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Integrating factor of \( x \frac{d y}{d x}-y=x^{4}-3 x \) is
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Verified Answer
The correct answer is:
\( \frac{1}{x} \)
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,
\[
\begin{array}{l}
\text { I.F. }=e^{\int P d x}=e^{\int-\frac{1}{x} d x} \\
=e^{-\log x}=e^{\log \left(\frac{1}{x}\right)}=\frac{1}{x}
\end{array}
\]
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