Differential equation

An equation involving derivatives ( i.e., dy/dx, d²y/dx²...etc) in x and y is a differential equation. For example 

i) x dy/dx = 3y

ii) xdy + ydx = 0

iii) xd²y/dx² - 6dy/dx + 3y = 0

Order and Degree of a Differential Equation

The order of the highest order derivative in the equation is the order of the equation and the degree of the highest order derivatives (when put in rational form ) is the degree of the equation .

For example - x² (d³y/dx³)² - 3y(dy/dx)³ + 2 = 0

This differential equation is of 3rd order because highest order derivatives is  d³y/dx³ and its degree is 2 because power of the highest order derivatives d³y/dx³ is 2.

Example 1: The differential equation of the family of curves y=Ae^x + Be^3x for different values of A and B is

(A) 3y + 4 dy/dx - d²y/dx²

(B) 3y-4 dy/dx - d²y/dx²

(C) 3y + 4 dy/dx + d²y/dx²

(D) none of these

Solution: y = A ex + Be3x ... (1)

y1 = Aex + 3Be3x ... (2) (y₁ = dy/dx)

y2 = Aex + 9B3x ... (3) (y₂ = d²y/dx²)

Eliminating A and B from the above three, we get

( 3y + 4y₁ – y₂ = 0

3y + 4 (dy/dx + d²y/dx² = 0)

Do Check: Matrix

Solution of Differential Equation

To solve differential equation of the first order and the first degree: Simple standard form of differential equation of the first order and first degree are as follows:

(i) Variable Separable

Form f(x) dx + ((y) dy = 0

Method: Integrate it i.e., find (∫ f(x) dx + (∫∮ ((y)dy = c

Example 2: Solution of the differential equation dy/dx = e^xy is

(A) ln y + e^x = c

(B) ln y – e^x = c

(C) ln y = c

(D) none of these

Solution: Given,

dy/dx - e^xy

dy/y = e^x.dx

dy/y - e^x.dx = 0

Integrating, we get ln y– e^x =c

Do Check: Determinant

(ii) Reducible into Variable Separable

Method: Sometimes differential equation of the first order cannot be solved directly by variable separation but by some substitution we can reduce it to a differential equation with separable variables. A differential equation of the [pic] is solved by writing ax + by + c = t

Example 3: Solution of the differential equation (x – y)² dy/dx = 1 is

(A) x =(x – y) + In x-y-1/x+y+1 = 1

(B) x =(x – y) + 1/2 In x-y-1/x+y+1 = 1

(C) x =(x + y) - 1/2 In x-y-1/x+y+1 = 1

(D) none of these

Solution: Put z = x –y

dz/dx =1- dy/dx

dy/dx = 1-dz/dx

Now z²  (1-dz/dx) =1

z²dz/dx =z²-1

dx = z²/ z²-1 dz

which is in the form of variable separable

Now integrating, we get x = z + 1/2 In z-1/z+1 + c

Solutions is x = (x-y) + 1/2 In x-y-1/ x+y+1 + 1

Exercise 1:

(i). The differential equation of all circles through the origin is

(A) (x² - y²)(d²y/dx²) = 2[x(dy/dx) - y][1 + (dy/dx)²]

(B) (x² + y²)(d²y/dx²) = 2[x(dy/dx) - y][1 + (dy/dx)²]

(C) (x² + y²)(d²y/dx²) = 2[x(dy/dx) + y][1 - (dy/dx)²]

(D) none of these

(ii) Solution of the differential equation dy/dx = (2√(1+x²) tan⁻¹y + x) = 0 is

(A) (1/2)ln(1+x²) + y tan⁻¹y - c/(1+y²)

(B) -(1/2)ln(1+x²) + y tan⁻¹y - c/(1+y²)

(C) (1+x²) ln y - tan⁻¹y = c/(1+y²)

(D) (1+x²) + 2ln y - tan⁻¹y = c/(1+y²)

Do Check: Statics and dynamics

(iii) Solution of the differential equation (eʸ + 1)cos x dx + eʸ sin x dy = 0 is

(A) (eˣ + 1) sin y = c

(B) (eˣ - 1) sin y = c

(C) (eʸ + 1) sin x = c

(D) none of these

(iv) Solution of the differential equation dy/dx = sin(x + y) is

(A) 1 + tan((x+y)/2) = (x+c)/2

(B) 1 - tan((x+y)/2) = (x+c)/2

(C) 1 + tan((x+y)/2) = -(x+c)/2

(D) none of these

(iii) Homogeneous Equation

When dy/dx is equal to a fraction whose numerator and the denominator both are homogeneous functions of x and y of the same degree then the differential equation is said to be homogeneous equation. i.e. when dy/dx = f(x,y), where f(kx, ky) = f(x, y) then this differential equation is said to be homogeneous differential equation .

Method: Put y= vx

Example 4: Solution of the differential equation dy/dx = x² + y²/2xyis

(A) k (x2 + y2) = x

(B) k (x2 ( y2) + x = 0

(C) k (x2 ( y2) = x

(D) k (x2 + y2) + x = 0

Solution:

(iv) Non-homogeneous Differential Equation

Form, dy/dx = a₁x + b₁y + c₁ / a₂x + b₂y + c₂ 

Method: If a₁/a₂ ≠ b₁/b₂, put x = X + h , y = Y + k such that a₁h +b₁k + c1 = 0, a₂h + b₂k +c₂ = 0 . In this way the equation becomes homogeneous in X, Y. then use the method for homogeneous equation. If a₁/a₂ = b₁/b₂. Put a₁x+b₁y=v or a₂x + b₂y = v. The equation in the form of variable separable in x, v.

Example 5: Solution of the differential equation dy/dx = x-y+3 / 2x-2y+5 is

(A) x – 2y + ln (x + y +2) = C

(B) x + 2y + ln (x + y +2) =C

(C) x + 2y + ln (x – y +2) = C

(D) x – 2y + ln (x – y +2) =C

Solution:

Exercise 2:

(i) Solution of the differential equation y² dx + x(x + y) dy = 0 is

(A) cxy² = 2y + x

(B) cxy² = 2y - x

(C) cxy² + 2y = x

(D) none of these

(ii) Solution of the differential equation (x dy)/(y dx) = (x - 2y)/(x - 3y) is

(A) y³ = c

(B) cy³ = x²e^(x/y)

(C) cy³ = c

(D) none of these

(iii) Solution of the differential equation dy/dx = (x + y + 3)/(3x + 3y + 1) is

(A) 3(x + y) + 2ln(x + y + 1) = x + c

(B) 3(x + y) - 2ln(x + y + 1) = x + c

(C) 3(x - y) + 2ln(x + y + 1) = x + c

(D) none of these

(iv) Solution of the differential equation (x + 2y)(dx - dy) = dx + dy

(A) 3(x + 2y) + 4 log(3x + 6y - 1) = 9x + c

(B) 3(x - 2y) + 4 log(3x + 6y - 1) = 9x + c

(C) 3(x + 2y) - 4 log(3x + 6y - 1) = 9x + c

(D) none of these

(v) Linear Equation 

Form dy/dx = P(x) y =Q(x), where P(x)  and  Q(x) are  functions  of x 

Method: Multiplying the equation by e∫P(x)dx, called integrating factor. Then the equation becomes

 

Exercise 3:

(i) Solution of the differential equation dy/dx + (y/x) = x² is

(A) 4xy + x⁴ = c

(B) 4xy + x⁴ + c = 0

(C) 4xy - x⁴ = c

(D) none of these

(ii) Solution of the differential equation dy/dx + y tan x = x² cos x is

(A) y sec x = c - x³/3

(B) y sec x = c + x³/3

(C) y tan x = c + x³/3

(D) none of these

Some Useful Results:

d(xy) = xdy + ydx 

Exercise 1:

(i) B         

(ii) A 

(iii) C

(iv)  A

Exercise 2.

(i) A    

(ii)  B

(iii) B

(iv) A

Exercise 3.

(i) C    

(ii) B

Formulae and Concepts at a Glance

1. The order of the highest order derivative in the equation is the order of the equation and the degree of the highest order derivatives (when put in rational form) is the degree of the equation.
2. Sometimes differential equation of the first order cannot be solved directly by variable separation but by some substitution we can reduce it to a differential equation with separable variables.
3. A differential equation of the dy/dx = f(ax + by + c) is solved by writing ax + by + c = t
4. When dy/dx = f(x, y), where f(kx, ky) = f(x, y) then this differential equation is said to be
5. The most general form of this category of differential equations is dy/dx + Py = Q, where P and Q are functions of x alone.
6. We use here an Integrating factor, namely e^∫P dx and the solution is obtained by y·e^∫P dx = ∫Q·e^∫P dx dx + c, where c is an arbitrary constant.
7. dy/dx + Py = Qy^n ... (1)
8. where P and Q are function of x alone, is called Bernoulli equation.

Solved Examples

1. A curve that passes through (2, 4) and having subnormal of constant length of 8 units can be;

(A) y² = 16x - 16

(B) y² = 16x + 24

(C) x² = 16y - 60

(D) x² = 16y + 68

Sol. (A) Let the curve be y = f(x). Subnormal at any point = y|dy/dx|

⇒ y(dy/dx) = 8 ⇒ y dy = 8dx

⇒ y²/2 = 8x + c

⇒ y² = 16x + 2c₁, c₁ = 8 or y² = 16x + 2c₂, c₂ = 24

2. Let m and n be the order and the degree of the differential equation whose solution is y = cx + c² - 3c^(3/2) + 2, where c is a parameter. Then,

(A) m = 1, n = 4

(B) m = 1, n = 3

(C) m = 1, n = 2

(D) none of these

Sol. (A) dy/dx = c ⇒ the differential equation is;

y = x·(dy/dx) + (dy/dx)² - 3·(dy/dx)^(3/2) + 2

Clearly its order is one and degree 4.

3. For any differential function y = f(x), the value of (d²y/dx²) + (dy/dx)³ · (d²x/dy²) is

(A) always zero

(B) always non-zero

(C) equal to 2y²

(D) equal to x²

Sol. (A) dy/dx = 1/(dx/dy) for a differential equation

or d²y/dx² = -1/(dx/dy)² · d/dy(dx/dy) = -(dx/dy)⁻² · d²x/dy² · dy/dx

or d²y/dx² + (dy/dx)³ · (d²x/dy²) = 0.

4. The degree and order of the differential equation of all parabolas whose axis is x-axis are

(A) 2, 1

(B) 1, 2

(C) 3, 2

(D) none of these

Sol. (B) Equation of required parabola is of the form y² = 4a(x - h)

Differentiating, we have 2y(dy/dx) = 4a ⇒ y(dy/dx) = 2a

Required differential equation y(d²y/dx²) + (dy/dx)² = 0.

Degree of the equation is 1 and order is 2.

5. Solution of equation dy/dx = (3x - 4y + 2)/(3x - 4y + 3) is

(A) (x - y)² + c = log(3x - 4y + 1)

(B) x - y + c = log(3x - 4y + 1)

(C) x - y + c = log(3x - 4y - 3)

(D) x - y + c = log(3x - 4y + 1)

Sol. (B) Let 3x - 4y = z

3 - 4(dy/dx) = dz/dx

dy/dx = (1/4)(3 - dz/dx)

Therefore the given equation (1/4)(3 - dz/dx) = (z + 2)/(z + 3)

⇒ -(z + 3)/(z + 1) dz = dx

⇒ -(1 + 4/(z + 1)) dz = dx

⇒ -z + 4 log(z + 1) = x + c

⇒ log(3x - 4y + 1) = x - y + c.

6. The differential equation representing the family of curves y² = 2c(x + c), where c is a positive parameter, is of

(A) Order 4

(B) Order 2

(C) Degree 3

(D) Degree 4.

Sol. (C) Differentiate it w.r.t x, we get

2yy₁ = 2c ⇒ c = yy₁ .......(1)

put c in the given equation,

y² = 2yy₁(x + yy₁) ⇒ y = 2xy₁ + 2yy₁²

⇒ y = 2xy₁ + 4yy₁³

which is a differential equation of order 1 and degree 3.

7. The family of curves represented by (d²y/dx²) = (x² + x + 1)/(y² + y + 1) and the family represented by (dy/dx) + (y² + y + 1)/(x² + x + 1) = 0

(A) touch each other

(B) are orthogonal

(C) are identical

(D) none of these

Sol. (B) Since dy/dx is equal to the slope by the tangent and if we denote the two families by c₁ and c₂ then,

(dy/dx)_{c₁} = (x² + x + 1)/(y² + y + 1)

(dy/dx)_{c₂} = -(y² + y + 1)/(x² + x + 1) ⇒ (dy/dx)_{c₁} × (dy/dx)_{c₂} = -1

8. The solution of dy/dx = (x(2log x + 1))/(sin y + y cos y) is

(A) y sin y = x² log x + c

(B) y sin y = -x² log x + c

(C) y = x² log x + c

(D) none of these

Sol. (A) (y cos y + sin y)dy = (2x log x + x) dx

⇒ y sin y - ∫sin y dy = ∫sin y dy

= x² log x - ∫x² · (1/x)dx + ∫x dx + c

⇒ y sin y = x² log x + c.

9. The solution of the differential equation (1 + x²)dy + (1 + y²)dx = 0 is

(A) x - y = c(1 + xy)

(B) x + y = c(1 - xy)

(C) xy = c(x + y)

(D) xy = c(x - y)

Sol. (B) dy/(1 + y²) + dx/(1 + x²) = 0 ⇒ tan⁻¹y + tan⁻¹x = tan⁻¹c

⇒ x + y = c(1 - xy).

10. If dy/dx = (xy + y)/(xy + x), then the solution of the differential equation is

(A) y = xe^x + c

(B) y = e^x + c

(C) y = Axe^(x-y)

(D) y = x + A

Sol. (C) dy/dx = (y(1 + x))/(x(1 + y)) ⇒ (1 + y)/(y) dy = (1 + x)/(x) dx

On integrating both sides, we get

∫log y + y = log x + x + log A

⇒ log(y/Ax) = x - y ⇒ y = Axe^(x-y)

ASSIGNMENT

1. (d²y/dx²)² + x(dy/dx)³ = 0 is a differential equation of

(A) degree 2, order 2

(B) degree 3, order 3

(C) order 2, degree 3

(D) none of these

2. I.F. for y ln(dx/dy) + x - ln y = 0 is

(A) ln x

(B) ln y

(C) ln xy

(D) none of these

3. Solution of (3x + 4y - 5)²(dy/dx) = a² is

(A) 4y + c = (2a/3)tan⁻¹((2a(3x + 4y - 5))/3(2a))

(B) 4y² + c = (2a/3)tan⁻¹((2a(2x + 4y - 5))/3(2a))

(C) 4y² + c = (a/3)cot⁻¹x

(D) none of these

4. Solution of (x + 2y³)(dy/dx) = y is

(A) x = y³ + cy

(B) x² = y³ + cy

(C) x = y³ - cy²

(D) none of these

5. The degree of the differential equation of all curves having normal of constant length k is

(A) 1

(B) 3

(C) 4

(D) 2

6. The solution of x sin(y/x) dy = [y sin(y/x) - x] dx is

(A) cos(y/x) = log cx

(B) cos(y/x) = log cx²

(C) cos(x/y) = log cx

(D) none of these

7. The solution of dy/dx + y cot x = y² sin² x cos² x is

(A) (1/y) cosec x = -(1/3)cos³ x + c

(B) (1/y) cosec x = (1/3)cos³ x + c

(C) (1/y) cosec x = (1/3)cos² x + c

(D) none of these

8. The solution of ds/dx + x² = x²e^(3x) is

(A) 1 - e^(-3x) = cx³/e^x

(B) 1 + e^(-3x) = ce^x

(C) 1 - e^(-3x) = ce^x

(D) 1 - e^(3x) = ce^(-x)

9. The degree of differential equation satisfying √(1+x²) + √(1+y²) = A(x√(1+y²) - y√(1+x²)) is

(A) 2

(B) 3

(C) 4

(D) none of these

10. Solution of dy/dx + 2y tan x = y² is

(A) cos² x + (1/2)x + (1/4)sin 2x) y = cy

(B) cos² x - (1/2)x + (1/4)sin 2x) y = cy

(C) sin² x + (1/2)x + (1/4)sin 2x) y = cy

(D) none of these

11. Solution of the equation x dy - y dx = √(x² - y²) dx subject to the condition y(1) = 0 is

(A) y = x² log(sin x)

(B) y = x log(sin x)

(C) y² = x(x - 1)²

(D) y² = 2x²(x - 1)

12. Solution of sin 2x(dy/dx) - y = tan x is

(A) y = tan x + c(y/x)

(B) xy tan x = c

(C) x - y sin x = c

(D) none of these

13. The function f(θ) = d/dθ ∫₀^θ dx/(1 - cos θ cos x) satisfies

(A) df/dθ + 2f(θ) cot θ = 0

(B) df/dθ - 2f(θ) cot θ = 0

(C) df/dθ + 2f(θ) = 0

(D) df/dθ - 2f(θ) = 0

14. Solution of (y dx - x dy) cot(x/y) = xy² dx is given by

(A) sin θ = ce^(nx)

(B) sin(x/y) = ce^(nx)

(C) tan(y/x) = ce^x

(D) none of these

15. The order and degree of the differential equation xy[(d²y/dx²)^(1/2) + x²(dy/dx - 1)^(3/4)] = 0 are

(A) 2, 3

(B) 2, 1

(C) 2, 2

(D) none of these

Answers to Assignment

1. A	2. B	3. A
4. A	5. D	6. A
7. A	8. A	9. D
10. A	11. B	12. A
13. A	14. B	15. C

Key Points to Remember:

• Order: The highest order of derivative present in the differential equation
• Degree: The power of the highest order derivative when the equation is polynomial in derivatives
• Variable Separable: Can be written as f(x)dx + g(y)dy = 0
• Homogeneous: When dy/dx = f(x,y) where f(kx,ky) = f(x,y)
• Linear: dy/dx + P(x)y = Q(x)
• Bernoulli: dy/dx + P(x)y = Q(x)y^n
• Integrating Factor: For linear equations, I.F. = e^∫P(x)dx

Solution Strategy:

1. Identify the type of differential equation
2. Apply appropriate method based on the type
3. Separate variables if possible
4. Use substitution for reducible forms
5. Apply integrating factor for linear equations
6. Integrate and find the general solution
7. Apply initial conditions if given to find particular solution