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

In Chapter 22 of RD Sharma's textbook for Class 12, which covers Differential Equations, Exercise 22.11 focuses on applying various methods to solve higher-order differential equations. These equations are critical in modeling dynamics and processes where the rate of change depends not only on the state but also on how the state changes over time.

Question 12. Experiments show that radium disintegrates at a rate 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,

= [(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

∫[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 angle 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|

e^x/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) = 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 + 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) + 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 + 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 - 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) + ce^3x

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

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

0 = 0 + 2/9 + ce⁰

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)

Formulas and Theorems

1. General Form of a First-Order Differential Equation:

P(x, y) dx + Q(x, y) dy = 0

2. Exact Differential Equation:

A differential equation M dx + N dy = 0 is exact if ∂M/∂y = ∂N/∂x

3. Solution of an Exact Differential Equation:

If M dx + N dy = 0 is exact, its solution is ∫M dx + ∫(N - ∂/∂y∫M dx) dy = C

4. Integrating Factor:

μ(x) = e^∫P(x)dx, where P(x) is the coefficient of dy/dx in the standard form

5. Linear Differential Equation:

dy/dx + P(x)y = Q(x)

Solution: y = e^(-∫P(x)dx) [∫Q(x)e^(∫P(x)dx)dx + C]

Practice Questions

1. Solve: (x² + y²) dx + xy dy = 0

2. Find the integrating factor and solve: (2x + y) dx + (x - y) dy = 0

3. Solve the exact differential equation: (2xy + y²) dx + (x² + 2xy) dy = 0

4. Verify that the following is an exact differential equation and solve it:

(y² + 2xy) dx + (2xy + x²) dy = 0

5. Solve the linear differential equation: dy/dx + y/x = x², x > 0

Conclusion

Exercise 22.11 | Set 2 of RD Sharma's Class 12 Differential Equations chapter delves into advanced techniques for solving first-order differential equations. This set focuses on exact differential equations, integrating factors, and linear differential equations. Students learn to identify exact equations, apply integrating factors to transform non-exact equations into exact ones, and solve linear differential equations using the integrating factor method. The exercise emphasizes the importance of recognizing different types of differential equations and selecting the appropriate solution method. By mastering these techniques, students develop crucial problem-solving skills applicable in various fields of science and engineering where differential equations model real-world phenomena. This exercise serves as a bridge between basic differential equation concepts and more complex applications in higher mathematics and applied sciences.

Related Articles

Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.12

Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.10