Match the differential equations in List I totheir integrating factors in List IIList-I List-IIDifferential equation Integrating factori) (x3+1)dydx+x2y=3x2 a) x3ii) x2dydx+3xy=x6 b) (x3+1)2iii) (x3+1)2dydx+6x2(x3+1)y=x2 c) (x2+1)2iv) (x3+1)dydx=4xy=1nx d) x2+1 e) (x3+1)1/3 f) (x3+1)1/2The correct match is

(i) dydx+(x2x3+1)y=3x2x3+1I.F=e∫x2x3+1 dx=e13 ∫3x2x3+1dx=e13 log(x3+1)=(x3+1)13i) →e(ii)dydx+3xy=x4I.F=e∫3xdx=e 3 logx=x3ii) →a(iii)dydx+6x2x3+1y=x2(x3+1)2I.F=e∫6x2x3+1dx=e2∫3x2x3+1dx=e2 log (x3+1)=(x3+1)2(iii)→ b

Match the differential equations in List I totheir integrating factors in List IIList-I List-IIDifferential equation Integrating factori) (x3+1)dydx+x2y=3x2 a) x3ii) x2dydx+3xy=x6 b) (x3+1)2iii) (x3+1)2dydx+6x2(x3+1)y=x2 c) (x2+1)2iv) (x3+1)dydx=4xy=1nx d) x2+1 e) (x3+1)1/3 f) (x3+1)1/2The correct match is

(i) dydx+(x2x3+1)y=3x2x3+1I.F=e∫x2x3+1 dx=e13 ∫3x2x3+1dx=e13 log(x3+1)=(x3+1)13i) →e(ii)dydx+3xy=x4I.F=e∫3xdx=e 3 logx=x3ii) →a(iii)dydx+6x2x3+1y=x2(x3+1)2I.F=e∫6x2x3+1dx=e2∫3x2x3+1dx=e2 log (x3+1)=(x3+1)2(iii)→ b

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Match the differential equations in List I to
their integrating factors in List II
List-I List-II
Differential equation Integrating factor
i) $(x^{3} + 1) \frac{dy}{dx} + x^{2} y = 3 x^{2}$ a) $x 3$
ii) $x^{2} \frac{dy}{dx} + 3 xy = x^{6}$ b) ${(x^{3} + 1)}^{2}$
iii) ${(x^{3} + 1)}^{2} \frac{dy}{dx} + 6 x^{2} (x^{3} + 1) y = x^{2}$ c) ${(x^{2} + 1)}^{2}$
iv) $(x^{3} + 1) \frac{dy}{dx} = 4 xy = 1 nx$ d) $x^{2} + 1$
e) ${(x^{3} + 1)}^{1 / 3}$
f) ${(x^{3} + 1)}^{1 / 2}$
The correct match is

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(i) (ii) (iii) (iv)
e a c d

(i) (ii) (iii) (iv)
e b c f

(i) (ii) (iii) (iv)
d a b c

(i) (ii) (iii) (iv)
e a b c

answer is B.

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Detailed Solution

$\begin{array}{l} (i) \frac{d y}{d x} + (\frac{x^{2}}{x^{3} + 1}) y = \frac{3 x^{2}}{x^{3} + 1} \\ I . F = e^{\int \frac{x^{2}}{x^{3} + 1} d x} = e^{\frac{1}{3} \int \frac{3 x^{2}}{x^{3} + 1} d x} \\ = e^{\frac{1}{3} \log (x^{3} + 1)} = {(x^{3} + 1)}^{\frac{1}{3}} \\ i) \to e \\ (i i) \frac{d y}{d x} + \frac{3}{x} y = x^{4} \\ I . F = e^{\int \frac{3}{x} d x} = e^{3 \log x} = x^{3} \\ i i) \to a \end{array}$

$\begin{array}{l} (i i i) \frac{d y}{d x} + \frac{6 x^{2}}{x^{3} + 1} y = \frac{x^{2}}{{(x^{3} + 1)}^{2}} \\ I . F = e^{\int \frac{6 x^{2}}{x^{3} + 1} d x} = e^{2 \int \frac{3 x^{2}}{x^{3} + 1} d x} \\ = e^{2 \log (x^{3} + 1)} = {(x^{3} + 1)}^{2} \\ (i i i) \to b \end{array}$