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# Previous year Questions (2016-20) - Differential Equations Notes | EduRev

## JEE : Previous year Questions (2016-20) - Differential Equations Notes | EduRev

The document Previous year Questions (2016-20) - Differential Equations Notes | EduRev is a part of the JEE Course Maths 35 Years JEE Mains & Advance Past year Papers Class 12.
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Q.1. If y = y (x) is the solution of the differential equationsuch that y (0) = 0, then y (1) is equal to    (2020)
(1) 1 + loge2
(2) 2 + loge2
(3) 2e
(4) loge2
Ans.
(1)
Solution.
We have,

Let ey = t â‡’ ex
Now, the differential equation becomes
Integrating factor of equation is
So, the solution of differential equation is
Now, at x = 0, y = 0, then c =1. Therefore,
Substituting x =1, we get y(1) = 1 + loge 2

Q.2. Let y = y (x) be the solution curve of the differential equationsatisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is    (2020)
(1) 2 â€“ e
(2) âˆ’e
(3) 2
(4) 2 + e
Ans.
(1)
Solution.

The given differential equation is
This is a linear differential equation.
Therefore,
Now, the solution of differential equation is given by

Given: y (0) =1
So, 0 = e + c â‡’ c = -e.
Therefore,
Hence, the abscissa of point of intersection with the x-axis is
x = (0 âˆ’ 0 + 2) - e = 2 - e

Q.3. Let y = y (x) be a solution of the differential equation  thenis equal to    (2020)
(1)
(2)
(3)
(4)
Ans.
(3)
Solution.

We have

Integrating both sides of equation, we get sin-1 x + sin-1 y = c
Now,
sin-1 x = cos-1y
At x = -1/âˆš2 , we get y = sin

Q.4. The differential equation of the family of curves, is    (2020)
(1)
(2)
(3) xy'' = y'
(4)
Ans.
(1)
Solution.

The equation of the family of curve is x= 4b (y+b) .... (i)
Differentiating Eq. (1) w.r.t. x, we get

Substituting the value of b in Eq. (1), we get

Q.5. If foris the solution of the differential equation then y(3) is equal to ________.    (2020)
Ans.
(4.00)
Solution.

We have

.....(1)
This is a linear differential equation of first order. So,

Therefore, the solution of differential equation is

Given, y (2) = 0 â‡’ c + 2 + 1 = 0 â‡’ c = -3.
Therefore,

Substituting x = 3 in above equation, we get

Q.6. Ify(1) = 1; then a value of x satisfying y(x) = e is    (2020)
(1)
(2)
(3)
(4)
Ans.
(4)
Solution.

We have..... (1)

Substitutingin Eq. (1), we get

Integrating both sides of the equation, we get

Now, y(1) = 1 â‡’
So,
For y(x) = e, we have

Q.7. If y = y(x) is the solution of the differential equation,  satisfying y(1) = 1, thenis equal to:    (2019)
(1) 7/64
(2) 1/2
(3) 49/16
(4) 13/16

Ans. (3)
Solution.

â‡’

Solution of differential equation is:

...(1)
âˆµ y(1) = 1
âˆ´ C = 3/4
Then, from equation (1)

âˆ´
âˆ´

Q.8. Let f : [0, 1] â†’ R be such that f(xy) = f(x).f(y), for all x, y âˆˆ [0, 1], and f(0) â‰  0. If y = y(x) satisfies the differential equation, dy/dx = f(x) with y(0) = 1, then      (2019)
(1) 3
(2) 4
(3) 2
(4) 5
Ans. (1)
Solution.
f(xy) = f(x).f(y)   ...(1)

By first principle derivative formula,

Q.9. and      (2019)

Ans. (1)
Solution.

...(i)

Q.10. The curve amongst the family of curves represented by the differential equation, (x2 - y2) dx + 2xydy = 0 which passes through (1, 1), is:     (2019)
(1) a circle with centre on the x-axis.
(2) an ellipse with major axis along they-axis.
(3) a circle with centre on the y-axis.
(4) a hyperbola with transverse axis along the x-axis.
Ans. (1)
Solution.
y2dx - 2xydy = x2dx
2xydy - y2dx = -x2dx
d(xy2) = -x2dx

...(1)
Since, the above curve passes through the point (1,1)

Now, the curve (1) becomes
y2 = -x2 + 2x
â‡’ y2 = -(x-1)2+1
(x - 1)2 + y2 = 1
The above equation represents a circle with centre (1, 0) and centre lies on x-axis.

Q.11. If y(x) is the solution of the differential equation

(2019)

Ans. (3)
Solution.
Given differential equation is,

Complete solution is given by

Differentiate both sides with respect to x,

Q.12. The solution of the differential equation, dy/dx = (x - y)2, when y(1) = 1, is:     (2019)

Ans. (2)
Solution. The given differential equation
...(1)

Now, from equation (1)

Q.13. Let y = y(x) be the solution of the differential equation,  If 2y(2) = loge 4 - 1, then y(e) is equal to:     (2019)

Ans. (3)
Solution. Consider the differential equation,

Q.14. Let y = y(x) be the solution of the differential equation,  such that y(0) = 0. If  , then the value of 'a' is:     (2019)
(1) 1/4
(2) 1/2
(3) 1
(4) 1/16
Ans. (4)
Solution.

Since, the above differential equation is a linear differential equation

Then, the solution of the differential equation

If x = 0 then y = 0 (given)
â‡’ 0 = 0 + c
â‡’ c = 0
Then, equation (1) becomes,
â‡’ y (1 + x2) = tan-1 x
Now put x = 1 in above equation, then

Q.15. The solution of the differential equation  (x â‰  0) with y(1) = 1, is:     (2019)

Ans. (3)
Solution.

Since, the above differential equation is the linear differential equation, then
Now, the solution of the linear differential equation

Q.16. and  then  is equal to:     (2019)

Ans. (3)
Solution.

Now, put y = Ï€/6 in the above equation,

Q.17. If y = y(x) is the solution of the differential equation dy/dx = (tan x - y)sec2 x,  such that y (0) = 0, then      (2019)

(1) e - 2

Ans. (1)
Solution.

Given equation is linear differential equation.

Hence, solution of differential equation,

Q.18. Let y = y(x) be the solution of the differential equation,  such that y (0) = 1.       (2019)
Then:

Ans. (4)
Solution.
Given differential equation is.

Q.19. Consider the differential equation, value of y is 1 when x = 1, then the value of x for which y = 2, is:     (2019)

Ans. (2)
Solution.
Consider the differential equation,

Q.20. The general solution of the differential equation (y2 - x3) dx - xydy = 0 (x â‰  0) is :     (2019)
(1) y2 - 2x2 + cx3 = 0
(2) y2 + 2x3 + cx2 = 0
(3) y2 + 2x2 + cx3 = 0
(4) y2 - 2x3 + cx2 = 0

(where c is a constant of integration)
Ans. (2)
Solution.
Given differential equation can be written as,

Q.21. Let y = y(x) be the solution of the differential equation
(2018)
(1)
(2)
(3)
(4)
Ans.
(3)
Solution:

Integrating both sides we get y sin x = 2x2+ C

Q.22. Let y = y(x) be the solution of the differential equation  where f(x) . If y(0) = 0, then y(3/2) is:    (2018)
(1) 1/2e
(2)
(3)
(4)
Ans.
Solution.
dy/dx + 2y = f(x)  is a linear differential equation

solution of the above equation is

Q.23. The differential equation representing the family of ellipses having foci either on the x-axis or on the y-axis, centre at the origin and passing through the point (0, 3) is:    (2018)
(1) xyy' - y2 + 9 = 0
(2) xyy" + x (y')2 - yy' = 0
(3) xyy' + y2 - 9 = 0
(4) x + yy" = 0
Ans.
(2)
Solution.

Equation of ellipse
since it passes through (0, 3)â‡’b2=9
âˆ´ Equation of ellipse becomes
â‡’
......(2)

Q.24. If ( 2 + sinx) dy/dx + (y + 1)cos x = 0 and y(0) = 1, then y(Ï€/2) is equal to    (2017)
(1) 4/3
(2) 1/3
(3) -2/3
(4) -1/3
Ans.
(2)

ln| y + 1 |+ ln (2 + sinx ) = lnC
(y + 1) (2+ sinx ) = C
Put x = 0, y = 1
(1 + 1)Ã— 2 = C â‡’ C = 4
Now, ( y +1) (2+ sinx ) = 4

( y +1) (2+ 1) = 4

Q.25. The curve satisfying the differential equation, ydx â€“ (x + 3y2)dy = 9 and passing through the point (1, 1), also passes through the point:    (2017)
(1)
(2)
(3)
(4)
Ans.
(2)
ydx - xdy - 3y2dy = 0

Passes through (1, 1) âˆ´ 1 = 3 + c ; c = -2
x = 3y2 - 2y

Q.26. If  then Î»+k is equal to:    (2017)
(1) 26
(2) -24
(3) -23
(4) -26
Ans.
(2)

Q.27. If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equation, y(1 + xy) dx = x dy, then is equal to:    (2016)
(1) -2/5
(2) -4/5
(3) 2/5
(4) 4/5
Ans.
(4)

We have to find
Put x = -1/2

Q.28. The solution of the differential equation   is given by    (2016)
(1)
(2)
(3)
(4)
Ans.
(4)

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