = (2/25)∫dt

Log|P| = (2/25)t + c            …(i)

At t = 0, P = P0

Log|P0| = 0 + c

c = Log|P0|

On putting the value of c in equation (i)

Log|P| = (2/25)t + Log|P0|

Log|P/P0| = (2t/25)

Amount after 1 year,

Log|P/P0| = (2/25)

e(2/25) = |P/P0|

e0.08 = |P/P0|

1.0833 = |P/P0|

P = 1.0833P0

Percentage increase = [(P – P0)/P0] × 100%

= [(1.0833P– P0)/P0] × 100%

= 0.0833 × 100%

= 8.33%

### Question 10. In a simple circuit of resistance R, self-inductance L and voltage E, the current i at any time is given by L(di/dt) + Ri = E. If E is constant and initially no current passes through the circuit, prove that i = (E/R){1 – e-(R/L)t}

Solution:

We have,

L(di/dt) + Ri = E

(di/dt) + (R/L)i = E/L

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (R/L), Q = E/L

So, I.F = e∫Pdi

= e∫(R/L)di

= e(R/L)t

The solution of a differential equation is,

i(I.F) = ∫Q(I.F)dt + c

e(R/L)t × i = (E/L)∫e(R/L)tdt + c

e(R/L)t × i = (E/L)(L/R)e(R/L)t + c

e(R/L)t × i = (E/R)e(R/L)t + c      …(i)

At t = 0, i = 0

e× 0 = (E/R)e+ c

c = -(E/R)

On putting the value of c in equation (i)

e(R/L)t × i = (E/R)e(R/L)t – (E/R)

i = (E/R) – (E/R)e-(R/L)t

### Question 11. The decay rate of the radius at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.

Solution:

Let us considered

and radius at a particular time ‘t’ = R

We have,

dR/dt ∝ R

dR/dt = -kR

dR/R = -kdt

On integrating both sides, we get

∫(dR/R) = -k∫dt

Log|R| = -kt + c         …(i)

At t = 0, R = R0

Log|R0| = 0 + c

c = Log|R0|

Log|R| = -kt + Log|R0|

kt = Log|R0/R|

At time ‘T’ mass becomes R0/2

Log|2| = kT

T = (1/k)Log|2|

### Question 12. Experiments show that radium disintegrates at a rue proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year.

Solution:

Let us considered

the original amount of radium = P0

and the amount of radium at a particular time ‘t’ = P

We have,

dP/dt ∝ P

(dP/dt) = -kP  (Where k is proportional constant)

(dP/P) = -kdt

On integrating both sides, we get

∫(dP/P) = -∫kdt

Log|P| = -kt + c       …(i)

At t = 0, P = P0

Log|P0| = 0 + c

c = log|P0|

Log|P| = -kt + Log|P0|

Log|P/P0| = -kt         …(ii)

According to the question,

At t = 1590, P = (P0/2)

Log|P0/2P0| = -1590t

-Log|2| = -1590k

k = Log(2)/1590

Log|P/P0| = -[Log(2)/1590] × t

|P/P0| = Find the radium after 1 year.

|P/P0| = P = 0.9996 × P0

Percentage of disappeared in 1 year,

= [(P– P)/P0] × 100

= [(1 – 0.9996)/1] × 100

= 0.04%

### Question 13. The slope of the tangent at a point P(x, y) on a curve is -(x/y). If the curve passes through the point (3, -4), Find the curve.

Solution:

Slope at a point is given by = (dy/dx)

According to the question,

(dy/dx) = -(x/y)

ydy = -xdx

On integrating both sides, we get

∫ydy = -∫xdx

(y2/2) = -(x2/2) + c      …(i)

Curve is passing through (3, -4)

16/2 = -(9/2) + c

c = 25/2

On putting the value of c in equation (i),

(y2/2) = -(x2/2) + 25/2

x+ y= (5)2

x+ y= 25

### Question 14. Find the equation of the curse which passes through the point (2, 2) and satisfies the differential equation y – x(dy/dx) = y2 + (dy/dx)

Solution:

We have,

y – x(dy/dx) = y+ (dy/dx)

(dy/dx)(x + 1) = y(1 – y)

[dy/y(1 – y)] = dx/(x + 1)

On integrating both sides, we get

∫[1/y + 1/(1 – y)]dy = ∫dx/(x + 1)

Log|y| – Log|1 – y| = Log|x + 1| + c    …(i)

At x = 2, y = 2

Log|2| – Log|1 – 2| = Log|3| + c

Log|2/3| = c

On putting the value of c in equation (i)

Log|y/(1 – y)| = Log|x + 1| + Log|2/3|

Log|y/(1 – y)| = Log|2(x + 1)/3|

|y/(1 – y)| = |2(x + 1)/3|

y/(1 – y) = ±(2x + 2)/3

y/(1 – y) = (2x + 2)/3 or -(2x + 2)/3

Point (2, 2) is not satisfy y/(1 – y) = (2x + 2)/3

It satisfies the equation y/(1 – y) = -(2x + 2)/3

So,

y/(1 – y) = -(2x + 2)/3

3y = -(2x + 2)(1 – y)

3y = -2x + 2xy – 2 + 2y

2xy – 2x – y – 2 = 0

### Question 15. Find the equation of the curve passing through the point (1, π/4) and tangent at any point of which makes an angel tan-1(y/x – cos2y/x) with x-axis.

Solution:

Slope of curve is given by, (dy/dx) = tanθ

We have,

(dy/dx) = tan{tan-1(y/x – cos2y/x)}

(dy/dx) = (y/x – cos2y/x)      …(i)

Let y = vx

On differentiating both sides we have,

(dy/dx) = v + x(dv/dx)

v + x(dv/dx) = v – cos2v

x(dv/dx) = -cos2v

sec2vdv = -(dx/x)

On integrating both sides, we get

∫sec2vdv = -∫(dx/x)

tanv = -log|x| + c

tan(y/x) = -log|x| + c          …(i)

Curve is passing through (1, π/4)

So,

tan(π/4) = -log|1| + c

c = 1

On putting the value of c in equation (i)

tan(y/x) = -log|x| + 1

tan(y/x) = -log|x| + loge

tan(y/x) = log|e/x|

### Question 16. Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

Solution:

Let us considered the point of contact of tangent = P(x, y)and the curve is y = f(x).

So, the equation of tangent of the curve is given by,

Y – y = (dy/dx)(X – x)

Where (X, Y) is arbitrary point on the tangent.

Putting Y = 0, we get

0 – y = (dy/dx)(X – x)

(X -x) = -y(dx/dy)

X = x – y(dx/dy)

We have,

According to the question,

x – y(dx/dy) = 4y

y(dx/dy) + 4y = x

(dx/dy) + 4 = x/y

(dx/dy) – (x/y) = -4

The above equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -1/y, Q = -4

So, I.F = e∫Pdy

= e∫-dy/y

= e-log|y|

= 1/y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + log|c|

x(1/y) = ∫(-4).(1/y)dy + log|c|

(x/y) = -4∫dy/y + log|c|

(x/y) = -4log|y| + log|c|

(x/y) = log|c/y4|

ex/y = c/y4

### Question 17. Show that the equation of the curve whole slope at any point is equal to y + 2x and which passes through the origin is y + 2(x + 1) = 2e2x.

Solution:

(dy/dx) = y + 2x

(dy/dx) – y = 2x           …(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = 2x

So, I.F = e∫Pdx

= e-∫dx

= e-x

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e-x) = ∫(e-x).(2x)dx + c

y(e-x) = 2x∫e-xdx – 2∫{(dx/dx)∫e-xdx}dx

e-x.y = -2xe-x + 2∫e-xdx + c

e-x.y = -2xe-x – 2e-x + c       …(ii)

Since the curve is passes though origin (0, 0)

0×e-0= -0 – 2e-0 + c

c = 2

On putting the value of c in equation (ii)

e-x.y = -2xe-x – 2e-x + 2

y = -2(x + 1) + 2ex

y + 2(x + 1) = 2ex

### Question 18. The tangent at any point (x, y) of a curve makes an angle tan-1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).

Solution:

Slope of curve is given by,

(dy/dx) = tanθ

θ = tan-1(2x + 3y)

(dy/dx) = tan[tan-1(2x + 3y)]

(dy/dx) = 2x + 3y

(dy/dx) – 3y = 2x

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -3, Q = 2x

So, I.F = e∫Pdx

= e-3∫dx

= e-3x

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e-3x) = ∫(e-3x).(2x)dx + c

y(e-3x) = 2x∫e-3xdx – 2∫{(dx/dx)∫e-3xdx}dx

ye-3x = -(2/3)xe-3x + (2/3)∫e-3xdx + c

ye-3x = -(2/3)xe-3x – (2/9)e-3x + c       …(i)

Since the curve passes through (1, 2)

2e-3 = -(2/3)e-3 – (2/9)e-3 + c

c = (26/9)e-3

On putting the value of c in equation (i)

ye-3x = -(2/3)xe-3x – (2/9)e-3x + (26/9)e-3

### Question 19. Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).

Solution:

Let us considered the point of contact of tangent = P(x, y)

and the curve is y = f(x).

So, the equation of tangent of the curve is given by,

Y – y = (dy/dx)(X – x)

Where (X, Y) is arbitrary point on the tangent.

Putting Y = 0,

0 – y = (dy/dx)(X – x)

(X – x) = -y(dx/dy)

X = x – y(dx/dy)

We have,

According to the question,

The tangent at a point is twice the abscissa (i.e. 2x)

x – y(dx/dy) = 2x

-x = y(dx/dy)

(dy/y) = -(dx/x)

On integrating both sides

∫(dy/y) = -∫(dx/x)

log|y| = -log|x| + log|c|

log|y| = log|c/x|

y = c/x

xy = c         …(i)

The curve is passing though the point (1, 2)

1 × 2 = c

c = 2

On putting the value of c in equation (i)

xy = 2

### Question 20. Find the equation to the curve satisfying x(x + 1)(dy/dx) – y = x(x + 1) and passing through (1, 0).

Solution:

We have,

x(x + 1)(dy/dx) – y = x(x + 1)

(dy/dx) – [y/x(x + 1)] = 1

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x(x + 1), Q = 1

So, I.F = e∫Pdx

= e-∫dx/x(x+1)

= (x + 1)/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[(x + 1)/x] = ∫[(x + 1)/x]dx + c

y[(x + 1)/x] = ∫(1 + 1/x)dx

y[(x + 1)/x] = x + log|x| + c          …(i)

Since line is passing through (1, 0)

0 = 1 + 0 + c

c = -1

y[(x + 1)/x] = x + log|x| – 1

y(x + 1) = x(x + log|x| – 1)

### Question 21. Find the equation of the curve which passes through the point (3, -4) and has the slope 2y/x at any points (x, y) on it.

Solution:

We have,

(dy/dx) = 2y/x

(dy/2y) = (dx/x)

On integrating both sides

∫(dy/2y) = ∫(dx/x)

(1/2)log|y| = log|x|+log|c|            …(i)

Since the curve is passing through (3,-4)

(1/2)log|-4| = log|3| + log|c|

log|2| – log|3| = log|c|

log|c| = log|2/3|

On putting the value of log|c| in equation (i)

log|y| = 2log|x| + 2log|2/3|

log|y| = log|4x2/9|

y = 4x2/9

9y – 4x= 0

### Question 22. Find the equation of the curve the slope which passes through the origin and has the slope x+3y-1 at any point on it.

Solution:

Slope of a curve is (dy/dx)

We have,

(dy/dx) = x + 3y – 1

(dy/dx) – 3y = (x – 1)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -3, Q = (x – 1)

So, I.F = e∫Pdx

= e-∫3dx

= e-3x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e-3x) = ∫(x – 1)e-3xdx + c

y(e-3x) = ∫xe-3xdx – ∫e-3xdx – ∫e-3xdx + c

y(e-3x) = x∫e-3xdx – ∫{(dx/dx)∫e-3xdx + (e-3x/3) + c

y(e-3x) = -(x/3)e-3x + ∫(e-3x/3) + (e-3x/3) + c

y(e-3x) = -(x/3)e-3x – (e-3x/9) + (e-3x/3) + c

y = -(x/3)-(1/9) + (1/3) + ce3x

y = -(x/3) + 2/9 + ce3x

Curve is passing through origin. x = 0 & y = 0

0 = 0 + 2/9 + ce0

c = -2/9

y = -(x/3) + (2/9) – (2/9)e3x

y + x/3 = (2/9)(1 – e3x)

(3y + x) = (2/3)(1 – e3x)

3(3y + x) = 2(1 – e3x)

### Question 23. At every point on a curve, the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1) Find the equation of the curve.

Solution:

Slope is given by, (dy/dx)

We have,

(dy/dx) = x + xy

(dy/dx) = x(y + 1)

dy/(y + 1) = xdx

On integrating both sides, we get

∫dy/(y + 1) = ∫xdx

Log|y + 1| = (x2/2) + c       …(i)

Since the curve is passing through (0, 1)

Log|2| = c

Log|y + 1| = (x2/2) + Log|2|

Log|(y + 1)/2| = x2/2

### Question 24. A curve is such that the length of the perpendicular from the origin on the tangent any point of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is y2 – 2xy(dy/dx) – x2 = 0, and hence find the curve.

Solution:

Tangent of the curve is given by,

Y – y = (dy/dx)(X – x)

If P be perpendicular from the origin, then

x2(dy/dx)– 2xy(dy/dx) + y= x+ x2(dy/dx)2

y– 2xy(dy/dx) – x= 0

Hence Proved.

Now we find the curve

So, weh have y– 2xy(dy/dx) – x= 0

(dy/dx) = (y– x2)/2xy

Above equation is homogeneous equation.

Let, y = vx

On differentiating both sides we have,

(dy/dx) = v + x(dv/dx)

v + x(dv/dx) = (v2x– x2)/2xvx

v + x(dv/dx) = (v– 1)/2v

x(dv/dx) = [(v– 1)/2v] – v

x(dv/dx) = (v– 1 – 2v2)/2v

x(dv/dx) = -(1 + v2)/2v

2vdv/(1 + v2) = -(dx/x)

On integrating both sides

∫2vdv/(1 + v2) = -∫(dx/x)

log|1 + v2| = -log|x| + log|c|

log|x(1 + v2)| = log|c|

x(1 + y2/x2) = c

(x+ y2) = cx

### Question 25. Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.

Solution:

Let us considered

the point of contact of tangent = P(x, y)

and the curve is y = f(x).

The equation of tangent of the curve is given by,

Y – y = (dy/dx)(X – x)

Where (X, Y) is arbitrary point on the tangent.

Putting Y = 0,

0 – y = (dy/dx)(X – x)

(X – x) = -y(dx/dy)

X = x – y(dx/dy)

Coordinates at contact of x-axis = [x – y(dx/dy), 0]

The distance between the foot of the ordinate of the point of

contact and the point of intersection of the tangent with x-axis is equal to 2x.

y(dx/dy) = 2x

(dx/x) = 2(dy/y)

On integrating both sides

∫(dx/x) = 2∫(dy/y)

log|x| = 2log|y| + log|c|        …(i)

Curve is passing through (1, 2)

log|1| = 2log|2| + log|c|

log|c| = -2log|2|

On putting the value of log|c| in equation (i)

log|x| = 2log|y| – 2log|2|

log|x| = log|y2/4|

x = (y2/4)

y= 4x

### Question 26. The normal to a given curve at each point is(x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4). Find its equation.

Solution:

Let us considered the point on the curve = P(x, y).

The equation of tangent of the curve is given by,

Y – y = -(dx/dy)(X – x)         …(i)

It passes through (3, 0) So,

0 – y = -(dx/dy)(3 – x)

ydy = 3dx – xdx

On integrating both sides

∫ydy = 3∫dx – ∫xdx

(y2/2) = 3x – (x2/2) + c

It passes through (3, 4)

(16/2) = 9 – (9/2) + c           …(ii)

c = (16/2) – (9/2)

c = (7/2)

(y2/2) = 3x – (x2/2) + (7/2)

or

y2 = 6x – x2 + 7

### Question 27. The rate of increase of bacteria in culture is proportional to the number of bacteria present, and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.

Solution:

Let us considered

the initial count of bacteria = P0

the count of bacteria at a particular time ‘t’ = P

and the growth of bacteria = g times.

We have,

dP/dt ∝ P

dP/dt = gP

dP/P = gdt

On integrating both sides

∫(dP/P) = g∫dt

Log|P| = gt + c

At t = 0, P = P0

log|P0| = c

Log|P| = gt + log|P0|

Log|P/P0| = gt

Count of bacteria becomes doubled in 6 hours.

At t = 6, P = 2P0

Log|2P0/P0| = 6g

g = Log|2|/6

Log|P/P0| = [Log|2|/6] × t

After t = 18 hours count of bacteria is equal to

Log|P/P0| = [Log|2|/6] × 18

Log|P/P0| = 3Log|2|

Log|P/P0| = Log|2|3

(P/P0) = 8

P = 8P0

Hence proved

### Question 28. Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of radium to decompose?

Solution:

Let us considered

the original amount of radium = P0

and the amount of radium at a particular time ‘t’ = P

We have,

dP/dt ∝ P

(dP/dt) = -kP  (Where k is proportional constant)

(dP/P) = -kdt

On integrating both sides

∫(dP/P) = -∫kdt

Log|P| = -kt + c       …(i)

At t = 0, P = P0

Log|P0| = 0 + c

c = log|P0|

Log|P| = -kt + Log|P0|

Log|P/P0| = -kt         …(ii)

According to the question,

In 25 years bacteria decomposes 1.1%.

So, P = (100 – 1.1)%P0

P = 0.989P0

Log|P/P0| = -kt

Log|0.989| = -25k

k = -(1/25)Log|0.989|

On putting the value of k in equation (ii)

Log|P/P0| = (1/25)Log|0.989| × t

Time ‘T’ for one-half of the original amount of radium(i.e., P= P/2)

Log|P0/(P0/2)| = (1/25)Log|0.989| × t

Log|2| = (1/25)Log|0.989| × t

t = (25Log|2|/Log|0.989|)

t = (25×0.69311)/(0.01106)

t = 1566.70

t = 1567 years

### Question 29. Show that all curves for which the slope at any point (x, y) on it is (x2 + y2)/2xy are rectangular hyperbola.

Solution:

We have,

(dy/dx) = (x+ y2)/2xy

The given equation is a homogeneous equation,

So, let us considered, y = vx

On differentiating both sides we have,

(dy/dx) = v + x(dv/dx)

(dy/dx) = (x+ v2x2)/2xvx

v + x(dv/dx) = (x+ v2x2)/2xvx

v + x(dv/dx) = (1 + v2)/2v

x(dv/dx) = [(1 + v2)/2v] – v

x(dv/dx) = (1 + v– 2v2)/2v

x(dv/dx) = (1 – v2)/2v

2vdv/(1 – v2) = (dx/x)

On integrating both sides

∫2vdv/(1 – v2) = ∫(dx/x)

-log|1 – v2| = log|x| – log|c|

-log|(1 – v2)| = -[log|c| – log|x|]

-log|(1 – v2)| = -log|c/x|

(1 – v2) = c/x

(1 – y2/x2) = c/x

(x– y2)/x= c/x

(x– y2) = cx

This is the required equation of a rectangular hyperbola.

### Question 30. The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.

Solution:

The equation of tangent of the curve is given by,

(dy/dx) = x + y

(dy/dx) – y = x

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -1, Q = -x

So, I.F = e∫Pdx

= e∫-dx

= e-x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx+c

y(e-x) = ∫(x).(e-x)dy+c

y(e-x) = x∫e-xdx-∫{(dx/dx)∫e-xdx}dx+c

y(e-x) = -xe-xx + ∫e-x + c

y(e-x) = -xe-x – e-x + c

y(e-x) = -e-x(x + 1) + c      (i)

Since the curve is passing through origin. So,

0 = -e-0(1) + c

c = 1

On putting the value of c in equation (i)

y(e-x) = -e-x(x + 1) + c

(x + y + 1) = c.ex

Put c = 1

x + y + 1 = ex

### Question 31. Find the equation of the curve passing through the point (0, 1) if the slope of tangent to the current each of its point is equal to the sum of the abscissa and product of the abscissa and the ordinate of the point.

Solution:

The equation of tangent of the curve is given by

(dy/dx) = x + xy

(dy/dx) – xy = x

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -x, Q = x

So, I.F = e∫Pdx

= e-∫xdx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

Let,

Let,x2/2 = z

On differentiating both sides,

xdx = dz

I = ∫e-zdz

I = -e-z

So,

Curve is passing through point (0, 1).

1e= -e+ c

c = 2

### Question 32. The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (-1, 1).

Solution:

According to the question,

(dy/dx) = x2

dy = x2dx

On integrating both sides

∫dy = ∫x2dx

y = (x3/3) + c         …(i)

Curve is passing through the point (-1, 1)

1 = -(1/3) + c

c = (4/3)

On putting the value of c in equation (i)

y = (x3/3) + (4/3)

3y = x+ 4

### Question 33. Find the equation of the curve that passes through the point (0, a) and as such at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.

Solution:

According to the question,

y(dy/dx) = x

dy = x2dx

On integrating both sides

∫ydy = ∫xdx

(y2/2) = (x2/2) + c         …(i)

Curve is passing through the point (0, a)

(a2/2) = c

c = a2/2

On putting the value of c in equation (i)

(y2/2) = (x2/2) + (a2/2)

x– y= -a2

### Question 34. The x-intercept of the tangent line to a curve is equal to the ordinate of the point contact. Find the particular curve through the point (1, 1).

Solution:

Slope at any point is given by P = (dy/dx)

According to the question,

Slope at any point is equal to ordinate

We have,

(dy/dx) = y

dy/y = dx

On integrating both sides

∫(dy/y) = ∫(dx)

log|y| = x + log|c|

log|y| = log|ex| + log|c|

y = c.ex            …(i)

Curve is passing through the point (1, 1)

1 = c.e

c = e-1

On putting the value of c in equation (i)

y = ex.e-1

y = e(x-1)

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