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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 + log_{e}2
(2) 2 + log_{e}2
(3) 2e
(4) log_{e}2
Ans. (1)
Solution.
We have,
Let e^{y} = t ⇒ e^{x}
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 + log_{e} 2
Q.2. Let y = y (x) be the solution curve of the differential equationsatisfying y(0) = 1. This curve intersects the xaxis at a point whose abscissa is (2020)
(1) 2 – e
(2) −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 ⇒ c = e.
Therefore,
Hence, the abscissa of point of intersection with the xaxis is
x = (0 − 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^{1}y
At x = 1/√2 , 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^{2 }= 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 ⇒ c + 2 + 1 = 0 ⇒ 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 ⇒
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.
⇒
Solution of differential equation is:
...(1)
∵ y(1) = 1
∴ C = 3/4
Then, from equation (1)
∴
∴
Q.8. Let f : [0, 1] → R be such that f(xy) = f(x).f(y), for all x, y ∈ [0, 1], and f(0) ≠ 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, (x^{2}  y^{2}) dx + 2xydy = 0 which passes through (1, 1), is: (2019)
(1) a circle with centre on the xaxis.
(2) an ellipse with major axis along theyaxis.
(3) a circle with centre on the yaxis.
(4) a hyperbola with transverse axis along the xaxis.
Ans. (1)
Solution.
y^{2}dx  2xydy = x^{2}dx
2xydy  y^{2}dx = x^{2}dx
d(xy^{2}) = x^{2}dx
...(1)
Since, the above curve passes through the point (1,1)
Now, the curve (1) becomes
y^{2} = x^{2} + 2x
⇒ y^{2} = (x1)^{2}+1
(x  1)^{2} + y^{2} = 1
The above equation represents a circle with centre (1, 0) and centre lies on xaxis.
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) = log_{e} 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)
⇒ 0 = 0 + c
⇒ c = 0
Then, equation (1) becomes,
⇒ y (1 + x^{2}) = tan^{1} x
Now put x = 1 in above equation, then
Q.15. The solution of the differential equation (x ≠ 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 = π/6 in the above equation,
Q.17. If y = y(x) is the solution of the differential equation dy/dx = (tan x  y)sec^{2} 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 (y^{2}  x^{3}) dx  xydy = 0 (x ≠ 0) is : (2019)
(1) y^{2}  2x^{2} + cx^{3} = 0
(2) y^{2} + 2x^{3} + cx^{2} = 0
(3) y^{2} + 2x^{2} + cx^{3} = 0
(4) y^{2}  2x^{3} + cx^{2} = 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 = 2x^{2}+ 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 xaxis or on the yaxis, centre at the origin and passing through the point (0, 3) is: (2018)
(1) xyy'  y^{2} + 9 = 0
(2) xyy" + x (y')^{2}  yy' = 0
(3) xyy' + y^{2}  9 = 0
(4) x + yy" = 0
Ans.(2)
Solution.
Equation of ellipse
since it passes through (0, 3)⇒b^{2}=9
∴ Equation of ellipse becomes
⇒
......(2)
Q.24. If ( 2 + sinx) dy/dx + (y + 1)cos x = 0 and y(0) = 1, then y(π/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)× 2 = C ⇒ C = 4
Now, ( y +1) (2+ sinx ) = 4
( y +1) (2+ 1) = 4
Q.25. The curve satisfying the differential equation, ydx – (x + 3y^{2})dy = 9 and passing through the point (1, 1), also passes through the point: (2017)
(1)
(2)
(3)
(4)
Ans. (2)
ydx  xdy  3y^{2}dy = 0
Passes through (1, 1) ∴ 1 = 3 + c ; c = 2
x = 3y^{2}  2y
Q.26. If then λ+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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