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Differential Equations

Multiple Choice Questions

311.

The integrating factor of the differential equation $x \frac{dy}{dx} - y = 2 x^{2}$ is

$\frac{1}{x}$
x
e^-x
e^-y

312.

If $\frac{d}{dx} f (x) = 4 x^{3} - \frac{3}{x^{4}}$ such that f(2) = 0. Then, f(x) is

$x^{3} + \frac{1}{x^{4}} - \frac{129}{8}$
$x^{4} + \frac{1}{x^{3}} + \frac{129}{8}$
$x^{3} + \frac{1}{x^{4}} + \frac{129}{8}$
$x^{4} + \frac{1}{x^{3}} - \frac{129}{8}$

313.

The order and degree of the differential equation ${(\frac{d^{3} y}{{dx}^{3}})}^{2} - 3 \frac{d^{2} y}{{dx}^{2}} + 2 {(\frac{dy}{dx})}^{4} = y^{4}$ are

1, 4
3, 4
2, 4
3, 2

314.

The solution of the differential equation (1 + y²) dx = (tan^-1((y) - x)dy is

${xe}^{\tan^{- 1} (y)} = (1 - \tan^{- 1} (y)) e^{\tan^{- 1} (y)} + C$
${xe}^{\tan^{- 1} (y)} = (\tan^{- 1} (y) - 1) e^{\tan^{- 1} (y)} + C$
$x = \tan^{- 1} (y) - 1 + {Ce}^{\tan^{- 1} (y)}$
None of the above

315.

The solution of the differential equation $x \frac{dy}{dx} = y - xtan (\frac{y}{x})$ is

$xsin (\frac{x}{y}) + C = 0$
$xsin (y) + C = 0$
$xsin (\frac{y}{x}) = C$
None of these

316.

The particular solution of $\cos (\frac{dy}{dx}) = a (where, a \in R)$ , (y = 2 when x = 0), is

$\cos (\frac{y - 2}{x}) = a$
$\sin (\frac{y - 2}{x}) = a$
$\cos^{- 1} (x) = y + a$
$y = {acos}^{- 1} (x)$

317.

The order and degree of the differential equation $\frac{d^{2} y}{{dx}^{2}} = {\{y + {(\frac{dy}{dx})}^{2}\}}^{\frac{1}{4}}$ are given by

4 and 2
1 and 2
1 and 4
2 and 4

318.

The differential equation of the family of circles touching the y-axis at the origin is

xy' - 2y = 0
y'' - 4y' + 4y = 0
2xyy' + x² = y²
2yy' + y² = x²

319.
Solution of the equation $\cos^{2} (x) \frac{dy}{dx} - (\tan (2 x)) y = \cos^{2} (x), |x| < \frac{π}{4}$ , where $y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ , is given by

$y \frac{\tan (2 x)}{1 - \tan^{2} (x)} = 0$

$y (1 - \tan^{2} (x)) = C$

$y = \sin (2 x) + C$

$y = \frac{1}{2} \frac{\sin (2 x)}{1 - \tan^{2} (x)}$

$y = \frac{1}{2} \frac{\sin (2 x)}{1 - \tan^{2} (x)}$

$Given differential equation is, \cos^{2} (x) \frac{dy}{dx} - (\tan (2 x)) y = \cos^{2} (x), |x| < \frac{π}{4} Given differential equation can be wntten as \frac{dy}{dx} = - (\frac{\tan (2 x)}{\cos^{2} (x)}) y = \cos^{2} (x) Here, P = - \frac{\tan (2 x)}{\cos^{2} (x)} = \frac{- \sin (2 x)}{\cos (2 x) (\frac{\cos (2 x + 1)}{2})} and Q = \cos^{2} (x) ∵ IF = e^{\int Pdx} = e^{- \int \frac{2 \sin (2 x)}{\cos (2 x) (\cos (2 x) + 1)} dx} Put \cos (2 x) = t \Rightarrow - 2 \sin (2 x) dx = dt ∴ IF = e^{\int \frac{1}{t} (\frac{1}{t + 1}) dt} = e^{\int (\frac{1}{t} + \frac{1}{t + 1})} = e^{[\log (t) - \log (t + 1)]} = e^{\log (\frac{t}{t + 1})} = e^{\log (\frac{\cos (2 x)}{\cos (2 x) + 1})} = \frac{\cos (2 x)}{\cos (2 x) + 1} Now, solution is$

$y \times \frac{\cos (2 x)}{\cos (2 x) + 1} = \int \frac{\cos (2 x)}{\cos (2 x) + 1} \times \cos^{2} (x) dx + C = \int \frac{\cos (2 x)}{2 \cos^{2} (x)} \times \cos^{2} (x) dx + C = \frac{1}{2} \int \cos (2 x) dx + C = \frac{1}{2} \times \frac{\sin (2 x)}{2} + C \Rightarrow y \frac{\cos (2 x)}{\cos (2 x) + 1} = \frac{1}{4} \sin (2 x) + C . . . (i) But, y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$

$∴ \frac{3 \sqrt{3}}{8} \times \frac{\cos (2 \times \frac{π}{6})}{\cos (2 \times \frac{π}{6}) + 1} = \frac{1}{4} \sin (2 \frac{π}{6}) + C \Rightarrow \frac{\frac{3 \sqrt{3}}{8} \times \frac{1}{2}}{\frac{1}{2} + 1} = \frac{1}{4} \times \frac{\sqrt{3}}{2} + C \Rightarrow \frac{3 \sqrt{3}}{2 \times 8 \times \frac{3}{2}} = \frac{\sqrt{3}}{8} + C \Rightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8} + C \Rightarrow C = 0 From Eq . (i), we get y \frac{\cos (2 x)}{\cos (2 x) + 1} = \frac{1}{4} \sin (2 x) + 0 \Rightarrow y = \frac{1}{4} \frac{\sin (2 x)}{\frac{\cos (2 x)}{\cos (2 x) + 1}} = \frac{1}{4} \frac{\sin (2 x)}{\frac{\cos^{2} (x) - \sin^{2} (x)}{2 \cos^{2} (x)}} = \frac{1}{2} \frac{\sin (2 x)}{1 - \tan^{2} (x)} ∴ y = \frac{1}{2} \frac{\sin (2 x)}{1 - \tan^{2} (x)}$