Differential Equation MCQ Level - 2 - IIT JAM MCQ

The given differential equation is,

The auxiliary equation is
m² – 1 = 0

or C₁e^–x + C₂e^x      (as C₁, C₂ are arbitrary constant)

Hence, y = CF + PI gives the required functions

The correct answer is:

Let x + 1 = e^z so that z = log (x + 1) and x = e^z – 1.

The given equation becomes

 [D(D – 1) – 3D + 4]y = (e^z –1)², D = d/dz
 or    (D² – 4D + 4)y = e_^2z – 2e^z + 1
The auxiliary equation is m² – 4m + 4 = 0
⇒   (m – 2)² = 0
∴   C.F. = [(C₁ + C₂z)e^2z = [C₁ + C₂log (x + 1)] (x +1)².

∴ 

The correct answer is: 

The auxiliary equation is 

Let  Their wronskian is

P.I. = u₁y₁ + u₂y₂, where

Hence y = C.F. + P.I. is the complete solution.

The correct answer is: y = e^x log (e^–x + 1) – xe^2x – e^x + e^2x log(e^x + 1) + C₁e^x + C₂e^2x

Multiplying the given differential equation with x^myⁿ and comparing the equation with

M dx + N dy = 0
We have,
M = x^myⁿ (axy² + by)
= ax^m+1yⁿ⁺² + bx^myⁿ⁺¹

∵  The equation is exact now,

Comparing coefficients we get,

a(n + 2) = b(m + 2)

and    b(n +1) = a(m +1)

abn – b²m = 2(b – a)b          ...(3)
abn – a²m = (a – b)a            ...(4)
Solving (3) and (4), we get

The correct answer is:  

We know that the slope is given by  

Integrating both sides, 

Given that the curve passes through  so it will satisfy the equation of the curve.
Putting x = 1 and 

1 = log c
⇒ c = e
Putting c = e in (1),

is the required curve.

The correct answer is:  

The auxiliary equation is m² – 6m + 8 = 0 or (m – 4)(m – 2) = 

Hence the required solution is y = C.F. + P.I.

The correct answer is:  

The given equation can be written as
(D – 1)x + (2D + 1)y = 0          ...(1)
2(D + 1)x + (D + 1)y = 3e^–t     ...(2)

Multiplying (1) by 2(D + 1), and (2) by (D – 1) and subtracting,
{2(D + 1)(2D + 1) – (D² – 1)}y = –3(D – 1)e^–t
= –3 (–e^–t – e^–t) = 6e^–t
or (D² + 2D + 1)y = 2e^–t          ...(3)
C.F. = (a + bt)e^–t

The solution of (3) is y = (a + bt + t²)e^–t
The correct answer is: (a + bt)e^t + te^–t

The given curve is,

which is a homogeneous equation

from (1) and (2)

Integrating  both sides,

The correct answer is: x² – 2xy – y² = k

This is a homogeneous linear equation.

Putting z = log x ⇔ x = e^z, the given equation becomes

[D (D – 1) (D – 2) + 2D – 2] = ze^2z + 3e^z

or    (D³ – 3D² + 4D – 2) = ze^2z + 3^z2

The auxiliary equation is m³ – 3m² + 4m – 2 = 0

⇒  (m – 1) (m² – 2m + 2) = 0
⇒ m =1, 1 ± i
∴   C.F.
= C₁e^z + e^z (C₂ cos z + C₃ sin z)
= x[C₁ + C₂ cos (log x) + C₃ sin (log x)]

The correct answer is: 

We have P = 1 – cot x, Q = –cot x and so 1 – P + Q = 0.

Thus e^–x is an integral of D²y + (1 – cot x) Dy – y cot x = 0

The given equation now becomes

Hence,  is the required solution.

The correct answer is: f(x) = 2 cos 2x – sin 2x and g(x) = sin x – cos x