ORDER AND DEGREE OF DIFFERENTIAL EQUATION PRACTICE PROBLEMS

Find the order and degree of the following differential equations.

(1)  (dy/dx) + y  =  x²

(2)  y' + y²  =  x

(3)  y'' + 3 (y')² + y³

(4)  d²y/dx² + x  =  √[y + (dy/dx)]

(5)  d²y/dx² - y + (dy/dx + d³y/dx³)^(3/2)  =  0

(6)  y''  =  (y - (y')³)^(2/3)

(7)  y' + (y'')²  =  (x + y'')²

(8)  (dy/dx)² + x = (dx/dy) + x²

Find the the following for the given differential equations.

(i) Order 

(ii) Degree

(iii) General solution

Problem 1 :

y' = 1 + x² + y + x²y

Solution :

y' = 1 + x² + y + x²y

i)  Order = 1

ii)  Degree = 1

iii) General solution :

(dy/dx) = (1 + x²) + y(1 + x²)

(dy/dx) = (1 + x²) (1 + y)

dy/(1 + y) = (1 + x²) dx

Integrating on both sides, we get

ln (1 + y) = x + (x³/3) + C

1 + y = e^{x + (x^3/3) + C}

1 + y = e^{x + (x^3/3)}  (e^C)

1 + y = Ce^{x + (x^3/3)}

y = Ce^{x + (x^3/3)} - 1

Problem 2 :

y' = x/(y² + 1)

Solution :

y' = x/(y² + 1)

i)  Order = 1

ii)  Degree = 1

iii) General solution :

y' = x/(y² + 1)

dy/dx = x/(y² + 1)

dy (y² + 1) = x dx

Integrating on both sides, we get

∫ dy (y² + 1) = ∫ x dx

y³/3 + y = x²/2 + C

Problem 3 :

The order and degree of the differential equation

d²y/dx² + (dy/dx)^1/3 + x^1/4 = 0

a) 2, 3    b) 3, 3    c) 2, 6     d)  2, 4

Solution :

d²y/dx² + (dy/dx)^1/3 + x^1/4 = 0

The maximum number of times they find the derivative of y with respect to x is 2. So, the order is 2. But the variable x is having the rational exponent, so we cannot find the order and degree now.

x^1/4 = -(d²y/dx² + (dy/dx)^1/3)

Raising power 4 on both sides,

x = [-(d²y/dx² + (dy/dx)^1/3]⁴

x = [(d²y/dx² + (dy/dx)^1/3]⁴

Order = 2 and degree = 4

So, option d is correct.

Problem 4 :

The differential equation representing the family of curves y = A cos (x + B), where A and B are parameters, is

a)  d²y/dx² - y = 0     b)  d²y/dx² + y = 0

c)  d²y/dx² = 0     d)  d²y/dx²  = 0

Solution :

y = A cos (x + B) ----(1)

Since we have two parameters A and B, we have to find the derivative two times.

y'= A (-sin (x + B))(1 + 0)

y' = -A sin (x + B) -----(2)

y'' = -A cos (x + B) ------(3)

Applying (1), we get

y'' = -y

y'' + y = 0

d²y/dx² + y = 0

So, option b is correct.

Problem 5 :

The order and degree of the differential equation 

√sin x (dx + dy) = √cos x (dx - dy) is

a)  1, 2      b)  2, 2     c)  1, 1     d)  2, 1

Solution :

√sin x (dx + dy) = √cos x (dx - dy) ----(1)

√sin x dx + √sin x dy = √cos x dx - √cos x dy

√sin x dy + √cos x dy = √cos x dx - √sin x dx

(√sin x + √cos x) dy = (√cos x - √sin x) dx

dy/dx = (√cos x - √sin x)/(√sin x + √cos x)

The maximum number of times derivative is done = 1

highest exponent = 1

So, option c is correct.

Problem 6 :

The order of the differential equation of all circles with center at (h, k) and radius a is 

a)  2    b)  3     c)  4    d)  1

Solution :

Equation of circle with center (h, k) and radius a is

(x - h)²  + (y - h)² = a²

The equation is differentiable twice since we have two parameters.

So, the order of the differential equation will be 2.

Problem 7 :

The differential equation of the family of curves 

y = Ae^x + Be^-x 

where A and B are arbitrary constants is 

a)  d²y/dx² + y = 0     b)  d²y/dx² - y = 0

c)  dy/dx + y = 0     d)  dy/dx - y = 0

Solution :

y = Ae^x + Be^-x  -----(1)

y' = Ae^x - Be^-x  -----(2)

y'' = Ae^x - Be^-x (-1)

y'' = Ae^x + Be^-x  -----(3)

Applying (1), we get

y'' = y

y'' - y = 0

 d²y/dx² - y = 0

So, option b is correct.

Problem 8 :

The solution of the differential equation 

2x (dy/dx) - y = 3

represents

a) straight lines   b) circles     c) parabola   d) ellipse

Solution :

2x (dy/dx) - y = 3

2x (dy/dx) = y + 3

dy/(y+3) = dx/2x

Integrating both sides, we get

log(y + 3) = (1/2) log x

(y + 3) = √x

Raising power 2 on both sides, we get

(y + 3)² = x 

So, the general solution must be a parabola.

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