Question

NCERT Solutions Class 12 Maths Chapter 9 Differential Equations is designed by the experts of the subject in a very concise manner to make it easy for students to through all different concepts and learn everyone in complete detail. NCERT Solutions have solutions to all kinds of queries that might arise during the student’s exam preparation.

Our experts have prepared the NCERT Solutions Class 12 for easy studying and revision of all different concepts.

• Differential Equations – the differential equation is such an equation that contains one or more than one function with its derivative. The derivative is defined as the function which changes according to some specific rate of a function at any given point. The differential equation has applications in different fields of physics, engineering, biology, and many others. One can solve the equation and the properties of the solutions. A differential equation contains the terms of the derivative of one variable concerning another variable.
• Order of Differential Equation – there are two types of differential equation
  • First-order differential equation – such differential equation which has order equal to 1 is called that first-order differential equation. All the derivative of the form of linear equation exists in the first order.
  • Second-order differential equation – such differential equation which has a second-order differential equation.
  • Basic Concepts of differential equations – calculus is a calculation of change and the rate of change is expressed in the derivatives. The differential equation is an unknown function \(y = f(x)\). Differential equations very often have information have quantities that change upon them.
• Particular Solutions of a Differential Equation – the unique solution of the differential equation satisfies some differential equation. In the differential equation, we have an arbitrary constant solution which is the particular solution of the differential equation.
• Formation of a Differential Equation whose General Solution is given – in an equation we have an independent and a dependent variable that involves an arbitrary constant.

NCERT Solutions Class 12 Maths has detailed solutions for a complete understanding of the subject matter.

Answer 1

NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

1. Determine order and degree(if defined) of differential equation \(\frac{d^4y}{dx^4} + sin(y^m) = 0\).

Answer:

\(\frac{d^4y}{dx^4} + sin(y^m) = 0\)

⇒ \(y ^{m'} + sin(y^m) = 0\)

The highest order derivative present in the differential equation is y^m'. 

Therefore, its order is four. 

The given differential equation is not a polynomial equation in its derivatives. 

Hence, its degree is not defined.

2. Determine order and degree(if defined) of differential equation y' + 5y = 0.

Answer:

The given differential equation is:

y' + 5y = 0

The highest order derivative present in the differential equation is y'. 

Therefore, its order is one. 

It is a polynomial equation in y'. The highest power raised to is 1. 

Hence, its degree is one.

3. Determine order and degree(if defined) of differential equation \(\left(\frac{ds}{dt}\right)^4 + 3s \frac{d^2s}{dt^2} = 0\).

Answer:

\(\left(\frac{ds}{dt}\right)^4 + 3s \frac{d^2s}{dt^2} = 0\) 

The highest order derivative present in the given differential equation is \(\frac{d^2s}{dt^2}\). 

Therefore, its order is two. 

It is a polynomial equation in \(\frac{d^2s}{dt^2}\) and \(\frac{ds}{dt}\). The power raised to \(\frac{d^2s}{dt^2}\) is 1. 

Hence, its degree is one.

4. Determine order and degree(if defined) of differential equation \(\left(\frac{d^2y}{dx^2}\right)^2 + cos\left(\frac{dy}{dx}\right) = 0\).

Answer:

\(\left(\frac{d^2y}{dx^2}\right)^2 + cos\left(\frac{dy}{dx}\right) = 0\) 

The highest order derivative present in the given differential equation is \(\frac{d^2y}{dx^2}\). 

Therefore, its order is 2. 

The given differential equation is not a polynomial equation in its derivatives. 

Hence, its degree is not defined.

5. Determine order and degree(if defined) of differential equation \(\frac{d^2y}{dx^2} = cos 3x + sin3x\) 

Answer:

\(\frac{d^2y}{dx^2} = cos 3x + sin3x\) 

⇒ \(\frac{d^2y}{dx^2} - cos 3x - sin3x = 0\) 

The highest order derivative present in the differential equation is \(\frac{d^2y}{dx^2}\). 

Therefore, its order is two.

It is a polynomial equation in \(\frac{d^2y}{dx^2}\) and the power raised to \(\frac{d^2y}{dx^2}\) is 1. 

Hence, its degree is one.

6. Determine order and degree(if defined) of differential equation (y’’’)² + (y’’)³ + (y’)⁴ + y⁵ = 0.

Answer:

(y’’’)² + (y’’)³ + (y’)⁴ + y⁵ = 0 

The highest order derivative present in the differential equation is y’’’. 

Therefore, its order is three.

 The given differential equation is a polynomial equation in y’’’, y’’ and y’. 

The highest power raised to y’’’ is 2. 

Hence, its degree is 2.

7. Determine order and degree(if defined) of differential equation y’’’ + 2y’’ + y’ = 0. 

Answer:

y’’’ + 2y’’ + y’ = 0

The highest order derivative present in the differential equation is y’’’. 

Therefore, its order is three. 

It is a polynomial equation in y’’’, y’’ and y’. 

The highest power raised to y’’’ is 1. 

Hence, its degree is 1.

8. Determine order and degree(if defined) of differential equation y' + y = e^x.

Answer:

y' + y = e^x

⇒ y' + y - e^x = 0

The highest order derivative present in the differential equation is y'. 

Therefore, its order is one. 

The given differential equation is a polynomial equation in y' and the highest power raised to y' is one. 

Hence, its degree is one.

9. Determine order and degree(if defined) of differential equation y’’ + (y’)² + 2y = 0.

Answer:

y’’ + (y’)² + 2y = 0

The highest order derivative present in the differential equation is y’’. 

Therefore, its order is two. 

The given differential equation is a polynomial equation in y’’ and y’ and the highest power raised to y’’ is one. 

Hence, its degree is one.

10. Determine order and degree(if defined) of differential equation y’’ + 2y’ + sin y = 0.

Answer:

y’’ + 2y’ + sin y = 0

The highest order derivative present in the differential equation is y’’. 

Therefore, its order is two. 

This is a polynomial equation in y’’ and y’ and the highest power raised to y’’ is one. 

Hence, its degree is one.

11. The degree of the differential equation  \(\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + sin\left(\frac{dy}{dx}\right) + 1 = 0\) is 

(A) 3 

(B) 2 

(C) 1 

(D) not defined

Answer:

\(\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + sin\left(\frac{dy}{dx}\right) + 1 = 0\) 

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined. 

Hence, the correct answer is D.

12. The order of the differential equation \(2x^2 \frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0\) is 

(A) 2 

(B) 1 

(C) 0 

(D) not defined

Answer:

\(2x^2 \frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0\) 

The highest order derivative present in the given differential equation is \(\frac{d^2y}{dx^2}\). Therefore, its order is two. 

Hence, the correct answer is A.

Answer 2

13. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

 y = e^x + 1 : y″ – y′ = 0

Answer:

y = e^x + 1

Differentiating both sides of this equation with respect to x, we get:

Now, differentiating equation (1) with respect to x, we get:

Substituting the values of y’ and y” in the given differential equations, we get,

y” – y’ = e^x – e^x = R.H.S. 

Thus, the given function is the solution of the corresponding differential equation.

14. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = x² + 2x + C : y′ – 2x – 2 = 0

Answer:

y = x² + 2x + C

Differentiating both sides of this equation with respect to x, we get:

Substituting the values of y’ in the given differential equations, we get,

L.H.S. = y’ – 2x -2

= 2x + 2 – 2x – 2

= 0

= R.H.S. 

Hence, the given function is the solution of the corresponding differential equation.

15. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = cos x + C : y′ + sin x = 0

Answer:

y = cos x + C

Differentiating both sides of this equation with respect to x, we get:

Substituting the values of y’ in the given differential equations, we get,

L.H.S. = y’ + sin x 

= – sin x + sin x 

= 0 

= R.H.S. 

Hence, the given function is the solution of the corresponding differential equation.

16. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = \sqrt{1 + x^2} : y' = \frac{xy}{1 + x^2}\)

Answer:

\(y = \sqrt{1 + x^2}\) 

Differentiating both sides of the equation with respect to x, we get:

∴ L.H.S. = R.H.S. 

Hence, the given function is the solution of the corresponding differential equation.

17. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax : xy′ = y (x ≠ 0)

Answer:

y = Ax

Differentiating both sides with respect to x, we get:

Substituting the values of y’ in the given differential equations, we get,

L.H.S. = xy’

= x × A

= Ax

= y  

= R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

18. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = x\, sin x : xy' = y + x \sqrt{x^2 - y^2} \) \((x \ne 0 \;\text{and}\;x > y \;\text{or}\; x < - y)\) 

Answer:

y = x sin x

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of y' in the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.

19. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(xy = log \,y + C\) \(: y' = \frac{y^2}{1 - xy} (xy \ne 1)\) 

Answer:

\(xy = log \,y + C\) 

Differentiating both sides of this equation with respect to x, we get:

∴ L.H.S. = R.H.S. 

Hence, the given function is the solution of the corresponding differential equation.

20. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y – cos y = x : (y sin y + cos y + x) y′ = y

Answer:

y – cos y = x    ....(1)

Differentiating both sides of the equation with respect to x, we get:

Substituting the value of y' in equation (1), we get:

Hence, the given function is the solution of the corresponding differential equation.

21. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

x + y = tan^-1y : y² y′ + y² + 1 = 0

Answer:

x + y = tan^-1y

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of y' in the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.

22. Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = \sqrt{a^2 - x^2}x\in (-a, a) : x+ y \frac{dy}{dx} = 0 (y \ne 0)\)

Answer:

\(y = \sqrt{a^2 - x^2}\) 

Differentiating both sides of this equation with respect to x, we get:

Substituting the value of \(\frac{dy}{dx}\) in the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.

Answer 3

23. The numbers of arbitrary constants in the general solution of a differential equation of fourth order are: 

(A) 0 

(B) 2 

(C) 3 

(D) 4

Answer:

We know that the number of constants in the general solution of a differential equation of order n is equal to its order. 

Therefore, the number of constants in the general equation of fourth order differential equation is four. 

Hence, the correct answer is D.

24. The numbers of arbitrary constants in the particular solution of a differential equation of third order are: 

(A) 3 

(B) 2 

(C) 1 

(D) 0

Answer:

In a particular solution of a differential equation, there are no arbitrary constants. 

Hence, the correct answer is D.

25. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

\(\frac xa + \frac yb = 1\)

Answer:

\(\frac xa + \frac yb = 1\)

Differentiating both sides of the given equation with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Hence, the required differential equation of the given curve is y'' = 0.

26. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y² = a (b² – x²)

Answer:

y² = a (b² – x²)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Dividing equation (2) by equation (1), we get:

This is the required differential equation of the given curve.

27. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = ae^3x + be^-2x

Answer:

y = ae^3x + be^-2x    ....(1)

Differentiating both sides with respect to x, we get:

y = 3ae^3x - 2be^-2x    ....(2)

Again, differentiating both sides with respect to x, we get:

y = 9ae^3x + 4be^-2x    ....(3)

Multiplying equation (1) with (2) and then adding it to equation (2), we get:

Now, multiplying equation (1) with equation (3) and subtracting equation (2) from it, we get:

Substituting the values of ae^3x and be^-2x in equation (3), we get:

This is the required differential equation of the given curve.

28. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = e^2x (a + bx)

Answer:

y = e^2x (a + bx)    ....(1)

Differentiating both sides with respect to x, we get:

Multiplying equation (1) with equation (2) and then subtracting it from equation (2), we get:

Differentiating both sides with respect to x, we get:

y'' k - 2y' = 2be^2x     .....(4)

Dividing equation (4) by equation (3), we get:

This is the required differential equation of the given curve.

29. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = e^x (a cos x + b sin x)

Answer:

y = e^x (a cos x + b sin x)    .....(1)

Differentiating both sides with respect to x, we get:

Again, differentiating with respect to x, we get:

Adding equations (1) and (3), we get:

This is the required differential equation of the given curve.

30. Form the differential equation of the family of circles touching the y-axis at the origin.

Answer:

The centre of the circle touching the y-axis at origin lies on the x-axis. 

Let (a, 0) be the centre of the circle. 

Since it touches the y-axis at origin, its radius is a.

 Now, the equation of the circle with centre (a, 0) and radius (a) is

Differentiating equation (1) with respect to x, we get:

Now, on substituting the value of a in equation (1), we get:

This is the required differential equation.

31. Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

x² = 4ay   ....(1)

Differentiating equation (1) with respect to x, we get:

2x = 4ay'     .....(2)

Dividing equation (2) by equation (1), we get:

This is the required differential equation.

Answer 4

32. Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Answer:

The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:

\(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)     .....(1)

Differentiating equation (1) with respect to x, we get:

Again, differentiating with respect to x, we get:

Substituting this value in equation (2), we get:

 

This is the required differential equation.

33. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Answer:

The equation of the family of hyperbolas with the centre at origin and foci along the xaxis is:

 \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)     .....(1)

Differentiating both sides of equation (1) with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Substituting the value of \(\frac1{a^2}\) in equation (2), we get:

This is the required differential equation.

34. Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Answer:

Let the centre of the circle on y-axis be (0, b). 

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

Differentiating equation (1) with respect to x, we get:

Substituting the value of (y – b) in equation (1), we get:

This is the required differential equation.

35. Find the general solution:

\(\frac{dy}{dx} = \frac{1 - cos\,x }{1 + cos \,x}\)

Answer:

The given differential equation is:

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

36. Find the general solution:

\(\frac{dy}{dx} = \sqrt{ 4 -y^2} (-2 < y < 2)\)

Answer:

The given differential equation is:

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

37. Find the general solution:

\(\frac{dy}{dx} + y = 1 (y \ne 1)\)

Answer:

The given differential equation is:

Now, integrating both sides, we get:

This is the required general solution of the given differential equation.

38. Find the general solution:

sec²x tan y dx + sec²y tan x dy = 0

Answer:

The given differential equation is:

Integrating both sides of this equation, we get:

Substituting these values in equation (1), we get:

This is the required general solution of the given differential equation.

39. Find the general solution:

\((e^x + e^{-x})dy - (e^x - e^{-x})dx = 0\)

Answer:

The given differential equation is:

Integrating both sides of this equation, we get:

Let (e^x + e^–x) = t.

 Differentiating both sides with respect to x, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

40. Find the general solution:

\(\frac{dy}{dx} = ( 1 +x^2) (1 + y^2)\)

Answer:

The given differential equation is:

Integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

41. Find the general solution:

y log y dx - x dy = 0

Answer:

The given differential equation is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

Answer 5

42. Find the general solution:

\(x^5 \frac{dy}{dx} = y^5\)

Answer:

The given differential equation is:

Integrating both sides, we get:

This is the required general solution of the given differential equation.

43. Find the general solution:

\(\frac{dy}{dx} = sin^{-1}x\)

Answer:

The given differential equation is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

44. Find a particular solution satisfying the given condition:

\(cos \left(\frac{dy}{dx}\right) = a(a \in R); y = 1 \;when \,x = 0\)

Answer:

Integrating both sides, we get:

Substituting C = 1 in equation (1), we get:

45. Find a particular solution satisfying the given condition:

\(cos \left(\frac{dy}{dx}\right) = y\, tan \, x; y = 1 \;when \,x = 0\) 

Answer:

Integrating both sides, we get:

Substituting C = 1 in equation (1), we get: 

y = sec x

46. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x sin x.

Answer:

The differential equation of the curve is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

Now, the curve passes through point (0, 0).

Substituting \(C =\frac12\) in equation (2), we get:

Hence, the required equation of the curve is 2y - 1 = e^x(sin x - cos x).

47. For the differential equation \(xy \frac{dy}{dx} = (x +2) (y + 2)\), find the solution curve passing through the point (1, –1).

Answer:

The differential equation of the given curve is:

Integrating both sides, we get:

Now, the curve passes through point (1, –1).

Substituting C = –2 in equation (1), we get:

This is the required solution of the given curve.

48. Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.

Answer:

Let x and y be the x-coordinate and y-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the \(\frac{dy}{dx}\)

According to the given information, we get:

\(y.\frac{dy}{dx}=x\)

⇒ y dy = x dx

Integrating both sides, we get:

Now, the curve passes through point (0, –2).

∴ (–2)² – 0² = 2C 

⇒ 2C = 4

Substituting 2C = 4 in equation (1), we get: 

y² – x² = 4 

This is the required equation of the curve.

49. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (–4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

It is given that (x, y) is the point of contact of the curve and its tangent. 

The slope (m₁) of the line segment joining (x, y) and (–4, –3) is \(\frac{y+3}{x+4}\).

We know that the slope of the tangent to the curve is given by the relation, \(\frac{dy}{dx}\)

∴ Slope (m₂) of the tangent = \(\frac{dy}{dx}\)

According to the given information:

Integrating both sides, we get:

This is the general equation of the curve. 

It is given that it passes through point (–2, 1).

Substituting C = 1 in equation (1), we get: 

y + 3 = (x + 4)² 

This is the required equation of the curve.

Answer 6

50. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Answer:

Let the rate of change of the volume of the balloon be k (where k is a constant).

Integrating both sides, we get:

⇒ 4π × 33 = 3 (k × 0 + C) 

⇒ 108π = 3C 

⇒ C = 36π

At t = 3, r = 6: 

⇒ 4π × 63 = 3 (k × 3 + C) 

⇒ 864π = 3 (3k + 36π) 

⇒ 3k = –288π – 36π = 252π

⇒ k = 84π 

Substituting the values of k and C in equation (1), we get:

Thus, the radius of the balloon after t seconds is \((63t + 27)^\frac13\).

51. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (log_e 2 = 0.6931).

Answer:

Let p, t, and r represent the principal, time, and rate of interest respectively. 

It is given that the principal increases continuously at the rate of r% per year.

Integrating both sides, we get:

It is given that when t = 0, p = 100. 

⇒ 100 = e k …(2)

Now, if t = 10, then p = 2 × 100 = 200. 

Therefore, equation (1) becomes:

Hence, the value of r is 6.93%.

52. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5 = 1.648).

Answer:

Let p and t be the principal and time respectively.

It is given that the principal increases continuously at the rate of 5% per year.

Integrating both sides, we get:

Now, when t = 0, p = 1000. 

⇒ 1000 = e^C … (2) 

At t = 10, equation (1) becomes:

Hence, after 10 years the amount will worth Rs 1648.

53. The general solution of the differential equation \(\frac{dy}{dx} = e^{x + y}\) is

A. e^x + e^-y = C

B. e^x + e^y = C

C. e^-x + e^-y = C

D. e^-x + e^-y = C

Answer:

Integrating both sides, we get:

Hence, the correct answer is A.

54. Show that the given differential equation is homogeneous and solve them.

(x² + x y) dy = (x² + y²) dx

Answer:

The given differential equation i.e., (x² + x y) dy = (x² + y²) dx can be written as:

This shows that equation (1) is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx 

Differentiating both sides with respect to x, we get:

\(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

Substituting the values of v and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

55. Show that the given differential equation is homogeneous and solve them.

\(y' = \frac{x + y}x\)

Answer:

The given differential equation is:

Thus, the given equation is a homogeneous equation. 

To solve it, we make the substitution as: y = vx 

Differentiating both sides with respect to x, we get:

\(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

56. Show that the given differential equation is homogeneous and solve them.

(x – y) dy – (x + y) dx = 0

Answer:

Thus, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer 7

57. Show that the given differential equation is homogeneous and solve them.

(x² – y²)dx + 2xy dy = 0

Answer:

The given differential equation is:

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

58. Show that the given differential equation is homogeneous and solve them.

\(x^2 \frac{dy}{dx} - x^2 - 2y^2 + xy\)

Answer:

The given differential equation is:

\(x^2 \frac{dy}{dx} - x^2 - 2y^2 + xy\) 

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as:

y = vx

⇒ \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution for the given differential equation.

59. Show that the given differential equation is homogeneous and solve them.

\(xdy- ydx = \sqrt{x^2 + y^2}dx\)

Answer:

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

60. Show that the given differential equation is homogeneous and solve them.

\(x \frac{dy}{dx} - y + x \, sin \left(\frac yx\right) = 0\)

Answer:

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

61. Show that the given differential equation is homogeneous and solve them.

\(\left(1 + e^\frac xy\right)dx + e^{\frac xy} \left(1 - \frac xy\right)dy = 0\)

Answer:

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

x = vy

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

62. Find the particular solution satisfying the given condition:

(x + y) dy + (x – y) dx = 0; y = 1 when x = 1

Answer:

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

Substituting the value of 2k in equation (2), we get:

This is the required solution of the given differential equation.

63. Find the particular solution satisfying the given condition:

 x²dy + (xy + y²)dx = 0; y = 1 when x = 1

Answer:

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

Substituting \(C^2 = \frac13\) in equation (2), we get:

This is the required solution of the given differential equation.

Answer 8

64. Find the particular solution satisfying the given condition:

\(\left[x\, sin^2\left(\frac xy - y\right)\right]dx + xdy = 0; y= \frac\pi4 \;\text{when}\; x = 1\)

Answer:

Therefore, the given differential equation is a homogeneous equation. 

To solve this differential equation, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

Substituting C = e in equation (2), we get: 

This is the required solution of the given differential equation.

65. Find the particular solution satisfying the given condition:

\(\frac{dy}{dx} - \frac yx + cosec\left(\frac yx\right) = 0; y = 0\; \text{when} \, x = 1\)

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation. 

Now, y = 0 at x = 1.

Substituting C = e in equation (2), we get: 

This is the required solution of the given differential equation.

66. Find the particular solution satisfying the given condition:

\(2xy + y^2 - 2x^2 \frac{dy}{dx} = 0; y = 2 \;\text{when} \; x = 1\)

Answer:

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as: 

y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

Integrating both sides, we get:

Substituting C = –1 in equation (2), we get:

This is the required solution of the given differential equation.

67. A homogeneous differential equation of the form \(\frac{dx}{dy} = h\left(\frac xy\right)\) can be solved by making the substitution 

A. y = vx 

B. v = yx 

C. x = vy 

D. x = v

Answer:

Correct option is C. x = vy 

For solving the homogeneous equation of the form \(\frac{dx}{dy} = h\left(\frac xy\right)\), we need to make the substitution as x = vy. 

68. Which of the following is a homogeneous differential equation?

A. (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0

B. (x y) dx – (x³ + y³) dy = 0

C. (x³ + 2y²) dx + 2xy dy = 0

D. y²dx + (x² – x y – y²) dy = 0

Answer:

Correct option is D. y²dx + (x² – x y – y²) dy = 0

Function F(x, y) is said to be the homogenous function of degree n, if 

F(λx, λy) = λⁿ F(x, y) for any non-zero constant (λ). 

Consider the equation given in alternative D:

Hence, the differential equation given in alternative D is a homogenous equation.

69. Find the general solution:

\(\frac{dy}{dx} + 2y = sin \, x\)

Answer:

The given differential equation is \(\frac{dy}{dx} + 2y = sin \, x\).

This is in the form of \(\frac{dy}{dx} + py = Q\) (where p = 2 and Q = sin x)

The solution of the given differential equation is given by the relation,

Therefore, equation (1) becomes:

This is the required general solution of the given differential equation.

70. Find the general solution:

\(\frac{dy}{dx} + 3y = e^{-2x}\)

Answer:

The given differential equation is \(\frac{dy}{dx} + 3y = e^{-2x}\)

This is in the form of \(\frac{dy}{dx} + py = Q\) (where p = 3 and Q = e^-2x)

The solution of the given differential equation is given by the relation,

This is the required general solution of the given differential equation.

Answer 9

71. Find the general solution:

\(\frac{dy}{dx} + \frac yx = x^2\)

Answer:

The given differential equation is:

The solution of the given differential equation is given by the relation,

This is the required general solution of the given differential equation.

72. Find the general solution:

\(\frac {dy}{dx} + sec\, xy = tan\,x\left(0\le x< \frac\pi2\right)\)

Answer:

The given differential equation is:

The general solution of the given differential equation is given by the relation,

73. Find the general solution:

\(\int\limits_0^\frac\pi2 cos\,2x\,dx\)

Answer:

By second fundamental theorem of calculus, we obtain

74. Find the general solution:

\(x\frac{dy}{dx} + 2y = x^2log \,x\)

Answer:

The given differential equation is:

This equation is in the form of a linear differential equation as:

The general solution of the given differential equation is given by the relation,

75. Find the general solution:

\((1 +x^2)dy + 2xy \,dx = cot\,x\,dx (x\ne 0)\)

Answer:

This equation is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

76. Find the general solution:

\(x\frac{dy}{dx} + y -x + xy\, cot\, x = 0(x \ne 0)\)

Answer:

This equation is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

77. Find the general solution:

\((x + y) \frac{dy}{dx} = 1\)

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

78. Find the general solution:

\(y\, dx + (x -y^2)dy =0\)

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

79. Find the general solution:

\((x + 3y^2)\frac{dy}{dx} = y(y>0)\)

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

 

80. Find the general solution:

\(\frac{dy}{dx} + 2y\,tan\,x = sin\,x; y= 0 \:\text{when}\; x = \frac\pi3\)

Answer:

The given differential equation is \(\frac{dy}{dx} + 2y\,tan\,x = sin\,x\).

This is a linear equation of the form:

The general solution of the given differential equation is given by the relation,

Substituting C = –2 in equation (1), we get:

Hence, the required solution of the given differential equation is y = cos x - 2cos²x.

Answer 10

81. Find the general solution:

\((1 + x^2)\frac{dy}{dx}+ 2xy = \frac1{1 +x^2}; y=0\;\text{when}\;x = 1\)

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Substituting \(C = -\frac\pi4\) in equation (1), we get: 

This is the required general solution of the given differential equation.

82. Find the general solution:

\(\frac{dy}{dx} - 3y\,cot\,x = sin\,2x; y = 2\;\text{when}\;x = \frac\pi2\)

Answer:

The given differential equation is \(\frac{dy}{dx} - 3y\,cot\,x = sin\,2x\).

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Substituting C = 4 in equation (1), we get: 

This is the required particular solution of the given differential equation.

83. 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.

Answer:

Answer Let F (x, y) be the curve passing through the origin. 

At point (x, y), the slope of the curve will be \(\frac{dy}{dx}\).

According to the given information:

\(\frac{dy}{dx}=x + y\)

⇒ \(\frac{dy}{dx}-y = x\)

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Substituting in equation (1), we get:

The curve passes through the origin. 

Therefore, equation (2) becomes: 

1 = C 

Substituting C = 1 in equation (2), we get:

⇒ x + y + 1 = e^x

Hence, the required equation of curve passing through the origin is x + y + 1 = e^x.

84. Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

Let F (x, y) be the curve and let (x, y) be a point on the curve. 

The slope of the tangent to the curve at (x, y) is \(\frac{dy}{dx}\).

According to the given information:

\(\frac{dy}{dx}+5=x + y\)

⇒ \(\frac{dy}{dx}-y = x-5\)

This is a linear differential equation of the form:

The general equation of the curve is given by the relation,

Therefore, equation (1) becomes:

The curve passes through point (0, 2).

Therefore, equation (2) becomes: 

0 + 2 – 4 = Ce⁰ 

⇒ – 2 = C 

⇒ C = – 2 

Substituting C = –2 in equation (2), we get:

This is the required equation of the curve.

85. The integrating factor of the differential equation \(x\frac{dy}{dx} -y = 2x^2\) is

A. e^–x

B. e^–y

C. \(\frac1x\)

D. x

Answer:

Correct option is C. \(\frac1x\)

The given differential equation is:

This is a linear differential equation of the form:

The integrating factor (I.F) is given by the relation,

86. The integrating factor of the differential equation.

\((1 -y^2)\frac{dx}{dy}+ yx = ay(-1< y<1)\) is

A. \(\frac1{y^2 - 1}\)

B. \(\frac1{\sqrt{y^2 - 1}}\)

C. \(\frac1{1-y^2 }\)

D. \(\frac1{\sqrt{1-y^2 }}\) 

Answer:

Correct option is D. \(\frac1{\sqrt{1-y^2 }}\)

The given differential equation is:

This is a linear differential equation of the form:

The integrating factor (I.F) is given by the relation,