IIT JAM Mathematics MCQ Test 1

Given differential equation is

Here  which shows that the transformation  t= 1/log z 

reduce the given differential equation into the form  

then its solution is 

Now,

Auxiliary equation is m² - m-2 = 0

m=-1,2

∴ C.F = c₁e^-x +c₂e^2x

Solving these, we get c₁= 1, c₂= -1

∴   the required solution is y = e^-x - e^2x+ xe^2x

Correct Answer :- d

Explanation :

The equation of the family of the circles touching x-axis at the origin is 

x² + (y-a)² = a²

differential w.r. to x

Given differential equation is 

(3a²x² + by cos x)dx + (2 sin x - 4ay³)dy = 0

since it si exact if 

Here M = 3a²x² + bycosx
N = 2sinx - 4ay³
then bcosx = 2cosx
=> cosx(b - 2) = 0

=> b= 2[∵ cos x ≠ 0]

which shows that given differential equation is exact for any value of a but b = 2

The given equation can be re-written as 

Integrating 

Putting y² = v so that 2y (dy/dx) = (dv/dx). We then get

which is linear. Its I.F = 

x=0, y=1/2   (1)  ⇒ -2 = -3 +c ⇒ c=1

We have,
Slope = dy/dx ⇒ dy/dx = 1/2y ⇒ 2 y dy = dx
Integrating both sides, we get y² = x + C
This passes through (4, 3)
∴ 9 = 4 + C ⇒ C = 5
So, the equation of the curve is y² = x + 5

Hence the given differential equation

It is clear by the definition of linear differential equation and order of differential equation, that is linear and of second order.

Given differential equation is

    --(1)

and it’s general solution is

(c₁ log x + c₂) x²     ..(2)

Since (1) is homogeneous differential  equation so put then (1) reduce to

It’s A.E is m² + m(k - 1) + 4 = 0 • • (3)
Now from (2) we have

By observation we can say that (3) has solution m = 2, 

 ... (4)

On comparing (3) and (4). we have

k-1 = -4

k=-3

Given differential equation is

Option (a), (b) are clearly correct

Now we have

Now from (4) and (5)

Given equation ca be written as,

not homogenous differential equation

compare with 

Here at x = 0, P and Q both are not defined.

⇒ x=0 is a singular point of given differential equation.

Now consider x P(x) = -1/2

and both are defined and differential at x = 0

⇒ x= 0 is a regular singular point of given equation.

we have x² y" - 2y = 0

Given differential equation is

(x⁷y² + 3y)dx + 3x⁸y - x) dy = 0   ..(1)

If Integrating factor is x^myⁿ then multiplying the given differential equation by I.F. , equation (1) becomes exact 
i.e

is exact differential equation.

and for exact 

on solving (1) and (2) we get

m=-7, n=1

m+n = -7+1 = -6

The equation of the families of parabolas whose centre (-a,0) is

y² = 4a(x+a)    ..(1)

Differentiating both sides with respect to x, we get

Dividing (i) by (ii) we get  

Substituting this value of a in (2), a is eliminated and we get

which shows that order of this differential equation is one and degree is 2. The sum of order of this differential equation and degree is 3.

Hence solution of equation (i) is,

But, y(1)=0, therefore 

From equation (ii), we have 

From equation (ii), we have 

where

v(x)= -1/6x²

∴ v(x)= -1/6(-1)² = 0.1666666

In the normalized form (6,3), this is 

Clearly the singular points of the differential equation are x = 0 and x = 2. We investigate them one at a time.
Consider x = 0 first. and form the functions defined by the products

of the form. The product function defined by x²P₂(x)f is analytical at x = 0, but that defined by xP₁(x) is not. Thus x = 0 is an irregular singular point of Now consider x = 2. Forming the products for this point, we have

Both of the product functions thus defined are analytic at x = 2. and hence x = 2 is a regular singular point.

The derivative of f at P₀(x₀,y_o) in the direction of the unit vector u = u_1i + u₂j is the number

the partial derivatives of f at P₀ are 

The gradient of f ata p₀ is 

The derivative of f at P₀ in the direction of v is therefore