Match the differential equaitons 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) $(x^{2} + 1) \frac{dy}{dx} + 4 xy = \ln x$ d) $x^{2} + 1$
e) ${(x^{3} + 1)}^{\frac{1}{3}}$
f) ${(x^{3} + 1)}^{\frac{1}{2}}$

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i ii iii iv

d a b c

i ii iii iv

e b c f

i ii iii iv

e a c d

i ii iii iv

e a b c

answer is B.

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Detailed Solution

Linear diff. eqn.

$\begin{array}{rcl} \frac{dy}{dx} + Py & = & Q \\ I . F & = & e^{\int Pdx} \end{array}$

(i) $\begin{array}{rcl} (x^{3} + 1) \frac{dy}{dx} + x^{2} y & = & 0 \\ \frac{dy}{dx} + \frac{x^{2}}{x^{3} + 1} y & = & 0 \\ P & = & \frac{x^{2}}{x^{3} + 1}, Q = 0 \end{array}$

$\begin{array}{rcl} I . F & = & e^{\int \frac{x^{2}}{x^{3} + 1} dx} \\ = & e^{\frac{1}{3} \log (x^{3} + 1)} \\ = & {(x^{3} + 1)}^{\frac{1}{3}} \end{array}$

(ii) $x^{2} \frac{dy}{dx} + 3 xy = x^{6}$

Divide with $x^{2}$ both sides

$\begin{array}{rcl} \frac{dy}{dx} + \frac{3}{x} y & = & x^{4} \\ P & = & \frac{3}{x}, Q = x^{4} \\ I . F & = & e^{\int \frac{3}{x} dx} = e^{3 \log x} = e^{\log x^{3}} \\ = & x^{3} \end{array}$

(iii) $\begin{array}{rcl} {(x^{3} + 1)}^{2} \frac{dy}{dx} + 6 x^{2} (x^{3} + 1) y & = & x^{2} \\ \frac{dy}{dx} + \frac{6 x^{2}}{(x^{3} + 1)} & = & \frac{x^{3}}{x^{3} + 1} \\ \Rightarrow & P = \frac{6 x^{2}}{(x^{3} + 1)}, Q = \frac{x^{2}}{x^{3} + 1} \\ I . F & = & e^{6 \int \frac{x^{2}}{x^{3} + 1} dx} = e^{2 \int \frac{3 x^{2}}{x^{3} + 1} dx} \\ = & e^{2 \log (x^{3} + 1)} = {(x^{3} + 1)}^{2} \end{array}$

(iv) $\begin{array}{rcl} \frac{dy}{dx} + \frac{4 xy}{x^{2} + 1} & = & \frac{\log x}{x^{3} + 1} \\ P & = & \frac{4 x}{x^{2} + 1}, Q = \frac{\log x}{x^{3} + 1} \\ I . F & = & e^{\int \frac{49}{x^{2} + 1} dx} \\ = & e^{2 \int \frac{2 x}{x^{2} + 1} dx} \\ = & {e^{2 \log (x^{2} + 1)}}^{2} \\ = & {(x^{2} + 1)}^{2} \end{array}$