The solution of \(x \frac{d y}{d x}=y(\log y-\log x+1)\) is
The solution of \(x \frac{d y}{d x}=y(\log y-\log x+1)\) is
- \(y=x e^{c x}\)
- \(y^2=c x^2\)
- \(y^2=c x \log (x)\)
- \(\log (y)=c x\)
Solution
\(\begin{aligned}
& x \frac{d x}{d x}=y(\log y-\log x+1) \\
& \frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right) \quad \ldots (i)
\end{aligned}\)
\(\begin{gathered}
\text{Let, } f(x, y)=\frac{y}{x}\left(\log \frac{y}{x}+1\right) \\
f(k x, k y)=\frac{k y}{k x}\left(\log \frac{k y}{k x}+1\right)
\end{gathered}\)
\(f(k x, k y)=\frac{y}{x}\left(\log \frac{y}{x}+1\right)=f(x, y)\)
\(\therefore\) Given differentiate Equation is homogeneous
\(\left\{\begin{array}{l}
\therefore p u t, y=v x \\
\frac{d y}{d x}=v+x \frac{d v}{d x}
\end{array}\right\}\)
From Eq. (i),
\(\begin{aligned}
& v+x \frac{d v}{d x}=\frac{v x}{x}\left(\log \frac{v x}{x}+1\right) \\
& v+x \frac{d v}{d x}=v(\log v+1) \\
& v+x \frac{d v}{d x}=v \log v+v \\
& \frac{1}{v \log v} d v=\frac{1}{x} d x
\end{aligned}\)
Integrating on both sides,
\(\int \frac{1}{v} \cdot \frac{1}{\log v} d v=\int \frac{1}{x} d x\)
Put, \(\log v=t\)
\(\begin{aligned}
\frac{1}{v} d v & =d t \\
\int \frac{1}{t} d t & =\int \frac{1}{x} d x \\
\log t & =\log x+\log c
\end{aligned}\)
\(\log (\log v)=\log x c\)
\(\log _e v=x c\)
\(v=e^{x c}\)
\(\frac{y}{x}=e^{x c}\)
\(y=x \cdot e^{x c}\)
Hence, option (a) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
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