Differential Equation JEE Advanced previous year questions with solutions are given on this page. Order and degree of differential equation, general and particular solution of the differential equation, linear differential equation of first order are the important topics in the differential equation. These solutions are prepared by our expert team. Students can easily download and learn these step-by-step solutions and thus improve their speed and accuracy.
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Question 1: A curve passes through the point (1, π/6). Let the slope of the curve at each point (x, y) be y/x + sec (y/x) , x> 0. Then the equation of the curve is
(a) sin (y/x) = log x + 1/2
(b) cosec (y/x) = log x + 2
(c) sec (2y/x) = log x + 2
(d) cos (2y/x) = log x + 1/2
Solution:
Given that dy/dx = y/x + sec (y/x)
This is a homogeneous differential equation.
Let y = vx
dy/dx = v + x (dv/dx)
v + x (dv/dx) = v + sec v
x (dv/dx) = sec v
dv/sec v = dx/x
Integrating
∫dv/sec v = ∫dx/x
∫ cos v dv = ∫dx/x
sin v = log |x| + C
sin (y/x) = log |x| + C
which passes through (1, π/6).
So C = sin (π/6) = ½
sin (y/x) = log x + ½ is the required equation.
Hence option a is the answer.
Question 2: Let y(x) be a solution of the differential equation (1+ex)y’ + yex = 1. If y(0) = 2, then which of the following statements are true?
(a) y(-4) = 0
(b) y(-2) = 0
(c) y(x) has critical point in the interval (-1, 0)
(d) y(x) has no critical point in the interval (-1, 0)
Solution:
Given that (1+ex)y’ + yex = 1
Divide by 1+ex
dy/dx + yex/(1+ex) = 1/(1+ex)
Integrating factor = 1+ex
Hence the solution is
y(1+ex) = ∫dx + C
y(1+ex) = x + C
Given y(0) = 2
=> x = 0, y = 2
2(1+1) = 0+C
=> C = 4
y(1+ex) = x + 4
y = (x + 4)/(1+ex)
y(-4) = 0
Also dy/dx = [(1+ex) – (x+4)ex]/(1+ex)2
For critical point dy/dx = 0
=> (1+ex) – (x+4)ex = 0
Let g(x) = (1+ex) – (x+4)ex
g(-1) = 1+(1/e) -(-1+ 4)/e
= 1+(1/e) – 3/e
= 1 – 2/e >0
g(0) = (1+1)- 4 = -2 < 0
So y(x) has critical point at (-1, 0).
Hence option a and c are correct.
Question 3: Consider the family of all circles whose centers lie on the straight line y = x. If this family of circles is represented by the differential equation Py” + Qy’ +1 = 0, where P, Q are functions of x, y and y’ (here y’ = dy/dx and y’’ = d2y/dx2) then which of the following statements is (are) true?
(a) P = y + x
(b) P = y – x
(c) P + Q = 1 – x + y + y’ + (y’)2
(d) P – Q = x + y – y’ – (y’)2
Solution:
Let the equation of circle be x2+y2+2gx+2gy+c = 0
Differentiating, we get
2x + 2y y’ + 2g + 2gy’ = 0
=> x + yy’ + g + gy’ = 0 ..(i)
Again differentiating
1 + yy’’ + (y’)2 + gy’’ = 0
=> g = -(1 + (y’)2 + yy’’)/y’’
Put g in (i) and solving we get
(x-y)y’’ – y’(1+y’+(y’)2) = 1
(y-x)y’’ + (1+y’+(y’)2)y’ + 1 = 0
=> Py’’ + Qy’ + 1 = 0
=> P = y-x, Q = 1+y’+(y’)2
P+Q = 1 – x + y + y’ + (y’)2
Hence options b and c are correct.
Question 4: The differential equation representing the family of curves y2 = 2c(x+√c), where c is a positive parameter, is of
(a) order 1, degree 3
(b) order 2, degree 2
(c) order 2, degree 1
(d) none of the above
Solution:
y2 = 2c(x+√c) ..(i)
Differentiating both sides
2yy’ = 2c
=> c = yy’
Put c in (i)
y2 = 2yy’(x+√(yy’))
=> (y2-2xyy’) = 2(yy’)3/2
(y2-2xyy’)2 = 2(yy’)3
So degree = 3 and order = 1
Hence option a is the answer.
Question 5: The order of the differential equation whose general solution is given by y = (c1+c2) cos (x + c3) – c4 ex + c5 where c1,c2, c3, c4,c5 are arbitrary constants, is
(a) 5
(b) 4
(c) 3
(d) 2
Solution:
Given that the solution of differential equation is y = (c1+c2) cos (x + c3) – c4 ex + c5
= (c1+c2) cos (x + c3) – c4 ex ec5
= A cos (x+c3) – Bex
Where c1+c2 = A and c4ec5 = B
Hence the number of arbitrary constants in the solution is 3.
The order of the differential equation is 3.
Hence option c is the answer.
Question 6: If x2 + y2 = 1, then
(a) yy’’ – 2(y’)2 + 1 = 0
(b) yy’’ + (y’)2 + 1 = 0
(c) yy’’ + (y’)2 – 1 = 0
(d) yy’’ + 2(y’)2 + 1 = 0
Solution:
Given x2 + y2 = 1
Differentiating we get
2x + 2yy’ = 0
Again differentiating
2 + 2(y’)2 + 2yy’’ = 0
=> 1 + (y’)2 + yy’’ = 0
Hence option b is the answer.
Question 7: The differential equation representing the family of ellipse having foci either on the x axis or on the y axis centre at the origin and passing through the point (0, 3) is
(a) xyy’’ + x(y’)2 – yy’ = 0
(b) x + y” = 0
(c) xyy’ + y2 – 9 = 0
(d) xyy’ – y2 + 9 = 0
Solution:
General equation of ellipse is given by x2/a2 + y2/b2 = 1 ..(i)
It passes through (0, 3)
=> 0 + 9/b2 = 1
=> b2 = 9
so(i) becomes
=> x2/a2 + y2/9 = 1
Differentiate w.r.t.x
2x/a2 + 2yy’/9 = 0
yy’/9 = -x/a2
y’ = -(x/y)(9/a2)
(y/x)y’ = -9/a2
Again differentiate w.r.t.x
(y/x)y’’ + [(xy’ – y)/x2]y’ = 0
=> xyy’’ + x(y’)2 – yy’ = 0
Hence option a is the answer.
Question 8: Let f: R→ R be a differential function with f(0) = 1 and satisfying the equation f(x+y) = f(x) f’(y) + f’(x) f(y) for all x, y belongs to R. then the value of loge (f(4)) is
Solution:
f(x+y) = f(x) f’(y) + f’(x) f(y)
Put x = 0, y = 0
f(0) = 1.f’(0) + f’(0)
=> 2f’(0) = f(0)
=> f’(0) = f(0)/2
=> f’(0) = ½
f(x+0) = f(x) f’(0) + f’(x) f(0)
f(x) = f(x)×½ + f’(x)
=> f’(x) = f(x)/2
∫ f’(x)/f(x) = ½ ∫dx
loge f(x) = ½ x + c
=> f(x) = cex/2
=> f(x) = ex/2
loge f(x) = x/2
loge f(4) = 4/2
= 2
Question 9: Let y = y(x) be the solution of the differential equation cos x (dy/dx) + 2y sin x = sin 2x, x belongs to (0, π/2). If y(π/3) = 0, then y(π/4) is equal to:
(a) 2 – √2
(b) 2 + √2
(c) √2 – 2
(d) 1/√2 – 1
Solution:
Given that cos x (dy/dx) + 2y sin x = sin 2x
Divide by cos x, we get
(dy/dx) + 2y tan x = 2 sin x
I.F = e∫2 tan x dx = sec2x
The solution of differential equation is
y. I.F = ∫I.F 2 sin x dx + c
=> y sec2x = ∫2 sin x sec2 x dx + c
=> y sec2 x = 2 sec x + c
When x = π/3, y = 0, then c = -4
From (i), y sec2 x = 2 sec x – 4
=> y = (2 sec x- 4)/sec2 x
=> y(π/4) = √2 – 2
Hence option c is the answer.
Question 10: Let f(x): (0, ∞) → R be a differential function such that f’(x) = 2 – f(x)/x for all x belongs to (0, ∞) and f(1) ≠ 1. Then
(a) limx→ 0+ f’(1/x) = 1
(b) limx→ 0+ xf’(1/x) = 2
(c) limx→ 0+ x2f’(1/x) = 0
(d) none of these
Solution:
Given f’(x) = 2 – f(x)/x
f’(x) + f(x)/x = 2
I.F = elog x = x
So f(x) x = ∫2x dx
= x2+c
=> f(x) = x+ c/x , c ≠ 0 as f(x) ≠ 1
Check the options
(a) limx→ 0+ f’(1/x) = limx→ 0+ (1-cx2) = 1
(b) limx→ 0+ xf’(1/x) = limx→ 0+ x(1/x + cx) = limx→ 0+ (1+ cx2) = 1
(c) limx→ 0+ x2f’(x) = limx→ 0+ x2(1 – c/x2)
= limx→ 0+ x2 – c = -c
Hence option a is the answer.
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