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This mock test of Test: Intro Differential Equations for JEE helps you for every JEE entrance exam.
This contains 10 Multiple Choice Questions for JEE Test: Intro Differential Equations (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The dif is a:

Solution:

We know to calculate degree or order, the powers of derivatives should be integer.

on squaring both sides, we get a differential equation with integers as the power of derivatives

so order is the highest derivative, which is 2

and degree is the power of highest degree which is again 2

QUESTION: 2

The differential equation for the equation is :

Solution:

y = Acos(αx) + Bsin(αx)

dy/dx = -Aαsin(αx) + Bαcos(αx)

d^{2}y/dx^{2} = -Aα^{2}cos(αx) - Bα^{2}sin(αx)

= -α^{2}(Acos(αx) + Bsin(αx))

= -α^{2} * y

d^{2}y/dx^{2} + α^{2}*y = 0

QUESTION: 3

Formation of the differential equation of the family of curves represented by y = Ae^{2x} + Be^{-2x} is :

Solution:

y = Ae^{2x} + Be^{-2x}

dy/dx = 2Ae^{2x} – 2Be^{-2x}

d^{2}y/dx^{2} = 4Ae^{2x} + 4Be^{-2x}

= 4*y

d^{2}y/dx^{2} – 4y = 0

QUESTION: 4

The degree of the differential equation

Solution:

Given equation is :

(dy/dx)^{2} + 1/(dy/dx) = 1

((dy/dx)^{3} + 1)/(dy/dx) = 1

(dy/dx)^{3} +1 = dy/dx

So, final equation is

(dy/dx)^{3} - dy/dx + 1 = 0

So, degree = 3

QUESTION: 5

Differential equation representing the family of curves given by y = ax + x^{2} is:

Solution:

**The answer is C. We eliminate constants.**

**We have**

**y=ax+x ^{2}**

**Differentiating with respect to x,**

QUESTION: 6

The order of the differential equation:

Solution:

Order of the D.E. is 3

Order of a differential equation is the order of the highest derivative present in the equation.

QUESTION: 7

Formation of the differential equation corresponding to the ellipse major axis 2a and minor axis 2b is:

Solution:

Equation of ellipse :

x^{2}/a^{2} + y^{2}/b^{2} = 1

Differentiation by x,

2x/a^{2} + (dy/dx)*(2y/b^{2}) = 0

dy/dx = -(b^{2}/a^{2})(x/y)

-(b^{2}/a^2) = (dy/dx)*(y/x) ----- eqn 1

Again differentiating by x,

d^{2}y/dx^{2} = -(b^{2}/a^{2})*((y-x(dy/dx))/y^{2})

Substituting value of -b^{2}/a^{2} from eqn 1

d^{2}y/dx^{2} = (dy/dx)*(y/x)*((y-x(dy/dx))/y^{2})

d^{2}y/dx^{2} = (dy/dx)*((y-x*(dy/dx))/xy)

(xy)*(d^{2}y/dx^{2}) = y*(dy/dx) - x*(dy/dx)^{2}

(xy)*(d^{2}y/dx^{2}) + x*(dy/dx)^{2}- y*(dy/dx) = 0

QUESTION: 8

The differential equation is a solution of the equation:

Solution:

Solving second order differential equation with variable coefficients becomes a bit lengthy and complicated. So, its better to check by options.

On checking option A :

y = A/x + B

dy/dx = -A/x^{2}

d^{2}y/dx^{2 }= (2A)/x^{3}

So,

d^{2}y/dx^{2} + (2/x)*(dy/dx) = 0

(2A)/x^{3} + (2/x)*((-A)/x^{2}) = 0

(2A - 2A)/x^{3} = 0

0 = 0

LHS = RHS

QUESTION: 9

The differential equation is a:

Solution:

3*(d^{2}y/dx^{2}) = [1+(dy/dx)^{2}]^{3}/^{2}

On squaring both side,

9*(d^{2}y/dx^{2})^{2} = [1+(dy/dx)^{2}]^{3}

The order of the equation is 2. The power of the term determining the order determines the degree.

So, the degree is also 2.

QUESTION: 10

The order and degree of the differential equation: (y”)^{2} + (y”)^{3} + (y’)^{4} + y^{5} = 0 is:

Solution:

The highest order derivative here is y’’. Therefore the order of the differential equation=2.

The highest power of the highest order derivative here is 3. Therefore the order of the differential equation=3.

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