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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.11 | Set 2

  • Last Updated : 05 Nov, 2021

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| = e^{-\frac{Log|2|}{1590}t}

Find the radium after 1 year. 

|P/P0| = e^{-\frac{Log|2|}{1590}}

P = 0.9996 × P0

Percentage of disappeared in 1 year,

= [(P0 – 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

x2 + y2 = (5)2

x2 + y2 = 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) = y2 + (dy/dx)

(dy/dx)(x + 1) = y(1 – y)

[dy/y(1 – y)] = dx/(x + 1)

On integrating both sides, we get

∫\frac{dy}{y(1-y)}=∫\frac{dx}{x+1}

∫[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)

=e^{-∫(\frac{1}{x}-\frac{1}{x+1})dx}

=e^{-log|x|+log|x+1|}

=e^{log|\frac{(x+1)}{x}|}

= (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 – 4x2 = 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)


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