# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.11 | Set 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 = P

_{0}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 = P

_{0}Log|P

_{0}| = 0 + cc = log|P

_{0}|Log|P| = -kt + Log|P

_{0}|Log|P/P

_{0}| = -kt …(ii)According to the question,

At t = 1590, P = (P

_{0}/2)Log|P

_{0}/2P_{0}| = -1590t-Log|2| = -1590k

k = Log(2)/1590

Log|P/P

_{0}| = -[Log(2)/1590] × t|P/P

_{0}| =Find the radium after 1 year.

|P/P

_{0}| =P = 0.9996 × P

_{0}Percentage of disappeared in 1 year,

= [(P

_{0 }– P)/P_{0}] × 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

(y

^{2}/2) = -(x^{2}/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),

(y

^{2}/2) = -(x^{2}/2) + 25/2x

^{2 }+ y^{2 }= (5)^{2}x

^{2 }+ y^{2 }= 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) = y^{2 }+ (dy/dx)

**Solution:**

We have,

y – x(dy/dx) = y

^{2 }+ (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 – cos^{2}y/x) with x-axis.

**Solution:**

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

We have,

(dy/dx) = tan{tan

^{-1}(y/x – cos^{2}y/x)}(dy/dx) = (y/x – cos

^{2}y/x) …(i)Let y = vx

On differentiating both sides we have,

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

v + x(dv/dx) = v – cos

^{2}vx(dv/dx) = -cos

^{2}vsec

^{2}vdv = -(dx/x)On integrating both sides, we get

∫sec

^{2}vdv = -∫(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/y

^{4}|e

^{x/y }= c/y^{4}

### 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) = 2e^{2x}.

**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 + cy(e

^{-x}) = 2x∫e^{-x}dx – 2∫{(dx/dx)∫e^{-x}dx}dxe

^{-x}.y = -2xe^{-x }+ 2∫e^{-x}dx + ce

^{-x}.y = -2xe^{-x }– 2e^{-x }+ c …(ii)Since the curve is passes though origin (0, 0)

0×e

^{-0}= -0 – 2e^{-0 }+ cc = 2

On putting the value of c in equation (ii)

e

^{-x}.y = -2xe^{-x }– 2e^{-x }+ 2y = -2(x + 1) + 2e

^{x}y + 2(x + 1) = 2e

^{x}

### 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 + cy(e

^{-3x}) = 2x∫e^{-3x}dx – 2∫{(dx/dx)∫e^{-3x}dx}dxye

^{-3x }= -(2/3)xe^{-3x }+ (2/3)∫e^{-3x}dx + cye

^{-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 }+ cc = (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|4x

^{2}/9|y = 4x

^{2}/99y – 4x

^{2 }= 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^{-3x}dx + cy(e

^{-3x}) = ∫xe^{-3x}dx – ∫e^{-3x}dx – ∫e^{-3x}dx + cy(e

^{-3x}) = x∫e^{-3x}dx – ∫{(dx/dx)∫e^{-3x}dx + (e^{-3x}/3) + cy(e

^{-3x}) = -(x/3)e^{-3x }+ ∫(e^{-3x}/3) + (e^{-3x}/3) + cy(e

^{-3x}) = -(x/3)e^{-3x }– (e^{-3x}/9) + (e^{-3x}/3) + cy = -(x/3)-(1/9) + (1/3) + ce

^{3x}y = -(x/3) + 2/9 + ce

^{3x}Curve is passing through origin. x = 0 & y = 0

0 = 0 + 2/9 + ce

^{0}c = -2/9

y = -(x/3) + (2/9) – (2/9)e

^{3x}y + x/3 = (2/9)(1 – e

^{3x})(3y + x) = (2/3)(1 – e

^{3x})3(3y + x) = 2(1 – e

^{3x})