JEE Advanced Level Test: Differential Equation - JEE MCQ

We have, IF = sin x

Clearly, it is a differential equation of degree 2.

The equation of the tangent at any point 
This cuts the coordinate axes at 
It is given that P (x, y) is the mid point of AB.


Clearly, it represents a rectangular hyperbola.

cos x dy = y (sin x- y) dx

Put y = vx

Put log v = z

Equation of the circle is x² + y² = 4 and equation of the line is x = 1.
Required area = area 

We have, xy = ce^x + be^-x + x²....(i)
On differentiating w.r.t. x, we get

On differentiating again,

Hence, the required differentiable equation is

The order of this differential equation is 2 and degree 1.
Hence, (B) is the correct answer  

Since, the above relation contains five arbitrary constants, so the order of the differential equation satisfying it, is 5.
Hence, (C) is the correct answer.

Put x² = sin α, y² = sin β
∴ Given equation reduces to cos α + cos β = a (sin α - sin β)


On differentiating w.r.t. x, we get

which is a differential equation of first order and first degree.
Hence, (A) is the correct answer

We have, y = e^x (A cos x + B sin x)....(i)
On differentiating w.r.t. x, we get


On differentiating again,

Let the directrix be x = - 2a and latusrectum be 4a. Then, the equation of the parabola is (distance from focus = distance from directrix),

On differentiating w.r.t, we get

On putting this value of a in Eq. (i), the differential equation is

Hence, (A) is the correct answer

We have, (2x - 2y + 5) dy = (x - y + 3) dx


∴ The given equation becomes



Hence, (A) is the correct answer.

Hence, (A) is the correct answer.

Hence, (B) is the correct answer

Hence, (A) is the correct answer.

Hence, (A) is the correct answer

Hence, (C) is the correct answer.

Put x² y² = z

Now, given expression transforms to



Hence, (A) is the correct answer

 Hence, the differential equation becomes




Hence, (A) is the correct answer.

We have, 
Cartesian sub tangent + abscissa = constant

Integrating, we get log y + log (x - a) = log c
∴ y(x - a) = c
As the curve passes through the point (2a, a), we have c = a² 
∴ The required curve is y (x - a) = a² 
Hence, (A) is the correct answer

The slope of the ray = y/x

Hence, (B) is the correct answer.

We have, f "(x) = 6 (x- 1)
⇒  f '(x) = 3(x -1)² + c .......(i)
It is given that, y = 3x- 5 is tangent to the curve y = f (x) at the point (2, 1)


On putting, x = 2, f ' (2) = 3 Eq. (i), we get c = 0
∴ f ' (x) = 3(x- 1)²
⇒ f (x) = (x - 1)³ + c₁ .....(ii)
The curve y = f (x) passes through (2, 1)
∴ f (2) = 1
On putting, x = 2, f (2) = 1 in Eq. (ii), we get
c₁ = 0
On putting c₁ = 0 in eq. (ii), we get f (x) = (x- 1)³