# MCQ : Differential Equations

Q:1. The degree and order of the differential equation of all the parabolas whose axis is x-axis are

(A) 2 , 1

(B) 1 , 2

(C) 3 , 2

(D) none of these

Q:2. The solution of differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(\frac{y}{x})}{\phi'(\frac{y}{x})}$ is

(A) $x \phi(\frac{y}{x}) = k$

(B) $\phi(\frac{y}{x}) = k x$

(C) $y \phi(\frac{y}{x}) = k$

(D) $\phi(\frac{y}{x}) = k y$

Q: 3. The differential equation for the family of curves y2 = a sinx + b cosx (a, b being parameters) is

(A) $y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + y = 0$

(B) $2y (\frac{d^2y}{dx^2}) + 2(\frac{dy}{dx})^2 + y = 0$

(C) $2y (\frac{d^2y}{dx^2}) + 2(\frac{dy}{dx})^2 – y = 0$

(D) none of these

Q:4. If x intercept of any tangent is 3 times the x-coordinate of the point of tangency , then the equation of the curve, given that it passes through (1, 1) is

(A) $y = \frac{1}{x}$

(B) $y = \frac{1}{x^2}$

(C) $y = \frac{1}{\sqrt{x}}$

(D) none of these

Q:5. If (x – a)2 + (y – b)2 = c2 , then $c \frac{d^2y}{dx^2}+ [1 + (\frac{dy}{dx})^2 ]^{3/2}$ is dependent on

(A) a

(B) b

(C) c

(D) none of these

Q:6. The curve for which the normal at every point passes through a fixed point (h, k) is

(A) (x – h)2 + y = c

(B) x + y = c

(C) (x – h)2 + (y – k)2 = c

(D) none of these

Q:7. The solution of $(x + 1)\frac{dy}{dx} + 1 = e^{(x-y)}$ is

(A) ey (x + 1) = c

(B) ey (x + 1) = ex + c

(C) ey (x + 1) = c ex

(D) None of these

Q: Solution of y (xy + 1) dx + x (1 + xy + x2y2) dy = 0 is

(A) $logy = \frac{1}{x y} + c$

(B) $logy = \frac{1}{2 x^2 y^2} + c$

(C) $logy = \frac{1}{x y} + \frac{1}{2 x^2 y^2} + c$

(D) None of these.

Q: 9. Solution of the differential equation, y dx – x dy + x y2 dx = 0 can be

(A) 2x + x2 y = λ y

(B) 2y + y2 x = λ y

(C) 2y – y2 x = λ y

(D) none of these

Q:10. If y = y(x) and $(\frac{2+ sinx}{y+1}) \frac{dy}{dx} = – cosx$ , y(0) = 1 , then y (π/2) equals,

(A) 1/3

(B) 2/3

(C) -1/3

(D) 1

Q:11. The equation of the curve whose subnormal is equal to a constant a is:

(A) y = ax + b

(B) y2 = 2ax + 2b

(C) ay2 – x3 = a

(D) None of these

where b is an arbitrary constant.

Q:12. Solution of the equation $\frac{dy}{dx} + \frac{1}{x} tany = \frac{1}{x^2} tany siny$ is

(A) x = siny (1 + 2cx2)

(B) 2x = siny (1 + cx2)

(C) 2x + siny (1 + cx2) = 0

(D) None of these

Q: 13. If y = e4x + 2e-x satisfies the relation y3 + Ay’ + By = 0 , then the value of A and B are respectively

(A) 12, 13

(B) – 12 , 13

(C) – 13 , – 12

(D) – 12 , – 13

Q:14. A solution of the differential equation $(\frac{dy}{dx})^2 – x \frac{dy}{dx} + y = 0$ is

(A) y = 2

(B) y = 2x

(C) y = 2x – 4

(D) y = 2x2 – 4

Q:15. Equation to the curve such that the y-intercept cut off by the tangent at any arbitrary point is proportional to the square of the ordinate of the point of tangency is of the form

(A) $\frac{a}{x} + \frac{b}{y^2} = 1$

(B) $\frac{a}{x^2} + \frac{b}{y^2} = 1$

(C) $\frac{a}{x} + \frac{b}{y} = 1$

(D) none of these

Q:16. The solution of the differential equation (1 + x2 y2)y dx + (x2 y2 – 1)x dy = 0 is

(A) $xy = ln(\frac{x}{y}) + c$

(B) $xy = 2ln\frac{y}{x} + c$

(C) $x2 y2 = 2ln \frac{y}{x} + c$

(D) none of these

Q:17. Tangent to a curve intersects the y-axis at a point P. A line perpendicular to this tangent through P passes through another point (1, 0). The differential equation of the curve is

(A) $y \frac{dy}{dx} – x (\frac{dy}{dx})^2 = 0$

(B) $x \frac{d_2 y}{dx^2} + (\frac{dy}{dx})^2 = 1$

(C) $y \frac{dx}{dy} + x = 1$

(D) none of these

Q:18. Solution to the differential equation $\frac{x + \frac{x^3}{3!}+ \frac{x^5}{5!}+ …}{1 + \frac{x^2}{2!}+ \frac{x^4}{4!}+ …}$ is

(A) 2y e2x = C.e2x + 1

(B) 2y e2x = C.ex – 1

(C) y e2x = C.e2x + 2

(D) none of these

Q:19. A curve is such that the mid point of the portion of the tangent intercepted between the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line y = x. If the curve passes through (1, 0), then the curve is

(A) 2y = x2 – x

(B) y = x2 – x

(C) y = x – x2

(D) y = 2(x – x2)

Q:20. The solution of the differential equation dx (x2 – t) = dt , is

(A) t = (x – 1)2ex + c

(B) t = (x + 1)2ex + c

(C) t = (x2 + 2x – 1)ex + c

(D) none of these