Test: Differential Equations- 1 - JEE MCQ

tangent of y at zero

maxima of y

derivatives of y

minima of y

contains arbitrary constants depending on the order of the differential equation

contains exactly one arbitrary constant

contains exactly two arbitrary constants

does not contain arbitrary constants

2y′ − x² = y²

2xyy′ + x² = y²

2xyy′ − x² = y²

2yy′ + x2 = y²

(xy) dx − (x³+ y³) dy = 0

(x³+ 2y²) dx + 2xy dy = 0

y²dx + (x²− xy − y²) dy = 0

(4x + 6y + 5) dy − (3y + 2x + 4) dx = 0

the number of constant terms

the order of the lowest order derivative ofthe dependent variable

the order of the highest order derivative ofthe dependent variable

the number of derivative terms

can contain exactly one arbitrary constant

does not contain arbitrary constants

can contain exactly two arbitrary constants

can contain arbitrary constants

y = sec x

y = tan x

y = sin x

y = cos x

xy′ – 2y = 0

xy′ + 2y = 0

y′ – 2y = 0

xy′² + 2y = 0

 + Py = Q

 + Py = 0

 + Px = Q

  = Q

Highest (positive integral index) of the lowest order derivative in the given differential equation.

Lowest power (positive integral index) of the highest order derivative in the given differential equation.

Lowest power (positive integral index) of the lowest order derivative in the given differential equation.

Highest power (positive integral index) of the highest order derivative in the given differential equation.

3

4

1

2

y + x + 2 = log (x²(y + 2)²)

y −− x − 2 = log (x²(y + 2)²)

y −− x + 2 = log (x²(y + 2)²)

y −− x − 2 = log (x²(y − 2)²)

xy′′ + x(y′)²  yy′ = 0

xyy′′ + x(y′)²  yy′ = 0

yy′′ + x(y′)²  yy′ = 0

xyy′′ −x(y′)² +yy′ = 0

x = sin 2y+C

x = sin y+C

x = cos y+C

None of these

1

4

3

2

2

0

3

1

4e^3x+3e^−4y+7 = 0

4e^3x−3e^−4y−7 = 0

4e^3x+3e^−4y−7 = 1

4e^3x+3e^−4y−7 = 0

4y − 1 = e^x ( sin x − cos 2x)

3y − 1 = e^x ( sin x − cos2x)

2y − 1 = e^x ( sin x − cos x)

2y + 1 = e^x ( sin2 x − cos x)

y = x log |3x| +Cx

y = log |x| +Cx

y = log |2x| +Cx

y = x log |x|+Cx

3

2

4

1

y = 2tan (x/2) − x+ c

y = 2tan²(x/2) − x+ c

y = 2tan (x) – x + c

y = 2tan (x/2) + x+ c

7.93%

8.93%

9.93%

6.93%

y³− x² = 4

y²− x² = 4

y³− x³ = 4

y²− x³ = 4