Q120 / 180 · KCET 2016 · Mathematics
Integrating factor of \( x \frac{d y}{d x}-y=x^{4}-3 x \) is
- A\( x \)
- B\( \log x \)
- Correct answerC\( \frac{1}{x} \)
- D\( -x \)
Solution
Given differential equation is,
\[ \begin{array}{l} x \frac{d y}{d x}-y=x^{4}-3 x \\ \Rightarrow \frac{d y}{d x}-\frac{1}{x} y=x^{3}-3 \end{array} \]
General differential equation of this type is given by
\( \frac{d y}{d x}+P(x) y=Q(x) \)
Here \( P=-\frac{1}{x} \) and \( Q=x^{3}-3 \) So, I.F. factor is given by,
\[ \begin{array}{l} \text { I.F. }=e^{\int P d x}=e^{\int-\frac{1}{x} d x} \\ =e^{-\log x}=e^{\log \left(\frac{1}{x}\right)}=\frac{1}{x} \end{array} \]
\[ \begin{array}{l} x \frac{d y}{d x}-y=x^{4}-3 x \\ \Rightarrow \frac{d y}{d x}-\frac{1}{x} y=x^{3}-3 \end{array} \]
General differential equation of this type is given by
\( \frac{d y}{d x}+P(x) y=Q(x) \)
Here \( P=-\frac{1}{x} \) and \( Q=x^{3}-3 \) So, I.F. factor is given by,
\[ \begin{array}{l} \text { I.F. }=e^{\int P d x}=e^{\int-\frac{1}{x} d x} \\ =e^{-\log x}=e^{\log \left(\frac{1}{x}\right)}=\frac{1}{x} \end{array} \]
Hence, the correct answer is option: 'C'.