Q87 / 160 · TS EAMCET 2015 · Mathematics

Match the differential equations in List I to their integrating factors in List II. Figure
The correct match is

  • A
    $\begin{array}{lllll} & 4 & 1 & 2 & R\end{array}$
  • Correct answer
    B
    $5 \quad 1 \quad 2 \quad 3$
  • C
    $5 \quad 2 \quad 3 \quad 6$
  • D
    $5 \quad 1 \quad 3 \quad 4$

Solution

(P) $$ \begin{aligned} & \left(x^3+1\right) \frac{d y}{d x}+x^2 y=3 x^2 \\ & \Rightarrow \frac{d y}{d x}+\frac{x^2}{x^3+1} y=\frac{3 x^2}{x^3+1} \\ & \therefore \text { Integrating factor }=e^{\int \frac{x^2}{x^3+1} d x} \\ & =e^{\frac{1}{3} \log \left(x^3+1\right)}=\left(x^3+1\right)^{1 / 3} \end{aligned} $$ (Q) $$ \begin{aligned} & x^2 \frac{d y}{d x}+3 x y=x^6 \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{3 x}{x^2} y=\frac{x^6}{x^2} \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{3}{x} y=x^4 \\ & \therefore \quad \text { Integrating factor }=e^{\int \frac{3}{x} d x} \\ & =e^{3 \log x} \\ & =x^3 \\ & \end{aligned} $$ (R) $$ \begin{aligned} & \text { R) }\left(x^3+1\right)^2 \frac{d y}{d x}+6 x^2\left(x^3+1\right) y=x^2 \\ & \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}+\frac{6 x^2\left(x^3+1\right)}{\left(x^3+1\right)^2} y=\frac{x^2}{\left(x^3+1\right)^2} \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{6 x^2}{x^3+1} y=\frac{x^2}{\left(x^3+1\right)^2} \\ & \therefore \text { Integrating factor }=e^{\int \frac{6 x^2}{x^3+1}} d x \\ &=e^{2 \log \left(x^3+1\right)}=\left(x^3+1\right)^2 \end{aligned} \end{aligned} $$ (S) $$ \begin{aligned} & \left(x^2+1\right) \frac{d y}{d x}+4 x y=\ln x \\ & \Rightarrow \frac{d y}{d x}+\frac{4 x}{x^2+1} y=\frac{\ln x}{x^2+1} \\ & \therefore \text { Integrating factor }=e^{\int \frac{4 x}{x^2+1} d x} \\ & =e^{2 \log \left(x^2+1\right)} \\ & =\left(x^2+1\right)^2 \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.

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