 # JEE Main Maths Differential Equations Previous Year Questions With Solutions

Get Differential Equations past year questions with solutions for JEE Main exams here. These problems are basically on the definition of the differential equation, the order of a differential equation, the degree of a differential equation, general solution, variable separable method, homogeneous differential equation and linear differential equation. The differential equations questions from the previous years of JEE Main are present in this page, along with a detailed solution for each question. These questions include all the important topics and formulae. JEE aspirants can also download all the relevant questions using below link.

## JEE Main Maths Differential Equations Previous Year Questions With Solutions

Question 1: The equation of the curve which passes through the point (1, 1) and whose slope is given by 2y / x, is __________.

Solution:

Slope dy / dx = 2y / x

2 ∫dx / x =∫dy / y

2 log x = log y + log c

x2 = yc

Since it passes through (1, 1), therefore, c = 1.

Hence, x2 − y = 0

y = x2.

Question 2: Equation of curve through the point (1, 0) which satisfies the differential equation (1 + y2) dx − xy dy = 0, is ________.

Solution:

We have [dx / x] = [y dy] / [1 + y2]

Integrating, we get log |x| = ([1 / 2] * log [1 + y2]) + log c

|x| = c √(1 + y2)−−−−−−−

But it passes through (1, 0), so we get c = 1

Therefore, the solution is x2 = y2 + 1 or x2 − y2 = 1.

Question 3: A function y = f (x) has the second-order derivatives f′′(x) = 6 (x − 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x − 5, then the function is __________.

Solution:

Given f′′(x) = 6 (x − 1)

f′(x) = 3 (x − 1)2 + c1 ..(i)

But at point (2, 1) the line y = 3x − 5 is tangent to the graph y = f(x).

Hence, dy / dx|x = 2 = 3 or

f′ (2) = 3.

Then from (i) f′ (2) = 3 (2 − 1)2 + c1

3 = 3 + c1

c1 = 0 i.e.,

f′ (x) = 3 (x − 1)2

Given f (2) = 1

f (x) = (x − 1)3 + c2

f (2) = 1 + c2

1 = 1 + c2

c2 = 0

Hence, f (x) = (x − 1)3.

Question 4: An integrating factor for the differential equation (1 + y2) dx − (tan−1 y − x) dy = 0

Solution:

(1 + y2) dx − (tan−1 y − x) dy = 0

dy / dx = (1 + y2) / (tan−1 y − x)

dx / dy = [tan−1 y / 1 + y2] − [x / 1 + y2] [dx / dy] + [x / 1 + y2] = [tan−1 y] / [1 + y2]

This is equation of the form dx / dy + Px = Q

So, I.F. = e∫Pdy

= e∫1 / 1 + y2.dy

= etan−1y

Question 5: If x dy = y (dx + y dy), y > 0 and y (1) = 1, then y (−3) is equal to _________.

Solution:

x dy = y (dx + y dy)

[x dy − ydx] / [y2] = dy

−d (x / y) = dy

Integrating both sides, we get x / y + y = c [Because y (1) = 1 ⇒ c = 2; Hence xy + y = 2 ]

For x = −3,

y2 − 2y − 3 = 0

⇒ y = −1 or 3

⇒y = 3 (Because y > 0)

Question 6: (x2 + y2) dy = xy dx. If y (x0) = e, y (1) = 1, then the value of x0 = __________.

Solution:

(x2 + y2) dy = xy dx

x (x dy − y dx) = −y2 dy

[x * (y dx − x dy)] / y2 = dy

[x / y] d (x / y) = dy / y

Integrating, x2 / 2y2 = loge y + c

Given y (1) = 1

c = 1 / 2

x2 / 2y2 = loge y + 1 / 2

Now y (x0) = e

x02 / 2e2 − loge e − 1 / 2 = 0

x02 = 3e2

x0 = ± √3e

Question 7: Solution of differential equation 2xy dy / dx = x2 + 3y2 is __________.

Solution:

It is homogeneous equation dy / dx = [x2 + 3y2] / 2xy

Put y = vx and dy / dx = v + x * [dv / dx]

So, we get x * [dv / dx] = [1 + v2] / 2v

2v dv / 1 + v2 = dx / x

On integrating, we get

x2 + y2 = px3. (where p is a constant)

Question 8: The general solution of the differential equation (x + y) dx + x dy = 0 is _______.

Solution:

(x + y) dx + x dy = 0

x dy = − (x + y) dx

dy / dx = −[x + y] / x

It is homogeneous equation, hence, put y = vx and dy / dx = v + x [dv / dx], we get

v + x dv / dx = −[x + vx] / x= −[1 + v] / 1

x dv / dx = −1 −2v

∫dv / [1 + 2v] = −∫dx / x

[1 / 2] log (1 + 2v) = −log x + log c

log (1 + 2 [y / x]) = 2 log [c / x] [x + 2y] / x = (c / x)2

x2 + 2xy = c.

Question 9: If y′= [x − y] / [x + y], then its solution is ____________.

Solution:

Given dy / dx = x −y / x + y.

Put y = vx

dy / dx = v + x * [dv / dx]

v + x * [dv / dx] = [x − vx] / [x + vx]

v + x [dv / dx] = [1 − v] / [1 + v] [1 + v] / [2 − (1 + v)2] dv = dx / x

Integrating both sides, ∫[1 + v] / [2 − (1 + v)2] dv = ∫dx / x

Put (1 + v)2 = t

⇒ 2 (1 + v) dv = dt

[1 / 2] ∫dt / [2 − t] =∫dx / x

[− 1 / 2] log (2 − t) = log xc

[−1 / 2] log [2 − (1 + v)2] = log xc

[−1 / 2] log [−v2 − 2v + 1] = log xc

log 1 / √[1 − 2v −v2] = log xc

x2c2 (1 − 2v −v2) = 1

y2 + 2xy − x2 = c1.

Question 10: What is the general solution of the differential equation (2x − y + 1) dx + (2y − x + 1) dy = 0?

Solution:

(2x −y + 1) dx + (2y − x + 1) dy = 0

dy / dx = 2x − y + 1x − 2y −1, put x = X + h, y = Y + k

dY / dX = [2X − Y + 2h − k + 1] / [X − 2Y + h −2k − 1]

2h − k + 1 = 0

h − 2k − 1 = 0

On solving h = −1, k = −1;

dY/ dX = [2X − Y] / [X − 2Y]

Put Y = vX;

dY / dX = v + [X dv / dX]

v + [X dv / dX] = [2X − vX] / [X − 2vX] = [2 − v] / [1 − 2v]

X dv / dX = [2 − 2v + 2v2] / [1 − 2v] = 2 (v2 −v + 1) / [1 − 2v]

dX / X = (1 − 2v) / 2 (v2 − v + 1) dv

Put v2 − v + 1 = t

(2v − 1) dv = dt

dX / X = −dt / 2t

log X = logt− ½ + log c

X = t−1/2 c

X = (v2 − v + 1)−1/2 * c

X2 (v2 − v + 1) = constant

(x + 1)2 (([y + 1)2 / (x + 1)2] − [(y+1) / (x + 1)] + 1) = constant

(y + 1)2 − (y + 1) (x + 1) + (x + 1)2 = c

y2 + x2 − xy + x + y = c

Question 11: The solution of the differential equation dy / dx = 1 + x + y + xy is __________.

Solution:

dy / dx = 1 + x + y + xy

dy / dx = (1 + x) (1 + y)

dy / dx + sin ([x + y] / 2)

= sin ([x − y] / 2)

On integrating, we get

log (1 + y) = x2 / 2 + x + c

Question 12: The differential equation for all the straight lines, which are at a unit distance from the origin is ___________.

Solution:

Since the equation of lines whose distance from the origin is unity, is given by

x cosα + y sinα = 1 …..(i)

Differentiate w.r.t. x, we get cos α + [dy / dx] sin α = 0 …..(ii)

On eliminating the ′α′with the help of (i) and (ii)

sin α (y − x [dy / dx]) = 1

(y − x [dy / dx]) = cosec α …..(iii)

dy / dx = −cot α

(dy / dx)2 = cot 2α …..(iv)

Therefore, by (iii) and (iv),

1 + (dy / dx)2 = (y − x [dy / dx])2

Question 13: The differential equation corresponding to primitive y = ecx is, or the elimination of the arbitrary constant m from the equation y = emx gives the differential equation ________.

Solution:

y = emx

logy = mx ⇒ m = [log y] / x

Now y = emx

dy / dx = memx

= ([log y] / x) * y

= (y / x) log y

Question 14: The differential equation of all parabolas whose axes are parallel to y-axis is ________________.

Solution:

The equation of a member of the family of parabolas having an axis parallel to y-axis is

y = Ax2 + Bx + C …..(i) where A, B, C are arbitrary constants.

Differentiating (i) w.r.t. x, we get

dy / dx = 2Ax + B …..(ii) which on differentiating w.r.t. x gives

d2 y / dx2 = 2A …..(iii)

Differentiating w.r.t. x again, we get

d3y / dx3 = 0.

Question 15: Order of the differential equation of the family of all concentric circles centred at (h, k) is ____________.

Solution:

(x − h)2 + (y − k)2 = r2

Here r is arbitrary constant.

Order of differential equation = 1.

Recommended video 