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Solving Linear Differential Equations of First Order

Solve the dif...

Question

Solve the differential equation: $\frac{d y}{d x} + \frac{4 x}{x^{2} + 1} y = \frac{1}{{(x^{2} + 1)}^{3}}$

A

y {(x^{2} + 1)}^{2} = {tan}^{- 1} x + c

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B

y {(x^{2} - 1)}^{2} = {tan}^{- 1} x + c

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C

y {(x^{2} - 1)}^{2} = - {tan}^{- 1} x + c

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D

y {(x^{2} + 1)}^{2} = - {tan}^{- 1} x + c

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Solution

The correct option is A $y {(x^{2} + 1)}^{2} = {tan}^{- 1} x + c$
$\frac{d y}{d x} + \frac{4 x}{x^{2} + 1} y = \frac{1}{{(x^{2} + 1)}^{3}}$ ...(1)
Here $P = \frac{4 x}{x^{2} + 1} \Rightarrow \int P d x = \int \frac{4 x}{x^{2} + 1} d x = 2 log (x^{2} + 1) = log {(x^{2} + 1)}^{2}$
$∴ I . F . = e^{log {(x^{2} + 1)}^{2}} = {(x^{2} + 1)}^{2}$
Multiplying (1) by $I . F .$ we get
${(x^{2} + 1)}^{2} \frac{d y}{d x} + 4 x (x^{2} + 1) y = \frac{1}{(x^{2} + 1)}$
Integrating both sides we get
${(x^{2} + 1)}^{2} y = \int \frac{1}{(x^{2} + 1)} d x + c = {tan}^{- 1} x + c$

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Similar questions

The general solution of the differential equation $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$ is

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

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{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{c o t}^{- 1} 9 + {c o s e c}^{1} \frac{\sqrt{41}}{4} = \frac{π}{4}

{t a n}^{- 1} \frac{1}{7} + {t a n}^{- 1} \frac{1}{13} = {t a n}^{- 1} \frac{2}{9}

{t a n}^{- 1} 2 + {t a n}^{- 1} 3 = 3 π / 4

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$ is
(a) $y (1 + x^{2}) = C + t a n^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + t a n^{- 1} x$
(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$

Q. Solve the following equations for x:

(i)

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1}

\sin

x

=

\tan

^{$-$ 1} (2

\sin

x

x \neq \frac{π}{2}

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

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