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Differential Equations

Multiple Choice Questions

681.

The differential equation of the family of parabola with focus as the origin and the axis as X-axis, is

$y {(\frac{dy}{dx})}^{2} + 4 x \frac{dy}{dx} = 4 y$
$- y {(\frac{dy}{dx})}^{2} = 2 x \frac{dy}{dx} - y$
$y {(\frac{dy}{dx})}^{2} + y = 2 xy \frac{dy}{dx}$
$y {(\frac{dy}{dx})}^{2} + 2 xy \frac{dy}{dx} + y = 0$

682.

Soution of $\frac{dy}{dx} = \frac{xlog (x^{2}) + x}{\sin (y) + ycos (y)}$ is

$ysin (y) = x^{2} \log (x) + C$
$ysin (y) = x^{2} + C$
$ysin (y) = x^{2} + \log (x)$
$ysin (y) = xlog (x) + C$

683.

The general solution of $y^{2} dx + (x^{2} - xy + y^{2}) dy = 0$ is :

$\tan^{- 1} (\frac{y}{x}) = \log (y) + C$
$2 \tan^{- 1} (\frac{x}{y}) + \log (x) + C = 0$
$\log (y + \sqrt{x^{2} + y^{2}}) + \log (y) + C = 0$
$\sinh^{- 1} (\frac{x}{y}) + \log (y) + C = 0$

684.

Integrating factor of (x + 2y³) $\frac{dy}{dx}$ = y² is

$e^{(\frac{1}{y})}$
$e^{- (\frac{1}{y})}$
y
$\frac{- 1}{y}$

685.

y = Ae^x + Be^2x + Ce^3x satisfies the differential equation

y''' - 6y'' + 11y' - 6y = 0
y''' + 6y'' + 11y' + 6y = 0
y''' + 6y'' - 11y' + 6y = 0
y''' - 6y'' - 11y' + 6y = 0

686.

Observe the following statements

A. Integrating factor of $\frac{dy}{dx} + y = x^{2}$ is e^x

R. Integrating factor of $\frac{dy}{dx} + P (x) y = Q (x)$ is $e^{\int P (x) dx}$

A is true, R is false
A is false, R is true
A is true, R is true, R $\Rightarrow$ A
Both are false

687.

$If dx + dy = (x + y) (dx - dy), then \log (x + y) = ?$

x + y + c
x + 2y + c
x - y + c
2x + y + c

688.
$If x^{2} y - x^{3} \frac{dy}{dx} = y^{4} cosx, then x^{3} y^{- 3} is equal to$

sin(x)

2sin(x) + c

- 3sin(x) + c

3cos(x) + c

- 3sin(x) + c

Given that

$x^{2} y - x^{3} \frac{dy}{dx} = y^{4} cosx On dividing by y^{4}, we get \frac{x^{2}}{y^{3}} - \frac{x^{3}}{y^{4}} \frac{dy}{dx} = \cos (x) \Rightarrow \frac{x^{3}}{y^{4}} \frac{dy}{dx} = \frac{x^{2}}{y^{3}} - \cos (x) \Rightarrow \frac{1}{y^{4}} \frac{dy}{dx} - \frac{1}{{xy}^{3}} = - \frac{1}{x^{3}} \cos (x) Let - \frac{1}{y^{3}} = t \Rightarrow \frac{1}{y^{4}} \frac{dy}{dx} = \frac{1}{3} \frac{dt}{dx} \Rightarrow \frac{1}{3} \frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^{3}} \cos (x) \Rightarrow \frac{dt}{dx} + \frac{3 t}{x} = \frac{3}{x^{3}} \cos (x) This is differential equation in t, On compairing with \frac{dt}{dx} + Pt = Q, we get P = \frac{3}{x}, Q = \frac{3}{x^{3}} \cos (x) ∴ I . F . = e^{\int pdx} = e^{\int \frac{3}{x} dx} = e^{\log (x^{3})} = x^{3} ∴ Complete solution is {tx}^{3} = 3 \int \frac{x^{3}}{x^{3}} \cos (x) dx + c_{1} \Rightarrow {tx}^{3} = 3 \sin (x) + c_{1} \Rightarrow - \frac{1}{y^{3}} x^{3} = 3 \sin (x) + c_{1} \Rightarrow y^{- 3} x^{3} = - 3 \sin (x) + c$

689.

Observe the following statements

$I . If dy + 2 xydx = 2 e^{- x^{2}} dx, then {ye}^{x^{2}} = 2 x + c, II . If {ye}^{- x^{2}} - 2 x = c, then dx = (2 e^{- x^{2}} - 2 xy) dy which of the following is correct statement$